MathsCBSE Class 10 · Chapter 4

Quadratic Equations Class 10: Concepts, Formulas and Solved Questions

Q: Is the method of completing the square in the 2026-27 syllabus?

No. Solving by completing the square is not in the 2026-27 syllabus. You solve quadratics by factorisation or by quoting and using the quadratic formula.

Q: Should I use factorisation or the quadratic formula in the exam?

Use the method the question names, if it names one. If it does not, factorisation is quicker when the numbers split easily; otherwise use the formula, which always works when real roots exist. Checking whether the discriminant is a perfect square helps you decide fast.

Q: What should I write if the discriminant is negative?

State clearly that the discriminant is negative, so the equation has no real roots. Only real roots are in your syllabus, so you stop there and do not try to find square roots of negative numbers.

Q: Do I really need to write why I rejected a root?

Yes. A one-line reason, such as speed cannot be negative or this value makes a denominator zero, is part of a complete answer. Leaving an impossible value in your final answer can cost you the final mark.

Q: Is the Quadratic Equations chapter the same for Mathematics Standard and Basic?

Yes. Both papers follow the same syllabus for this chapter, so the concepts and methods are identical. The difference is only in the difficulty of the questions set in the paper.

Updated 13 October 2026 · Fact-checked

A quadratic equation is an equation of degree two, written in standard form as ax squared plus bx plus c equals zero with a not zero. In the CBSE Class 10 board exam you solve such equations by factorisation or the quadratic formula, use the discriminant to judge the nature of roots, and model day-to-day word problems.

What Quadratic Equations is about

This chapter is about equations where the highest power of the unknown is 2, written in the standard form $a x^{2} + b x + c = 0$ with $a \neq = 0$ . You learn to find their real roots in two ways, by factorisation (splitting the middle term) and by the quadratic formula. You also learn to predict the kind of roots, without solving, from the discriminant $b^{2} - 4 a c$ . Most importantly, you learn to turn everyday situations about speed, work, area and numbers into a quadratic equation and solve it.

The chapter builds directly on Polynomials, because the roots of $a x^{2} + b x + c = 0$ are the zeroes of the polynomial $a x^{2} + b x + c$ . It is used again in Arithmetic Progressions (finding the number of terms from a sum often gives a quadratic), in Coordinate Geometry (distance conditions), and in Triangles and mensuration problems where Pythagoras or an area formula leads to an $x^{2}$ term.

Board questions take a few clear shapes: MCQs and assertion-reason items on identifying a quadratic or on the nature of roots; short answers asking you to find $k$ for equal roots or to solve an equation with fractions; long answers where you model and solve a word problem and reject a meaningless root; and case studies set in real situations such as gardens, journeys or sales.

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 theory marks together with Polynomials, Pair of Linear Equations in Two Variables and Arithmetic Progressions.Source: CBSE Mathematics syllabus 2026-27, Unit II Algebra

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.3 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.5 marks per paper. See the questions

Key concepts

Standard form and identifying a quadratic equation

NCERT section 4.2 Quadratic Equations

A quadratic equation in $x$ is any equation that can be rearranged into the standard form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a \neq = 0$ .

Why must $a \neq = 0$ ? If $a = 0$ the $x^{2}$ term disappears and you are left with $b x + c = 0$ , which is linear. So the test is simple: after you expand every bracket, cancel like terms and bring everything to one side, the highest power of $x$ left must be exactly 2.

This matters because equations can look cubic and turn out quadratic, or look quadratic and turn out linear once the $x^{2}$ terms cancel. Never judge by appearance. Standard form is also the first line of every method in this chapter, because you read $a$ , $b$ and $c$ , with their signs, only from it.

Worked example

Which of the following is a quadratic equation?
(a) $(x + 2)^{3} = x^{3} - 4$
(b) $x^{2} + 3 x = (x - 1)^{2}$
(c) $x (x^{2} + 1) = x^{3} + 2$
(d) $(2 x - 1)^{2} = 4 x^{2} + x$

Try it first, then show the solutionSolution

(a) Expand: $(x + 2)^{3} = x^{3} + 6 x^{2} + 12 x + 8$ .
So $x^{3} + 6 x^{2} + 12 x + 8 = x^{3} - 4$ , which gives $6 x^{2} + 12 x + 12 = 0$ , i.e. $x^{2} + 2 x + 2 = 0$ . Highest power is 2, so it is quadratic.
(b) $x^{2} + 3 x = x^{2} - 2 x + 1$ gives $5 x - 1 = 0$ : linear, because the $x^{2}$ terms cancel.
(c) $x^{3} + x = x^{3} + 2$ gives $x - 2 = 0$ : linear.
(d) $4 x^{2} - 4 x + 1 = 4 x^{2} + x$ gives $- 5 x + 1 = 0$ : linear.

Answer: (a) $(x + 2)^{3} = x^{3} - 4$ , which simplifies to $x^{2} + 2 x + 2 = 0$ .

Roots of a quadratic equation

NCERT section 4.3 Solution of a Quadratic Equation by Factorisation

A real number $α$ is a root (solution) of $a x^{2} + b x + c = 0$ if putting $x = α$ makes the left side zero, that is $a α^{2} + b α + c = 0$ . The roots of the equation are exactly the zeroes of the polynomial $a x^{2} + b x + c$ from the Polynomials chapter, so a quadratic equation has at most two roots.

This definition gives you two board skills. First, you can test whether a given number is a root just by substituting it. Second, if a question tells you that a number is a root but one coefficient is unknown, substitution gives a linear equation in that unknown. Once you know the coefficient, the equation is complete and you can solve it to get the other root. Always finish by checking the second root too.

Worked example

If $x = 2$ is a root of the equation $2 x^{2} + k x - 6 = 0$ , find the value of $k$ and the other root.

Try it first, then show the solutionSolution

Since $x = 2$ is a root, it satisfies the equation: $2 (2)^{2} + k (2) - 6 = 0$ .
$8 + 2 k - 6 = 0 \Rightarrow 2 k = - 2 \Rightarrow k = - 1$ .
The equation is now $2 x^{2} - x - 6 = 0$ .
Split the middle term: $a c = - 12$ , and $- 4 + 3 = - 1$ , so $2 x^{2} - 4 x + 3 x - 6 = 0$ .
$2 x (x - 2) + 3 (x - 2) = 0 \Rightarrow (x - 2) (2 x + 3) = 0$ .
So $x = 2$ or $x = - \frac{3}{2}$ .
Check: $2 (\frac{9}{4}) + \frac{3}{2} - 6 = \frac{9}{2} + \frac{3}{2} - 6 = 0$ . Correct.

Answer: $k = - 1$ , and the other root is $- \frac{3}{2}$ .

Solving by factorisation (splitting the middle term)

NCERT section 4.3 Solution of a Quadratic Equation by Factorisation

Factorisation rests on one fact: if a product of two numbers is zero, at least one of them is zero. So if you can write $a x^{2} + b x + c$ as $(p x + q) (r x + s)$ , the roots come from $p x + q = 0$ and $r x + s = 0$ .

To find the factors, split the middle term. Multiply $a$ and $c$ , then look for two numbers whose product is $a c$ and whose sum is $b$ . Rewrite $b x$ as the sum of these two terms, group in pairs and take out the common factor. The two brackets left must match; if they do not, recheck your numbers.

Sign guide: if $a c$ is positive, both numbers have the sign of $b$ ; if $a c$ is negative, they have opposite signs and the larger one takes the sign of $b$ . The method works with surds too, as long as you use $3 \times 3 = 3$ carefully.

Worked example

Solve by factorisation: $3 x^{2} + 10 x + 73 = 0$ .

Try it first, then show the solutionSolution

Product $a c = 3 \times 73 = 21$ ; we need two numbers with product 21 and sum 10: they are 7 and 3.
Split: $3 x^{2} + 7 x + 3 x + 73 = 0$ .
Write $3 x = 3 \cdot 3 x$ : $x (3 x + 7) + 3 (3 x + 7) = 0$ .
$(3 x + 7) (x + 3) = 0$ .
$3 x + 7 = 0 \Rightarrow x = - \frac{7}{3} = - \frac{7 3}{3}$ , or $x + 3 = 0 \Rightarrow x = - 3$ .
Check: product of roots $= (- \frac{7}{3}) (- 3) = 7 = \frac{c}{a} = \frac{7 3}{3}$ . Correct.

Answer: $x = - \frac{7 3}{3}$ or $x = - 3$ .

Equations with fractions that reduce to quadratics

NCERT section 4.3 Solution of a Quadratic Equation by Factorisation

Many board questions do not start as quadratics. They have $x$ in denominators, and only after you combine the fractions and cross-multiply do you get $a x^{2} + b x + c = 0$ .

The method: first write the restrictions, the values of $x$ that make any denominator zero, because the equation has no meaning there. Then take the LCM on the side with fractions, simplify the numerator, cross-multiply and bring everything to standard form. Solve by factorisation or by the formula.

Finally, compare each answer with the restrictions. A value that makes a denominator zero must be rejected, with the reason written in one line. Look for shortcuts too: when two fractions are subtracted, the $x$ terms in the numerator often cancel and leave just a number, which keeps the algebra short.

Worked example

Solve for $x$ : $\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}$ , where $x \neq = 1, - 5$ .

Try it first, then show the solutionSolution

Restrictions: $x \neq = 1$ and $x \neq = - 5$ , since denominators cannot be zero.
LHS $= \frac{( x + 5 ) - ( x - 1 )}{( x - 1 ) ( x + 5 )} = \frac{6}{( x - 1 ) ( x + 5 )}$ .
So $\frac{6}{( x - 1 ) ( x + 5 )} = \frac{6}{7} \Rightarrow (x - 1) (x + 5) = 7$ .
$x^{2} + 4 x - 5 = 7 \Rightarrow x^{2} + 4 x - 12 = 0$ .
$(x + 6) (x - 2) = 0 \Rightarrow x = - 6$ or $x = 2$ .
Neither value is a restricted value, so both are accepted.
Check $x = 2$ : $1 - \frac{1}{7} = \frac{6}{7}$ . Check $x = - 6$ : $- \frac{1}{7} + 1 = \frac{6}{7}$ . Both correct.

Answer: $x = 2$ or $x = - 6$ .

The quadratic formula

NCERT section 4.4 Nature of Roots

Some quadratics do not factorise neatly, for example when the roots are irrational. For these, use the quadratic formula: the roots of $a x^{2} + b x + c = 0$ are $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ . It works for every quadratic with real roots, so it is your safe method when splitting the middle term is not quick. You may quote and use it directly; you are not asked to derive it.

Work in a fixed order: write the equation in standard form, list $a$ , $b$ , $c$ with signs, compute $D = b^{2} - 4 a c$ on its own line, then substitute. Computing $D$ first catches the case $D < 0$ early and keeps the square root simple. When the coefficients contain letters, the same steps work, and $D$ usually turns out to be a perfect square such as $16 b^{2}$ .

Worked example

Using the quadratic formula, solve for $x$ : $4 x^{2} - 4 a x + (a^{2} - b^{2}) = 0$ .

Try it first, then show the solutionSolution

Compare with $A x^{2} + B x + C = 0$ : $A = 4$ , $B = - 4 a$ , $C = a^{2} - b^{2}$ .
$D = B^{2} - 4 A C = 16 a^{2} - 16 (a^{2} - b^{2}) = 16 b^{2}$ .
$D \geq 0$ , so real roots exist, and $D = 4 b$ (the $\pm$ in the formula covers the sign).
$x = \frac{- B \pm D}{2 A} = \frac{4 a \pm 4 b}{8}$ .
$x = \frac{a + b}{2}$ or $x = \frac{a - b}{2}$ .
Second-method check: $4 x^{2} - 4 a x + a^{2} - b^{2} = (2 x - a)^{2} - b^{2} = (2 x - a - b) (2 x - a + b)$ , which gives the same two roots.

Answer: $x = \frac{a + b}{2}$ or $x = \frac{a - b}{2}$ .

Discriminant and the nature of roots

NCERT section 4.4 Nature of Roots

The expression $D = b^{2} - 4 a c$ sits under the square root in the formula, and its sign decides what kind of roots you get. It is called the discriminant.

If $D > 0$ , $D$ is a positive real number, so the $\pm$ gives two different values: two distinct real roots.
If $D = 0$ , you add and subtract zero, so both roots equal $- \frac{b}{2 a}$ : two equal real roots, and the left side is a perfect square.
If $D < 0$ , there is no real square root, so there are no real roots.

Board questions use this in reverse: they give an equation with an unknown $k$ and a condition such as equal roots. You write $D$ in terms of $k$ , set it equal to zero and solve. If $k$ also appears in the coefficient of $x^{2}$ , reject any value that makes it zero, since then the equation is not quadratic.

Worked example

Find the value(s) of $k$ for which the equation $k x^{2} - 2 k x + 6 = 0$ has two equal real roots. Also find the roots.

Try it first, then show the solutionSolution

$A = k$ , $B = - 2 k$ , $C = 6$ .
Equal roots means $D = 0$ : $(- 2 k)^{2} - 4 (k) (6) = 0$ .
$4 k^{2} - 24 k = 0 \Rightarrow 4 k (k - 6) = 0 \Rightarrow k = 0$ or $k = 6$ .
Reject $k = 0$ : it makes the coefficient of $x^{2}$ zero and the equation becomes $6 = 0$ .
With $k = 6$ : $6 x^{2} - 12 x + 6 = 0 \Rightarrow 6 (x - 1)^{2} = 0$ .
Equal roots: $x = - \frac{B}{2 A} = \frac{12}{12} = 1$ .

Answer: $k = 6$ , and both roots are $1$ .

Using the discriminant for ranges and proofs

NCERT section 4.4 Nature of Roots

The discriminant also answers questions about ranges of a constant and proofs about the roots.

Real roots means $D \geq 0$ (distinct or equal), so you solve an inequality, not an equation.
Distinct real roots means $D > 0$ .
No real roots means $D < 0$ .

For a proof question, the coefficients contain letters such as $a$ and $b$ . Find $D$ , expand fully and try to write it as a square, or as minus a square. Any real square is never negative, so $(a - b)^{2} \geq 0$ always and $- (a - b)^{2} \leq 0$ always. This tells you the sign of $D$ without knowing the actual values of $a$ and $b$ . State clearly when the equality case happens, for example when $a = b$ , because a complete answer covers both the strict case and the boundary case.

Worked example

Prove that the equation $2 x^{2} - 2 (a + b) x + (a^{2} + b^{2}) = 0$ has no real roots when $a \neq = b$ . What are the roots when $a = b$ ?

Try it first, then show the solutionSolution

$A = 2$ , $B = - 2 (a + b)$ , $C = a^{2} + b^{2}$ .
$D = B^{2} - 4 A C = 4 (a + b)^{2} - 8 (a^{2} + b^{2})$ .
$D = 4 (a^{2} + 2 ab + b^{2} - 2 a^{2} - 2 b^{2}) = 4 (- a^{2} + 2 ab - b^{2})$ .
$D = - 4 (a - b)^{2}$ .
If $a \neq = b$ , then $(a - b)^{2} > 0$ , so $D < 0$ : the equation has no real roots. Hence proved.
If $a = b$ , then $D = 0$ : two equal roots $x = - \frac{B}{2 A} = \frac{2 ( a + b )}{4} = \frac{2 a + 2 a}{4} = a$ .

Answer: No real roots when $a \neq = b$ ; when $a = b$ the two roots are equal, both equal to $a$ .

Word problems: speed, time and work

NCERT section 4.3 Solution of a Quadratic Equation by Factorisation

Situational problems are where the chapter meets daily life. The skill being tested is modelling: turning words into one equation in one unknown.

For speed, time and distance, use time $= \frac{distance}{speed}$ . Let the unknown speed be $x$ , write the two times as fractions, and use the given difference in time to form the equation. For boats and streams, the speed upstream is $(u - v)$ and downstream is $(u + v)$ , where $u$ is the speed in still water and $v$ the speed of the stream. For work problems, if someone finishes a job in $x$ hours, they do $\frac{1}{x}$ of it in one hour.

After solving, check each root against the situation. Speeds, times and lengths cannot be negative, so reject such values with a reason. Then answer exactly what was asked, with units.

Worked example

A cyclist covers 60 km at a uniform speed. If her speed had been 5 km/h more, she would have taken 1 hour less for the same journey. Find her original speed.

Try it first, then show the solutionSolution

Let the original speed be $x$ km/h, where $x > 0$ .
Time at original speed $= \frac{60}{x}$ h; time at increased speed $= \frac{60}{x + 5}$ h.
Given: $\frac{60}{x} - \frac{60}{x + 5} = 1$ .
$\frac{60 ( x + 5 ) - 60 x}{x ( x + 5 )} = 1 \Rightarrow \frac{300}{x ( x + 5 )} = 1$ .
$x^{2} + 5 x - 300 = 0$ .
Split: $20 \times (- 15) = - 300$ and $20 - 15 = 5$ , so $(x + 20) (x - 15) = 0$ .
$x = 15$ or $x = - 20$ ; reject $x = - 20$ because speed cannot be negative.
Check: $\frac{60}{15} = 4$ h and $\frac{60}{20} = 3$ h, a difference of 1 h. Correct.

Answer: Her original speed is 15 km/h.

Word problems: area, numbers and ages

NCERT section 4.3 Solution of a Quadratic Equation by Factorisation

Area, number and age problems follow the same plan: choose one variable, express every other quantity in terms of it, and use the one fact that connects them.

Rectangles: area $= l b$ , perimeter $= 2 (l + b)$ , and the diagonal $d$ satisfies $l^{2} + b^{2} = d^{2}$ .
Right triangles: use Pythagoras with the sides written in terms of $x$ .
Two-digit numbers: a number with tens digit $t$ and units digit $u$ is $10 t + u$ .
Ages: if the present age is $x$ , the age $n$ years ago is $x - n$ .

The equation often has a common factor, so divide through before factorising to keep the numbers small. Reject negative lengths or ages. Case-study questions often add one more step, such as a cost of fencing, so read the last sub-part before you stop.

Worked example

A rectangular garden has its length 4 m more than its breadth, and its diagonal is 20 m long. Find the length and breadth of the garden, and the cost of fencing it all round at Rs 50 per metre.

Try it first, then show the solutionSolution

Let the breadth be $x$ m; then the length is $(x + 4)$ m.
The diagonal and two adjacent sides form a right triangle, so by Pythagoras $x^{2} + (x + 4)^{2} = 2 0^{2}$ .
$2 x^{2} + 8 x + 16 = 400 \Rightarrow 2 x^{2} + 8 x - 384 = 0 \Rightarrow x^{2} + 4 x - 192 = 0$ .
Split: $16 \times (- 12) = - 192$ and $16 - 12 = 4$ , so $(x + 16) (x - 12) = 0$ .
$x = 12$ or $x = - 16$ ; reject $x = - 16$ because a length cannot be negative.
Breadth $= 12$ m, length $= 16$ m. Check: $1 2^{2} + 1 6^{2} = 144 + 256 = 400 = 2 0^{2}$ .
Perimeter $= 2 (16 + 12) = 56$ m.
Cost of fencing $= 56 \times 50$ = Rs 2800.

Answer: Length 16 m, breadth 12 m; cost of fencing is Rs 2800.

Maths in real life: Quadratic Equations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Solving by factorisation (splitting the middle term)

Imagine your school is setting up chairs in the auditorium for the annual day. The teacher wants every row to have the same number of chairs. She wants each row to have 6 more chairs than there are rows. Exactly 280 chairs are available, and all of them must be used.

How many rows should there be, and how many chairs go in each row? Form a quadratic equation and solve it by splitting the middle term.

Try it first, then show the solutionSolution

Let the number of rows be $x$ . Then chairs in each row $= x + 6$ .
Total chairs: $x (x + 6) = 280$ .
Standard form: $x^{2} + 6 x - 280 = 0$ .
We need two numbers with product $a c = - 280$ and sum $b = 6$ . These are $20$ and $- 14$ .
Split the middle term: $x^{2} + 20 x - 14 x - 280 = 0$ .
Group in pairs: $x (x + 20) - 14 (x + 20) = 0$ .
Take out the common factor: $(x + 20) (x - 14) = 0$ .
So $x = - 20$ or $x = 14$ .
The number of rows cannot be negative, so reject $x = - 20$ . Hence $x = 14$ .
Chairs in each row $= 14 + 6 = 20$ .
Check: $14 \times 20 = 280$ , which is correct.

Answer: 14 rows with 20 chairs in each row.

Memory hook: Rows times chairs per row gives the total: the product becomes your quadratic.

Equations with fractions that reduce to quadratics

Suppose your class books a bus for a school trip at a fixed hire charge of Rs 7200, shared equally by the students going. At the last minute, 8 more students join the trip. Because of this, each student's share drops by Rs 30.

How many students had planned to go at first, and what did each student finally pay?

Try it first, then show the solutionSolution

Let the original number of students be $x$ . Restrictions: $x \neq = 0$ and $x \neq = - 8$ .
Original share $= \frac{7200}{x}$ rupees. New share $= \frac{7200}{x + 8}$ rupees.
Given: $\frac{7200}{x} - \frac{7200}{x + 8} = 30$ .
Take the LCM: $\frac{7200 ( x + 8 ) - 7200 x}{x ( x + 8 )} = 30$ , which gives $\frac{57600}{x ( x + 8 )} = 30$ .
Cross-multiply and divide by 30: $x (x + 8) = 1920$ .
Standard form: $x^{2} + 8 x - 1920 = 0$ .
Here $D = 8^{2} - 4 (1) (- 1920) = 64 + 7680 = 7744$ , and $7744 = 88$ .
So $x = \frac{- 8 \pm 88}{2}$ , which gives $x = 40$ or $x = - 48$ .
The number of students cannot be negative, so $x = 40$ . This does not break any restriction.
Final share $= \frac{7200}{48} = 150$ rupees.
Check: $\frac{7200}{40} = 180$ , and $180 - 150 = 30$ , which is correct.

Answer: 40 students had planned to go, and each student finally paid Rs 150.

Memory hook: More friends share the bus, so the share shrinks: old share minus new share gives the equation.

The quadratic formula

Imagine you are framing a 20 cm by 15 cm photograph of your family for Diwali. The frame forms a border of the same width all around the photo. The photo and frame together must cover exactly 500 square centimetres of wall.

Find the width of the frame. Use the quadratic formula.

Try it first, then show the solutionSolution

Let the width of the border be $x$ cm. The outer dimensions are then $(20 + 2 x)$ cm and $(15 + 2 x)$ cm.
Total area: $(20 + 2 x) (15 + 2 x) = 500$ .
Expand: $300 + 70 x + 4 x^{2} = 500$ , so $4 x^{2} + 70 x - 200 = 0$ .
Divide by 2: $2 x^{2} + 35 x - 100 = 0$ . So $a = 2$ , $b = 35$ , $c = - 100$ .
$D = 3 5^{2} - 4 (2) (- 100) = 1225 + 800 = 2025$ , and $2025 = 45$ .
$x = \frac{- 35 \pm 45}{2 \times 2} = \frac{- 35 \pm 45}{4}$ .
So $x = \frac{10}{4} = 2.5$ or $x = \frac{- 80}{4} = - 20$ .
A width cannot be negative, so reject $- 20$ . Hence $x = 2.5$ .
Check: $(20 + 5) (15 + 5) = 25 \times 20 = 500$ , which is correct.

Answer: The frame is 2.5 cm wide, so the outer size is 25 cm by 20 cm.

Memory hook: Border on both sides adds $2 x$ to each side; when splitting is slow, the formula frames it.

Discriminant and the nature of roots

Suppose your housing society is making a rectangular rangoli for a festival. The volunteers have exactly 36 m of coloured ribbon to run along its border. Two volunteers argue about how large the rangoli can be.

(a) Can the rangoli have an area of 90 square metres? (b) Can it have an area of 81 square metres? If it can, find its length and breadth. Use the discriminant to decide.

Try it first, then show the solutionSolution

The perimeter is 36 m, so length $+$ breadth $= 18$ m. Let the length be $l$ m. Then the breadth $= (18 - l)$ m.
(a) Area: $l (18 - l) = 90$ , which gives $l^{2} - 18 l + 90 = 0$ .
$D = (- 18)^{2} - 4 (1) (90) = 324 - 360 = - 36$ .
Since $D < 0$ , there are no real roots. So no such rectangle exists.
(b) Area: $l (18 - l) = 81$ , which gives $l^{2} - 18 l + 81 = 0$ .
$D = 324 - 4 (1) (81) = 324 - 324 = 0$ .
Since $D = 0$ , there are two equal real roots: $l = - \frac{b}{2 a} = \frac{18}{2} = 9$ .
Breadth $= 18 - 9 = 9$ m, so the rangoli is a square.
Check: $9 \times 9 = 81$ and $2 (9 + 9) = 36$ , both correct.

Answer: (a) No, an area of 90 square metres is not possible, because the discriminant is negative. (b) Yes, an area of 81 square metres is possible, with length 9 m and breadth 9 m (a square).

Memory hook: The discriminant is the gatekeeper: negative means the rangoli cannot be drawn, zero means only a square fits.

Word problems: speed, time and work

Imagine the water tank on your building's terrace is filled by two taps. Working alone, the smaller tap takes 5 hours longer than the larger tap to fill the empty tank. With both taps open together, the tank fills in 6 hours.

How long does each tap take to fill the tank on its own?

Try it first, then show the solutionSolution

Let the larger tap fill the tank alone in $x$ hours. Then the smaller tap takes $(x + 5)$ hours, with $x > 0$ .
Work done in 1 hour: $\frac{1}{x}$ by the larger tap and $\frac{1}{x + 5}$ by the smaller tap. Together they do $\frac{1}{6}$ of the job in 1 hour.
Equation: $\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}$ .
Take the LCM: $\frac{2 x + 5}{x ( x + 5 )} = \frac{1}{6}$ .
Cross-multiply: $6 (2 x + 5) = x^{2} + 5 x$ , which gives $x^{2} + 5 x - 12 x - 30 = 0$ .
Standard form: $x^{2} - 7 x - 30 = 0$ .
Split the middle term: $x^{2} - 10 x + 3 x - 30 = 0$ , so $(x - 10) (x + 3) = 0$ .
So $x = 10$ or $x = - 3$ . Time cannot be negative, so $x = 10$ .
The smaller tap takes $10 + 5 = 15$ hours.
Check: $\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{1}{6}$ , which is correct.

Answer: The larger tap takes 10 hours and the smaller tap takes 15 hours.

Memory hook: Taps add their hourly shares: one over x plus one over (x + 5) equals one over 6.

Topics in this chapter

Quadratic Equations formula sheet

Standard form: $a x^{2} + b x + c = 0, a \neq = 0$; Rewrite every equation in this form first; read a, b and c with their signs only from here.
Condition for a root: $a α^{2} + b α + c = 0$; Use to check a given value, or to find an unknown coefficient when one root is given.
Zero product rule: $pq = 0 \Rightarrow p = 0 or q = 0$; The reason factorisation works. Never divide both sides by x; factor it out instead.
Splitting the middle term: $p + q = b, pq = a c$; Find p and q, write bx as px + qx, then group in pairs.
Discriminant: $D = b^{2} - 4 a c$; Compute it on its own line before using the formula or deciding the nature of roots.
Quadratic formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}, b^{2} - 4 a c \geq 0$; Works for every quadratic with real roots. The whole numerator is divided by 2a.
Two distinct real roots: $D > 0 \Rightarrow x = \frac{- b + D}{2 a}, \frac{- b - D}{2 a}$; Used for 'distinct' or 'unequal' real roots.
Two equal real roots: $D = 0 \Rightarrow x = - \frac{b}{2 a}$; Used in 'find k for equal roots'; the expression is then a perfect square.
No real roots: $D < 0$; State 'no real roots'; do not try to find square roots of negative numbers.
Real roots (equal or distinct): $D \geq 0$; Used when the question only says 'real roots'; gives an inequality for k.
Sum and product check (from Polynomials): $α + β = - \frac{b}{a}, α β = \frac{c}{a}$; A quick way to check your two roots after solving.
Algebraic identities: $(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}, a^{2} - b^{2} = (a - b) (a + b)$; Needed for expanding brackets and for writing D as a perfect square in proofs.
Speed, distance and time: $time = \frac{distance}{speed}$; Journey problems: write two times as fractions and use the given difference.
Boat and stream speeds: $upstream = u - v, downstream = u + v$; u is the speed in still water, v the speed of the stream; u must be greater than v.
Work rate: $work done in 1 hour = \frac{1}{x}$; If a job takes x hours alone; rates of people or machines working together add.
Two-digit number: $10 t + u$; t is the tens digit, u the units digit; the reversed number is 10u + t.
Rectangle and right triangle: $A = l b, P = 2 (l + b), l^{2} + b^{2} = d^{2}$; Area, perimeter and diagonal relations for geometric word problems.

How the board asks Quadratic Equations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Identify a quadratic equation	An MCQ or assertion-reason item gives equations with brackets on both sides and asks which one is quadratic.	Expand every bracket, cancel like terms, bring everything to one side. It is quadratic only if the $x^{2}$ term survives and nothing higher remains.
Given one root, find k and the other root	A 2- or 3-mark question states that a number is a root of an equation with an unknown coefficient.	Substitute the root to get a linear equation in k. Put k back, solve the complete quadratic, and check the second root by substitution.
Solve by factorisation, including surds and fractions	Solve an equation such as one with $2$ or $3$ in the coefficients, or one with x in the denominators and stated restrictions.	For surds, multiply a and c and split normally. For fractions, write restrictions, combine, cross-multiply, reach standard form, factorise, then reject any restricted value.
Solve using the quadratic formula	Equations whose roots are irrational, or with letters like a and b in the coefficients, often stated as 'using the quadratic formula'.	List A, B, C with signs, compute D on its own line, simplify D (often a perfect square in letter questions), substitute, and simplify each root separately.
Find k for equal, real, distinct or no real roots	A 1- to 3-mark question gives an equation with k and a condition on the roots; sometimes it also asks for the roots.	Write D in terms of k. Equal: D = 0. Real: D ≥ 0. Distinct: D > 0. No real roots: D < 0. Reject values that make the $x^{2}$ coefficient zero.
Prove the nature of roots	Show that an equation with letter coefficients has no real roots, or has equal roots under a condition.	Expand D fully and rewrite it as a square or minus a square. Use the fact that a square is never negative and state the equality case.
Situational word problems	3- to 5-mark problems on journeys, boats, work rates, areas, two-digit numbers, ages or prices.	Define the variable with units, express other quantities in terms of it, form one equation, reduce to standard form, solve, reject impossible roots with a reason, and answer what was asked with units.
Case study and assertion-reason	A real-life passage (a garden, a trip, a shop) with sub-parts: form the equation, find the nature of roots, then solve; or an A-R pair about roots and the discriminant.	Treat each sub-part as a small question. Form the equation carefully in the first part since later parts depend on it. In A-R, judge each statement separately before asking whether R explains A.

Tricks to remember Quadratic Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Product and sum for splitting
Product = ac, Sum = b. If ac is positive, both numbers share the sign of b. If ac is negative, the signs differ and the bigger number takes the sign of b.
Use it when: Find the two numbers quickly when factorising, without trial and error on signs.
Positive pair, zero twins, negative none
D positive: a pair of different real roots. D zero: twin (equal) roots. D negative: no real roots.
Use it when: Nature-of-roots MCQs and 'find k' questions.
Everything over 2a
Say the formula as 'minus b, plus or minus root D, ALL over 2a', and draw the fraction bar under the whole numerator first.
Use it when: Stops the common error of dividing only the square root by 2a.
Perfect square D means it factorises
With rational coefficients, if D is a perfect square the roots are rational, so splitting the middle term will work. If not, go straight to the formula.
Use it when: Decide your method in seconds before you start.
Substitute or use sum-product to check
After solving, check one root by substitution, or check that the roots add to $- \frac{b}{a}$ and multiply to $\frac{c}{a}$ .
Use it when: Catch sign slips before you move on, especially in long answers.
The three rejections
Reject a value of k that makes a = 0, a value of x that makes a denominator zero, and a negative length, speed, time or age.
Use it when: Final line of fraction, discriminant and word problems.

Quadratic Equations: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Spot the equation with distinct real roots · 1 mark · MCQBoard level

Which of the following quadratic equations has two distinct real roots?
(a) $x^{2} + x + 1 = 0$
(b) $2 x^{2} - 4 x + 2 = 0$
(c) $3 x^{2} + 2 x - 1 = 0$
(d) $x^{2} + 4 = 0$

Try it first, then show the solutionSolution

(a) $D = 1^{2} - 4 (1) (1) = - 3 < 0$ : no real roots.
(b) $D = (- 4)^{2} - 4 (2) (2) = 16 - 16 = 0$ : two equal real roots.
(c) $D = 2^{2} - 4 (3) (- 1) = 4 + 12 = 16 > 0$ : two distinct real roots.
(d) $D = 0^{2} - 4 (1) (4) = - 16 < 0$ : no real roots.

Answer: (c) $3 x^{2} + 2 x - 1 = 0$ , since its discriminant is $16 > 0$ .

Example 2: Opposite signs of a and c · 1 mark · Assertion-reasonBoard level

Assertion (A): The equation $x^{2} + p x - 4 = 0$ has two distinct real roots for every real value of $p$ .
Reason (R): If $a$ and $c$ have opposite signs in $a x^{2} + b x + c = 0$ , then $b^{2} - 4 a c > 0$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Check R: if $a$ and $c$ have opposite signs, $a c < 0$ , so $- 4 a c > 0$ .
Since $b^{2} \geq 0$ , we get $b^{2} - 4 a c > 0$ . So R is true.
Check A: $a = 1$ , $c = - 4$ have opposite signs, so $D = p^{2} + 16 > 0$ for every real $p$ . So A is true.
A follows directly from R, so R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Use one root to unlock another equation · 3 marks · Short answerHarder board level

If $x = - 2$ is a root of the equation $3 x^{2} + 7 x + p = 0$ , find the values of $k$ for which the equation $x^{2} + k (4 x + k - 1) + p = 0$ has two equal real roots.

Try it first, then show the solutionSolution

Put $x = - 2$ in $3 x^{2} + 7 x + p = 0$ : $12 - 14 + p = 0$ , so $p = 2$ . (1 mark)
Second equation becomes $x^{2} + 4 k x + (k^{2} - k + 2) = 0$ . (½ mark)
For equal roots, $D = 0$ : $(4 k)^{2} - 4 (1) (k^{2} - k + 2) = 0$ . (½ mark)
$16 k^{2} - 4 k^{2} + 4 k - 8 = 0$ , so $12 k^{2} + 4 k - 8 = 0$ , that is $3 k^{2} + k - 2 = 0$ .
Factorise: $3 k^{2} + 3 k - 2 k - 2 = 0$ , so $(3 k - 2) (k + 1) = 0$ .
So $k = \frac{2}{3}$ or $k = - 1$ . (1 mark)
Check $k = - 1$ : equation is $x^{2} - 4 x + 4 = 0 = (x - 2)^{2}$ , equal roots. Correct.

Answer: $p = 2$ and $k = \frac{2}{3}$ or $k = - 1$ .

Example 4: Fraction equation with surd roots · 3 marks · Short answerHarder board level

Solve for $x$ using the quadratic formula: $\frac{1}{x + 2} - \frac{1}{x - 6} = \frac{8}{9}$ , where $x \neq = - 2, 6$ .

Try it first, then show the solutionSolution

Combine the left side: $\frac{( x - 6 ) - ( x + 2 )}{( x + 2 ) ( x - 6 )} = \frac{- 8}{( x + 2 ) ( x - 6 )}$ . (1 mark)
So $\frac{- 8}{( x + 2 ) ( x - 6 )} = \frac{8}{9}$ , giving $(x + 2) (x - 6) = - 9$ .
$x^{2} - 4 x - 12 = - 9$ , so $x^{2} - 4 x - 3 = 0$ . (½ mark)
$a = 1$ , $b = - 4$ , $c = - 3$ , so $D = 16 + 12 = 28$ and $28 = 27$ . (½ mark)
$x = \frac{4 \pm 2 7}{2} = 2 \pm 7$ . (1 mark)
Neither value is $- 2$ or $6$ , so both are accepted.

Answer: $x = 2 + 7$ or $x = 2 - 7$ .

Example 5: Motorboat in a stream · 5 marks · Long answerHarder board level

The speed of a stream is 2 km/h. A motorboat goes 24 km upstream and returns to its starting point in a total of 5 hours. Find the speed of the motorboat in still water.

Try it first, then show the solutionSolution

Let the speed of the boat in still water be $u$ km/h, with $u > 2$ . (½ mark)
Upstream speed $= (u - 2)$ km/h, downstream speed $= (u + 2)$ km/h.
Time upstream $= \frac{24}{u - 2}$ h, time downstream $= \frac{24}{u + 2}$ h. (1 mark)
Total time: $\frac{24}{u - 2} + \frac{24}{u + 2} = 5$ . (1 mark)
$24 (u + 2) + 24 (u - 2) = 5 (u^{2} - 4)$ , so $48 u = 5 u^{2} - 20$ .
$5 u^{2} - 48 u - 20 = 0$ . (1 mark)
$D = (- 48)^{2} - 4 (5) (- 20) = 2304 + 400 = 2704$ , and $2704 = 52$ .
$u = \frac{48 \pm 52}{10}$ , so $u = 10$ or $u = - \frac{2}{5}$ . (1 mark)
Speed cannot be negative, so $u = 10$ . Check: $\frac{24}{8} + \frac{24}{12} = 3 + 2 = 5$ h. Correct. (½ mark)

Answer: The speed of the motorboat in still water is 10 km/h.

Example 6: Ages of mother and daughter · 5 marks · Long answerHarder board level

The sum of the present ages of a mother and her daughter is 50 years. Four years ago, the product of their ages (in years) was 272. Find their present ages. Also find after how many years the mother's age will be twice her daughter's age.

Try it first, then show the solutionSolution

Let the mother's present age be $x$ years. Then the daughter's age is $(50 - x)$ years. (½ mark)
Four years ago: mother $(x - 4)$ years, daughter $(46 - x)$ years. (½ mark)
Given: $(x - 4) (46 - x) = 272$ . (1 mark)
$46 x - x^{2} - 184 + 4 x = 272$ , so $x^{2} - 50 x + 456 = 0$ . (1 mark)
Numbers with product 456 and sum 50: 38 and 12. So $(x - 38) (x - 12) = 0$ , giving $x = 38$ or $x = 12$ . (1 mark)
The mother is older, so $x = 38$ : mother 38 years, daughter 12 years. Check: $34 \times 8 = 272$ . (½ mark)
After $t$ years: $38 + t = 2 (12 + t)$ , so $t = 14$ . Check: $52 = 2 \times 26$ . (½ mark)

Answer: Mother is 38 years and daughter is 12 years old. After 14 years, the mother will be twice as old as her daughter.

Example 7: Find the error in the formula · 3 marks · Short answerHOTS / competency

Riya solved $2 x^{2} - 5 x - 3 = 0$ using the quadratic formula and wrote:
$x = \frac{- 5 \pm 25 - 24}{4} = \frac{- 5 \pm 1}{4}$ , so $x = - 1$ or $x = - \frac{3}{2}$ .
Identify her mistakes and find the correct roots. Verify one root by substitution.

Try it first, then show the solutionSolution

Here $a = 2$ , $b = - 5$ , $c = - 3$ . Mistake 1: $- b = + 5$ , not $- 5$ . Mistake 2: $- 4 a c = - 4 (2) (- 3) = + 24$ , so $D = 25 + 24$ , not $25 - 24$ . (1 mark)
Correct $D = 49$ , so $D = 7$ . (1 mark)
$x = \frac{5 \pm 7}{4}$ , so $x = 3$ or $x = - \frac{1}{2}$ . Check $x = 3$ : $18 - 15 - 3 = 0$ . Correct. (1 mark)

Answer: Riya used the wrong sign for $- b$ and for $- 4 a c$ . The correct roots are $x = 3$ and $x = - \frac{1}{2}$ .

Example 8: How high can the ball go? · 5 marks · Long answerHOTS / competency

A ball is thrown straight up. Its height above the ground after $t$ seconds is $h = 20 t - 5 t^{2}$ metres.
(i) At what times is the ball at a height of 15 m? Explain why there are two answers.
(ii) Using the discriminant, show that the ball never reaches a height of 25 m.
(iii) Using the discriminant, find the greatest height the ball reaches and the time at which it reaches it.

Try it first, then show the solutionSolution

(i) $20 t - 5 t^{2} = 15$ gives $5 t^{2} - 20 t + 15 = 0$ , so $t^{2} - 4 t + 3 = 0$ . (1 mark)
$(t - 1) (t - 3) = 0$ , so $t = 1$ s or $t = 3$ s: once going up and once coming down. (1 mark)
(ii) $20 t - 5 t^{2} = 25$ gives $t^{2} - 4 t + 5 = 0$ . (½ mark)
$D = 16 - 20 = - 4 < 0$ , so no real time exists. The ball never reaches 25 m. (1 mark)
(iii) For height $h$ : $5 t^{2} - 20 t + h = 0$ . A real time exists only if $D = 400 - 20 h \geq 0$ , so $h \leq 20$ . (1 mark)
Greatest height is 20 m, when $D = 0$ , at $t = - \frac{b}{2 a} = \frac{20}{10} = 2$ s. (½ mark)

Answer: (i) At 1 s and 3 s. (ii) For 25 m, $D = - 4 < 0$ , so not possible. (iii) Greatest height 20 m, reached at 2 s.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Eco Club sapling drive in Bhopal

The Eco Club of a school in Bhopal has Rs 3600 to spend on neem saplings for Van Mahotsav. At the nursery, every sapling costs the same, Rs $x$ each. The club secretary finds that a government scheme gives a discount of Rs 30 on each sapling. With the discount, the same Rs 3600 buys 20 more saplings than it would at the full price. The club wants to know the full price of a sapling and how many saplings it can now plant along the school boundary.

(i) · 1 mark
Form a quadratic equation in $x$ to represent this situation.
Answer
Answer: $x^{2} - 30 x - 5400 = 0$
$\frac{3600}{x - 30} - \frac{3600}{x} = 20$ gives $\frac{3600 \times 30}{x ( x - 30 )} = 20$ , so $x (x - 30) = 5400$ , that is $x^{2} - 30 x - 5400 = 0$ .
(ii) · 1 mark
Find the discriminant of this equation and state the nature of its roots.
Answer
Answer: $D = 22500$ ; two distinct real roots.
$D = (- 30)^{2} - 4 (1) (- 5400) = 900 + 21600 = 22500 > 0$ , so the roots are real and distinct.
(iii) · 2 marks
Find the full price of one sapling and the number of saplings the club can buy with the discount.
Answer
Answer: Full price Rs 90; 60 saplings with the discount.
$22500 = 150$ , so $x = \frac{30 \pm 150}{2} = 90$ or $- 60$ . (1 mark) Price cannot be negative, so $x = 90$ . Discounted price Rs 60, so saplings $= \frac{3600}{60} = 60$ . Check: at full price $\frac{3600}{90} = 40$ , and $60 - 40 = 20$ . (1 mark)

Kitchen garden against a society wall in Pune

A residents' welfare association in Pune has 40 m of fencing wire to make a rectangular kitchen garden against the long boundary wall of the society. The wall forms one side of the garden, so wire is needed only for the other three sides: two sides of width $x$ m that are perpendicular to the wall, and one side parallel to the wall. The committee first wants a garden of area 192 square metres. Some members would like the garden to be as large as possible.

(i) · 1 mark
Form a quadratic equation in $x$ for a garden of area 192 square metres.
Answer
Answer: $x^{2} - 20 x + 96 = 0$
The side parallel to the wall is $(40 - 2 x)$ m. Area: $x (40 - 2 x) = 192 \Rightarrow 2 x^{2} - 40 x + 192 = 0 \Rightarrow x^{2} - 20 x + 96 = 0$ .
(ii) · 1 mark
Find the possible dimensions of the garden.
Answer
Answer: 8 m by 24 m, or 12 m by 16 m.
$(x - 8) (x - 12) = 0$ , so $x = 8$ or $x = 12$ . Both give a positive length: $40 - 16 = 24$ m and $40 - 24 = 16$ m. Check: $8 \times 24 = 12 \times 16 = 192$ .
(iii) · 2 marks
A member wants an area of 250 square metres. Use the discriminant to decide whether this is possible, and find the greatest area the 40 m of fencing can enclose.
Answer
Answer: 250 square metres is not possible; the greatest area is 200 square metres (10 m by 20 m).
$x (40 - 2 x) = 250$ gives $x^{2} - 20 x + 125 = 0$ , and $D = 400 - 500 = - 100 < 0$ , so no real $x$ exists: not possible. (1 mark) For an area $A$ : $2 x^{2} - 40 x + A = 0$ has real roots only if $D = 1600 - 8 A \geq 0$ , so $A \leq 200$ . When $A = 200$ , $D = 0$ and $x = \frac{40}{4} = 10$ m, with the side along the wall $40 - 20 = 20$ m. (1 mark)

Uniform order at two tailoring units in Tiruppur

Two tailoring units in Tiruppur stitch school uniforms. Working together, they complete a standard order in 12 days. Unit B has older machines: working alone, it takes 10 days more than Unit A working alone. Let Unit A alone take $x$ days to complete the order.

(i) · 1 mark
Form a quadratic equation in $x$ to represent this situation.
Answer
Answer: $x^{2} - 14 x - 120 = 0$
Work per day: $\frac{1}{x} + \frac{1}{x + 10} = \frac{1}{12}$ . So $12 (2 x + 10) = x (x + 10)$ , giving $24 x + 120 = x^{2} + 10 x$ , that is $x^{2} - 14 x - 120 = 0$ .
(ii) · 1 mark
Find the discriminant of this equation and state the nature of its roots.
Answer
Answer: $D = 676$ ; two distinct real roots.
$D = (- 14)^{2} - 4 (1) (- 120) = 196 + 480 = 676 > 0$ , so the roots are real and distinct. Since $676 = 2 6^{2}$ is a perfect square, the roots are also rational.
(iii) · 2 marks
Find the number of days each unit takes alone. If Unit A works alone for the first 5 days and then Unit B joins it, in how many more days is the order finished?
Answer
Answer: Unit A 20 days, Unit B 30 days; 9 more days.
$x = \frac{14 \pm 26}{2} = 20$ or $- 6$ ; days cannot be negative, so Unit A takes 20 days and Unit B takes 30 days. Check: $\frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}$ . (1 mark) In 5 days Unit A does $\frac{5}{20} = \frac{1}{4}$ of the order. The remaining $\frac{3}{4}$ at $\frac{1}{12}$ per day takes $\frac{3}{4} \times 12 = 9$ days. (1 mark)

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the values of $k$ for which the quadratic equation $(k + 1) x^{2} - 2 (k - 1) x + 1 = 0$ has two equal real roots. (2 marks)

Step	Marks
Here $A = k + 1$ , $B = - 2 (k - 1)$ , $C = 1$ . Condition for equal roots: $D = B^{2} - 4 A C = 0$ , that is $4 (k - 1)^{2} - 4 (k + 1) = 0$	0.5
Expand $4 (k - 1)^{2}$ : $4 (k^{2} - 2 k + 1) - 4 k - 4 = 0 \Rightarrow 4 k^{2} - 8 k + 4 - 4 k - 4 = 0 \Rightarrow 4 k^{2} - 12 k = 0$	0.5
Factorise: $4 k (k - 3) = 0$	0.5
$k = 0$ or $k = 3$ ; both keep $k + 1 \neq = 0$ , so both are valid	0.5

Solve for $x$ : $\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15}$ , where $x \neq = 0, - 1$ . (3 marks)

Step	Marks
Combine the fractions: $\frac{x ^{2} + ( x + 1 ) ^{2}}{x ( x + 1 )} = \frac{34}{15}$	0.5
Cross-multiply: $15 (2 x^{2} + 2 x + 1) = 34 (x^{2} + x)$	0.5
Standard form: $4 x^{2} + 4 x - 15 = 0$	0.5
Split the middle term: $4 x^{2} + 10 x - 6 x - 15 = 0$	0.5
Factorise: $(2 x + 5) (2 x - 3) = 0$	0.5
$x = \frac{3}{2}$ or $x = - \frac{5}{2}$ ; neither is a restricted value, so both are accepted	0.5

Two machines working together can pack an order in 6 hours. The faster machine alone takes 5 hours less than the slower machine alone. Find the time each machine takes to pack the order alone. (5 marks)

Step	Marks
Let the slower machine take $x$ hours; the faster takes $(x - 5)$ hours, with $x > 5$	0.5
Work in 1 hour: $\frac{1}{x}$ and $\frac{1}{x - 5}$ ; together $\frac{1}{6}$	0.5
Form the equation $\frac{1}{x} + \frac{1}{x - 5} = \frac{1}{6}$	1
Simplify: $6 (2 x - 5) = x (x - 5) \Rightarrow x^{2} - 17 x + 30 = 0$	1
Factorise: $(x - 15) (x - 2) = 0 \Rightarrow x = 15$ or $x = 2$	1
Reject $x = 2$ , since the faster machine would then take $- 3$ hours	0.5
Slower machine 15 hours, faster machine 10 hours; check $\frac{1}{15} + \frac{1}{10} = \frac{1}{6}$	0.5

Where students lose marks in Quadratic Equations

The slips examiners see most often, and what to write instead.

Mistake: Reading a, b, c before writing standard form
Students copy the numbers as they appear, forgetting that terms change sign when moved across the equals sign.
Wrong working
For $2 x^{2} = 3 x - 5$ : $a = 2, b = 3, c = - 5$ , so $D = 9 + 40 = 49 > 0$ , two distinct real roots.
Right working
$2 x^{2} - 3 x + 5 = 0$ , so $a = 2, b = - 3, c = 5$ ; $D = 9 - 40 = - 31 < 0$ , so no real roots.
Do this: Always write the equation as $a x^{2} + b x + c = 0$ on its own line, then list a, b and c with signs.
Usually costs: The whole mark in an MCQ; ½ to 1 mark in a written answer, plus every step that follows.
Mistake: Losing the sign in -b
When b is negative, students write -b as negative again instead of positive.
Wrong working
$x^{2} - 7 x + 10 = 0$ : $x = \frac{- 7 \pm 3}{2} = - 2, - 5$
Right working
$b = - 7$ , so $- b = 7$ and $D = 49 - 40 = 9$ : $x = \frac{7 \pm 3}{2} = 5, 2$
Do this: Write the line ' $- b = \dots$ ' separately before substituting into the formula.
Usually costs: Usually 1 mark or more, since both roots come out wrong.
Mistake: Dividing only the square root by 2a
The formula is written in one line without a full fraction bar.
Wrong working
$x^{2} - 6 x + 8 = 0$ : $x = 6 \pm \frac{2}{2} = 7, 5$
Right working
$D = 36 - 32 = 4$ , so $x = \frac{6 \pm 2}{2} = 4, 2$
Do this: Draw the fraction bar under the whole numerator $- b \pm D$ before writing anything else.
Usually costs: Usually 1 mark for the roots; the method mark may still be given if the formula was quoted correctly.
Mistake: Dividing both sides by x
It looks like simplifying, but it throws away the root $x = 0$ .
Wrong working
$3 x^{2} = 12 x \Rightarrow 3 x = 12 \Rightarrow x = 4$
Right working
$3 x^{2} - 12 x = 0 \Rightarrow 3 x (x - 4) = 0 \Rightarrow x = 0$ or $x = 4$
Do this: Bring everything to one side and take x out as a common factor; never cancel a variable.
Usually costs: ½ to 1 mark for the missing root.
Mistake: Keeping a value of k that makes the equation non-quadratic
Students solve $D = 0$ and report every value without checking the coefficient of $x^{2}$ .
Wrong working
$k x^{2} - 2 k x + 6 = 0$ : $4 k^{2} - 24 k = 0 \Rightarrow k = 0$ or $k = 6$
Right working
$4 k (k - 6) = 0$ gives $k = 0$ or $6$ ; $k = 0$ makes the equation $6 = 0$ , not quadratic, so $k = 6$ only.
Do this: Whenever k appears in the $x^{2}$ coefficient, test it is non-zero and write the rejection reason.
Usually costs: ½ mark for the final answer.
Mistake: Using D > 0 when the question says real roots
Students forget that equal roots are also real roots.
Wrong working
$x^{2} - 6 x + k = 0$ has real roots: $36 - 4 k > 0 \Rightarrow k < 9$
Right working
Real roots need $D \geq 0$ : $36 - 4 k \geq 0 \Rightarrow k \leq 9$
Do this: Match the words exactly: real means $\geq$ , distinct means $>$ , equal means $=$ , no real roots means $<$ .
Usually costs: ½ to 1 mark.
Mistake: Accepting an impossible root
Students stop as soon as factorisation gives two values.
Wrong working
$(x + 20) (x - 15) = 0$ , so the speed is 15 km/h or $- 20$ km/h.
Right working
$x = 15$ or $x = - 20$ ; $x = - 20$ is rejected since speed cannot be negative, so the speed is 15 km/h.
Do this: Before the final answer, test each root against the context and any denominator restrictions, and write one line of reason for each rejection.
Usually costs: ½ mark in a 3- or 5-mark word problem.

NCERT exercise guide: Quadratic Equations

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 4.1

What it tests: Checking whether a given equation is quadratic, and turning a described real-life situation into a quadratic equation in standard form. You are not asked to solve the equations here.

How to approach it: First expand and simplify both sides fully, then bring every term to one side. The equation is quadratic only if the highest power left is 2 with a non-zero coefficient. Watch out for $x^{3}$ terms that cancel out. For situations, name the unknown clearly (with its unit), translate each sentence into algebra, and simplify to the form $a x^{2} + b x + c = 0$ .

Like the questions in Exercise 4.1 that ask you to represent a situation as a quadratic equation

The altitude of a right triangle is 7 cm less than its base. The hypotenuse is 13 cm. Represent this situation as a quadratic equation in the base of the triangle.

Show the solutionSolution

Let the base be $x$ cm. Then the altitude is $(x - 7)$ cm.
By Pythagoras theorem: $x^{2} + (x - 7)^{2} = 1 3^{2}$ .
Expand: $x^{2} + x^{2} - 14 x + 49 = 169$ .
Bring all terms to one side: $2 x^{2} - 14 x - 120 = 0$ .
Divide by 2: $x^{2} - 7 x - 60 = 0$ .
This is of the form $a x^{2} + b x + c = 0$ with $a = 1 \neq = 0$ , so it is a quadratic equation.

Answer: $x^{2} - 7 x - 60 = 0$ , where $x$ cm is the base of the triangle.

Exercise 4.2

What it tests: Finding the roots of quadratic equations by factorisation (splitting the middle term), including equations with surd or fraction coefficients, and word problems on numbers, areas and products that lead to a quadratic.

How to approach it: Write the equation in standard form, then find two numbers whose product is $a c$ and whose sum is $b$ . Split the middle term, group in pairs and take out the common factor. Set each factor equal to zero. In word problems, always check each root against the situation and reject ones that make no sense, such as negative lengths or non-integers where integers are needed.

Like the number word problems in Exercise 4.2

The product of two consecutive positive integers is 306. Form a quadratic equation and find the integers.

Show the solutionSolution

Let the integers be $x$ and $x + 1$ , where $x$ is a positive integer.
Product: $x (x + 1) = 306$ , so $x^{2} + x - 306 = 0$ .
We need two numbers with product $- 306$ and sum $1$ : these are $18$ and $- 17$ .
Split the middle term: $x^{2} + 18 x - 17 x - 306 = 0$ .
Group: $x (x + 18) - 17 (x + 18) = 0$ , so $(x + 18) (x - 17) = 0$ .
So $x = - 18$ or $x = 17$ .
The integers are positive, so reject $x = - 18$ . Hence $x = 17$ and $x + 1 = 18$ .
Check: $17 \times 18 = 306$ . Correct.

Answer: The integers are 17 and 18.

Exercise 4.3

What it tests: Using the discriminant to decide the nature of roots, finding the value of an unknown constant k so that the roots are equal, finding the roots when they exist, and deciding whether a real-life design is possible.

How to approach it: Identify $a$ , $b$ and $c$ carefully with their signs and compute $D = b^{2} - 4 a c$ . Use $D > 0$ for two distinct real roots, $D = 0$ for two equal real roots and $D < 0$ for no real roots. For 'is it possible' questions, form the quadratic from the situation and look at the sign of $D$ : a negative $D$ means no real dimensions exist.

Like the 'is it possible to design' questions in Exercise 4.3

Is it possible to design a rectangular hall whose perimeter is 60 m and area is 250 square metres? If so, find its length and breadth.

Show the solutionSolution

Let the length be $l$ m. Perimeter $= 60$ gives $l + b = 30$ , so breadth $= (30 - l)$ m.
Area: $l (30 - l) = 250$ .
So $30 l - l^{2} = 250$ , that is $l^{2} - 30 l + 250 = 0$ .
Here $a = 1$ , $b = - 30$ , $c = 250$ .
$D = b^{2} - 4 a c = 900 - 1000 = - 100$ .
Since $D < 0$ , the equation has no real roots, so no real length exists.
(Extra insight: $D \geq 0$ needs $900 - 4 \times area \geq 0$ , so the area can be at most 225 square metres, which happens for a 15 m square.)

Answer: No, such a hall is not possible, because the discriminant is $- 100 < 0$ .

Quadratic Equations: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.3 marks per paper; 56 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.5 marks per paper; 65 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Quadratic Equations

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q3 · 1 mark · MCQ

The discriminant of the quadratic equation $x^{2} - 3 x - 2 = 0$ is :

(A) $1$
(B) $17$
(C) $17$
(D) $- 17$

Try it first, then show our solutionSolution

Here $a = 1$ , $b = - 3$ , $c = - 2$ .
$D = b^{2} - 4 a c = 9 - 4 (1) (- 2) = 9 + 8 = 17$ . The answer is option (B).
1 mark

Answer: 17, option (B)

Where the marks are

Computes $D = b^{2} - 4 a c = 17$ (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q32(a) · 5 marks · Long answer

The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.

Try it first, then show our solutionSolution

Let the greater number be $x$ and the smaller number be $y$ . Then $x^{2} - y^{2} = 180$ ...(1)
1 mark
Also $y^{2} = 8 x$ ...(2)
1 mark
Substitute (2) in (1): $x^{2} - 8 x = 180 \Rightarrow x^{2} - 8 x - 180 = 0$ .
1 mark
$x^{2} - 18 x + 10 x - 180 = 0 \Rightarrow (x - 18) (x + 10) = 0$ , so $x = 18$ or $x = - 10$ . Reject $x = - 10$ because the numbers are positive, so $x = 18$ .
1 mark
$y^{2} = 8 \times 18 = 144 \Rightarrow y = 12$ (positive). The numbers are 18 and 12. Check: $324 - 144 = 180$ .
1 mark

Answer: 18 and 12

Where the marks are

Forms $x^{2} - y^{2} = 180$ (1 mark)
Forms $y^{2} = 8 x$ (1 mark)
Gets the quadratic $x^{2} - 8 x - 180 = 0$ (1 mark)
Solves to get $x = 18$ and rejects $- 10$ (1 mark)
Finds the smaller number 12 and states both numbers (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/4/1, Q36 · 4 marks · Case-basedCompetency-based

Observe the figure given above. It shows six identical rectangular enclosures made by using fencing wire mesh. These enclosures are used to protect baby animals in a zoo. Dimensions of each enclosure is $x$ feet $\times$ $y$ feet. The total length of fencing required is $152$ feet and area of each enclosure is $80$ square feet.

Based on the above, answer the following questions :

(i)
Write an expression for length of fencing required in terms of $x$ and $y$ .
1 mark
(ii)
Write the area of each enclosure in terms of $x$ .
1 mark
(iii)
(a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method.
2 marks
OR
(iii)
(b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula.
2 marks

Figure in the paper: Left: photo of an animal behind wire mesh. Right: a rectangle of width x (top, marked with arrows) and height y (right side), divided into six identical rectangular enclosures arranged in 2 columns and 3 rows, with fencing lines and tick marks on the sides. See it in the official paper.

Try it first, then show our solutionSolution

(i) The six enclosures sit in 2 columns and 3 rows. There are 4 horizontal fence lines, each of length $2 x$ , which give $8 x$ . There are 3 vertical fence lines, each of length $3 y$ , which give $9 y$ .
Length of fencing: $8 x + 9 y = 152$
1 mark
(ii) From (i), $y = \frac{152 - 8 x}{9}$ .
Area of each enclosure $= x y = \frac{x ( 152 - 8 x )}{9}$ , so $\frac{x ( 152 - 8 x )}{9} = 80$ .
1 mark
(iii)(a) Rearrange into quadratic form:
$152 x - 8 x^{2} = 720 \Rightarrow 8 x^{2} - 152 x + 720 = 0 \Rightarrow x^{2} - 19 x + 90 = 0$
½ mark
Factorise by splitting the middle term:
$x^{2} - 10 x - 9 x + 90 = 0 \Rightarrow (x - 10) (x - 9) = 0$
½ mark
So $x = 10$ or $x = 9$ .
½ mark
Find $y$ from $y = \frac{80}{x}$ :
- When $x = 10$ , $y = 8$ .
- When $x = 9$ , $y = \frac{80}{9}$ .
Dimensions of each enclosure: $10$ ft $\times$ $8$ ft, or $9$ ft $\times \frac{80}{9}$ ft.
½ mark
(iii)(b) The quadratic is $x^{2} - 19 x + 90 = 0$ , with $a = 1$ , $b = - 19$ , $c = 90$ .
Discriminant: $D = b^{2} - 4 a c = 361 - 360 = 1$
½ mark
By the quadratic formula:
$x = \frac{- b \pm D}{2 a} = \frac{19 \pm 1}{2} = \frac{19 \pm 1}{2}$
½ mark
So $x = 10$ or $x = 9$ .
½ mark
Then $y = \frac{80}{x}$ gives $y = 8$ or $y = \frac{80}{9}$ .
Dimensions: $10$ ft $\times$ $8$ ft, or $9$ ft $\times \frac{80}{9}$ ft.
½ mark

Answer: (i) $8 x + 9 y = 152$ ; (ii) $\frac{x ( 152 - 8 x )}{9} = 80$ ; (iii) $x^{2} - 19 x + 90 = 0$ , so $x = 10$ , $y = 8$ (or $x = 9$ , $y = \frac{80}{9}$ ): each enclosure is 10 ft × 8 ft (or 9 ft × 80/9 ft)

Where the marks are

Writes the fencing equation $8 x + 9 y = 152$ (1 mark)
Writes the area as $\frac{x ( 152 - 8 x )}{9} = 80$ (1 mark)
Forms $x^{2} - 19 x + 90 = 0$ (½ mark)
Factorises as $(x - 10) (x - 9) = 0$ in (a), or substitutes correctly into the quadratic formula in (b) (½ mark)
Gets $x = 10$ or $9$ (½ mark)
Gets $y = 8$ or $\frac{80}{9}$ (½ mark)

See it in the full 2026 board paper, set 430/4/1

Every Quadratic Equations question, by year

Standard (65 questions, sample papers included)

For 2027

Sample paper 2026-27: Q4 (1 mark, solved), Q32 (5 marks, solved)

2026

Set 30/1/1: Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

Set 30/2/1: Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/3/1: Q1 (1 mark, solved), Q10 (1 mark, solved), Q29 (3 marks, solved)

Set 30/4/1: Q1 (1 mark, solved), Q21 (2 marks, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 30/5/1: Q1 (1 mark, solved), Q9 (1 mark, solved)

Sample paper 2025-26: Q6 (1 mark, solved), Q32 (5 marks, solved)

2025

Set 30/1/1: Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 30/2/1: Q3 (1 mark, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/3/1: Q5 (1 mark, solved), Q24(a) (2 marks, solved), Q36 (4 marks, solved)

Set 30/4/1: Q3 (1 mark, solved), Q30 (3 marks, solved), Q32(A) (5 marks, solved), Q32(B) (5 marks, solved)

Set 30/5/1: Q4 (1 mark, solved), Q5 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 30/6/1: Q3 (1 mark, solved), Q32(b) (5 marks, solved)

Sample paper 2024-25: Q11 (1 mark, solved), Q27 (3 marks, solved)

2024

Set 30/1/1: Q3 (1 mark, solved), Q36 (4 marks, solved)

Set 30/2/1: Q4 (1 mark, solved)

Set 30/3/1: Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 30/4/1: Q4 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 30/5/1: Q5 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2023-24: Q4 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

2023

Set 30/1/1: Q8 (1 mark, solved), Q36 (4 marks, solved)

Set 30/2/1: Q1 (1 mark, solved)

Set 30/4/1: Q2 (1 mark, solved), Q23(A) (2 marks, solved), Q23(B) (2 marks, solved), Q31 (3 marks, solved)

Set 30/5/1: Q5 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 30/6/1: Q9 (1 mark, solved), Q19 (1 mark, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

Basic (80 questions, sample papers included)

For 2027

Sample paper 2026-27: Q5 (1 mark, solved), Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

2026

Set 430/1/1: Q3 (1 mark, solved), Q4 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/2/1: Q4 (1 mark, solved), Q5 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/3/1: Q5 (1 mark, solved), Q20 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/4/1: Q2 (1 mark, solved), Q36 (4 marks, solved)

Set 430/5/1: Q2 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Sample paper 2025-26: Q4 (1 mark, solved), Q24 (2 marks, solved), Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

2025

Set 430/1/1: Q3 (1 mark, solved), Q4 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/2/1: Q4 (1 mark, solved), Q5 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/3/1: Q5 (1 mark, solved), Q20 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/4/1: Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/5/1: Q2 (1 mark, solved), Q7 (1 mark, solved), Q27 (3 marks, solved)

Set 430/6/1: Q21(a) (2 marks, solved), Q27 (3 marks, solved)

Sample paper 2024-25: Q3 (1 mark, solved), Q16 (1 mark, solved), Q32(A) (5 marks, solved), Q32(B) (5 marks, solved)

2024

Set 430/1/1: Q6 (1 mark, solved), Q36 (4 marks, solved)

Set 430/2/1: Q3 (1 mark, solved), Q32 (5 marks, solved)

Set 430/3/1: Q2 (1 mark, solved), Q27 (3 marks, solved)

Set 430/4/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/5/1: Q2 (1 mark, solved), Q14 (1 mark, solved), Q23 (2 marks, solved)

Sample paper 2023-24: Q3 (1 mark, solved), Q5 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

2023

Set 430/1/1: Q4 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/2/1: Q10 (1 mark, solved)

Set 430/4/1: Q15 (1 mark, solved), Q16 (1 mark, solved), Q20 (1 mark, solved), Q23(A) (2 marks, solved), Q23(B) (2 marks, solved)

Set 430/5/1: Q5 (1 mark, solved), Q10 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/6/1: Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q26 (3 marks, solved)

Practice questions on Quadratic Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Solve by factorisation: $2 x^{2} + 7 x + 52 = 0$ .
Hint
Here $a c = 2 \times 52 = 10$ ; find two numbers with product 10 and sum 7.
Answer
Split: $2 x^{2} + 5 x + 2 x + 52 = x (2 x + 5) + 2 (2 x + 5)$ .
So $(2 x + 5) (x + 2) = 0$ , giving $x = - \frac{5 2}{2}$ or $x = - 2$ .
Q2 · 1 mark · MCQBoard level
If $x = 3$ is a root of $x^{2} - 2 k x + 6 = 0$ , then the other root is
(a) $- 2$
(b) $2$
(c) $\frac{5}{2}$
(d) $- 3$
Hint
Find $k$ first by substituting $x = 3$ .
Answer
(b) $2$ . $9 - 6 k + 6 = 0$ gives $k = \frac{5}{2}$ , so the equation is $x^{2} - 5 x + 6 = 0 = (x - 2) (x - 3)$ . The other root is 2.
Q3 · 2 marks · Very short answerBoard level
Find all values of $k$ for which the quadratic equation $k x^{2} + 4 x + 1 = 0$ has real roots.
Hint
Real roots need $D \geq 0$ , and remember the equation must stay quadratic.
Answer
$D = 16 - 4 k \geq 0$ gives $k \leq 4$ .
For a quadratic, $k \neq = 0$ . So $k \leq 4$ and $k \neq = 0$ .
Q4 · 3 marks · Short answerBoard level
Solve for $x$ using the quadratic formula: $\frac{1}{x} - \frac{1}{x - 2} = 3$ , where $x \neq = 0, 2$ .
Hint
Combine the fractions first; the numerator becomes a constant.
Answer
$\frac{( x - 2 ) - x}{x ( x - 2 )} = 3$ gives $- 2 = 3 x^{2} - 6 x$ , so $3 x^{2} - 6 x + 2 = 0$ .
$D = 36 - 24 = 12$ , $12 = 23$ , so $x = \frac{6 \pm 2 3}{6} = \frac{3 \pm 3}{3}$ .
Q5 · 3 marks · Short answerHarder board level
Prove that the roots of the equation $(x - a) (x - b) = k^{2}$ are real for all real values of $a$ , $b$ and $k$ .
Hint
Write it in standard form and try to write $D$ as a sum of squares.
Answer
Standard form: $x^{2} - (a + b) x + (ab - k^{2}) = 0$ .
$D = (a + b)^{2} - 4 ab + 4 k^{2} = (a - b)^{2} + 4 k^{2}$ .
A sum of squares is never negative, so $D \geq 0$ and the roots are always real.
Q6 · 3 marks · Short answerBoard level
A two-digit number is such that the product of its digits is 14. When 45 is added to the number, its digits are reversed. Find the number.
Hint
Write the number as $10 t + u$ ; the second condition gives a linear relation between $t$ and $u$ .
Answer
$(10 u + t) - (10 t + u) = 45$ gives $u - t = 5$ .
$t (t + 5) = 14$ gives $t^{2} + 5 t - 14 = 0 = (t + 7) (t - 2)$ , so $t = 2$ , $u = 7$ .
The number is 27. Check: $27 + 45 = 72$ .
Q7 · 5 marks · Long answerHarder board level
A rectangular park is 30 m long and 20 m wide. A path of uniform width runs inside the park along all four sides. The lawn left inside the path has an area of 336 square metres. Find the width of the path and the cost of paving the path at Rs 40 per square metre.
Hint
If the path is $w$ m wide, the lawn measures $(30 - 2 w)$ m by $(20 - 2 w)$ m.
Answer
$(30 - 2 w) (20 - 2 w) = 336$ gives $4 w^{2} - 100 w + 264 = 0$ , so $w^{2} - 25 w + 66 = 0 = (w - 3) (w - 22)$ .
$w = 22$ is impossible (lawn length would be negative), so $w = 3$ m.
Path area $= 600 - 336 = 264$ square metres; cost $= 264 \times 40 =$ Rs 10,560.
Q8 · 5 marks · Long answerHarder board level
A delivery van has to cover 240 km. It starts 1 hour late, so to arrive on time it has to travel 20 km/h faster than its usual speed. Find its usual speed and the time it actually took for the journey.
Hint
Usual time minus actual time equals 1 hour.
Answer
$\frac{240}{x} - \frac{240}{x + 20} = 1$ gives $4800 = x^{2} + 20 x$ , so $x^{2} + 20 x - 4800 = 0 = (x + 80) (x - 60)$ .
Speed is positive, so usual speed $= 60$ km/h.
Actual time $= \frac{240}{80} = 3$ hours (usual time 4 hours).
Q9 · 1 mark · Assertion-reasonBoard level
Assertion (A): The equation $x^{2} = 5 x$ has only one root, $x = 5$ .
Reason (R): Dividing both sides of an equation by an expression containing the variable can lose a root.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Do not divide by $x$ ; bring everything to one side and factorise.
Answer
(d). $x^{2} - 5 x = 0$ gives $x (x - 5) = 0$ , so the roots are $0$ and $5$ ; A is false.
R is true: dividing by $x$ silently assumes $x \neq = 0$ , which loses the root $0$ .
Q10 · 4 marks · Case-basedHOTS / competency
For an annual day in a Lucknow school, 240 chairs are arranged in rows. Each row has 8 more chairs than the number of rows.
(i) Taking the number of rows as $r$ , form the quadratic equation. (1 mark)
(ii) Find the number of rows and chairs in each row. (1 mark)
(iii) The principal asks for 200 chairs arranged so that each row has 3 more chairs than the number of rows. Use the discriminant to explain whether this is possible. (2 marks)
Hint
In (iii), a positive $D$ only tells you the roots are real. The number of rows must be a positive whole number, so also check whether $D$ is a perfect square.
Answer
(i) $r (r + 8) = 240$ , so $r^{2} + 8 r - 240 = 0$ .
(ii) $(r + 20) (r - 12) = 0$ ; rows cannot be negative, so 12 rows of 20 chairs.
(iii) $r (r + 3) = 200$ gives $r^{2} + 3 r - 200 = 0$ , with $D = 9 + 800 = 809 > 0$ . So the roots are real, but that alone does not settle the question, because the number of rows must be a positive whole number. The coefficients are integers and 809 is not a perfect square ( $2 8^{2} = 784$ , $2 9^{2} = 841$ ), so $809$ is irrational and both roots are irrational. No whole number of rows works, so the arrangement is not possible.
Q11 · 5 marks · Long answerHarder board level
One root of the equation $x^{2} + p x + 12 = 0$ is 4.
(i) Find $p$ and the other root of this equation. (2 marks)
(ii) The other root found in (i) is also a root of $x^{2} + k x - 2 k = 0$ . Find $k$ and the second root of this equation. (2 marks)
(iii) With the value of $p$ from (i), find $q$ so that $x^{2} + p x + q = 0$ has two equal roots. (1 mark)
Hint
Each part unlocks the next: substitute the known root to get a linear equation in the unknown coefficient.
Answer
(i) $16 + 4 p + 12 = 0$ gives $p = - 7$ . Then $x^{2} - 7 x + 12 = (x - 3) (x - 4) = 0$ , so the other root is 3.
(ii) $9 + 3 k - 2 k = 0$ gives $k = - 9$ . The equation is $x^{2} - 9 x + 18 = (x - 3) (x - 6) = 0$ , so the second root is 6.
(iii) $x^{2} - 7 x + q = 0$ has equal roots when $D = 49 - 4 q = 0$ , so $q = \frac{49}{4}$ (both roots equal $\frac{7}{2}$ ).
Q12 · 5 marks · Long answerHarder board level
A train covers 360 km at a uniform speed. If its speed were 5 km/h more, it would take 1 hour less for the journey.
(i) Find the usual speed of the train. (3 marks)
(ii) A planner claims that on the same 360 km route there is a whole-number speed $x$ km/h for which raising the speed by 10 km/h saves exactly 2 hours. Use the discriminant to show that the claim is false. (2 marks)
Hint
In (ii), real roots are not enough; check whether $D$ is a perfect square.
Answer
(i) $\frac{360}{x} - \frac{360}{x + 5} = 1$ gives $\frac{1800}{x ( x + 5 )} = 1$ , so $x^{2} + 5 x - 1800 = 0 = (x + 45) (x - 40)$ . Speed is positive, so the usual speed is 40 km/h. Check: $\frac{360}{40} - \frac{360}{45} = 9 - 8 = 1$ h.
(ii) $\frac{360}{x} - \frac{360}{x + 10} = 2$ gives $3600 = 2 x (x + 10)$ , so $x^{2} + 10 x - 1800 = 0$ . $D = 100 + 7200 = 7300 > 0$ , so real roots exist. But $8 5^{2} = 7225$ and $8 6^{2} = 7396$ , so 7300 is not a perfect square. With integer coefficients the roots are then irrational, so no whole-number speed works and the claim is false.
Q13 · 3 marks · Short answerHarder board level
If the roots of the equation $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ are equal, prove that $\frac{a}{b} = \frac{c}{d}$ , where $b \neq = 0$ and $d \neq = 0$ .
Hint
Expand $D$ fully; it collapses to minus a perfect square.
Answer
$D = 4 (a c + b d)^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2})$ .
$D = 4 (a^{2} c^{2} + 2 ab c d + b^{2} d^{2} - a^{2} c^{2} - a^{2} d^{2} - b^{2} c^{2} - b^{2} d^{2}) = - 4 (a^{2} d^{2} - 2 ab c d + b^{2} c^{2}) = - 4 (a d - b c)^{2}$ .
Equal roots means $D = 0$ , so $a d - b c = 0$ , that is $a d = b c$ . Dividing by $b d$ gives $\frac{a}{b} = \frac{c}{d}$ . Hence proved.
Q14 · 3 marks · Short answerHarder board level
A shopkeeper buys some books for Rs 1200. Had he bought 10 more books for the same amount, each book would have cost Rs 20 less. Find the number of books he bought. If he sells every book at Rs 75, find his profit per cent.
Hint
Cost per book is $\frac{1200}{x}$ ; the two costs per book differ by 20.
Answer
Let the number of books be $x$ . $\frac{1200}{x} - \frac{1200}{x + 10} = 20$ gives $12000 = 20 x (x + 10)$ , so $x^{2} + 10 x - 600 = 0 = (x + 30) (x - 20)$ . The number of books cannot be negative, so $x = 20$ .
Cost per book $= \frac{1200}{20}$ = Rs 60, so the profit per book is Rs 15. Profit per cent $= \frac{15}{60} \times 100 = 25$ , that is 25 per cent.
Q15 · 3 marks · Short answerHOTS / competency
Prove that for every real value of $k$ , at least one of the equations $x^{2} - 2 x + k = 0$ and $x^{2} - 2 x - k = 0$ has real roots. Also find all values of $k$ for which both equations have real roots.
Hint
Add the two discriminants.
Answer
$D_{1} = 4 - 4 k$ and $D_{2} = 4 + 4 k$ . Their sum is $D_{1} + D_{2} = 8 > 0$ , so both cannot be negative. At least one discriminant is positive, so at least one equation has real roots.
Both have real roots when $4 - 4 k \geq 0$ and $4 + 4 k \geq 0$ , that is $- 1 \leq k \leq 1$ .
Q16 · 3 marks · Short answerHOTS / competency
From each corner of a rectangular sheet of cardboard 20 cm long and 15 cm wide, a square of side $x$ cm is cut off, and the flaps are folded up to make an open box. The base of the box must have an area of 126 square cm.
(i) Form the quadratic equation and solve it. (2 marks)
(ii) One root is a real number but cannot be used. Explain why, and find the volume of the box. (1 mark)
Hint
The base measures $(20 - 2 x)$ cm by $(15 - 2 x)$ cm, and both must be positive.
Answer
(i) $(20 - 2 x) (15 - 2 x) = 126$ gives $4 x^{2} - 70 x + 300 = 126$ , so $4 x^{2} - 70 x + 174 = 0$ , that is $2 x^{2} - 35 x + 87 = 0$ . $D = 1225 - 696 = 529$ and $529 = 23$ , so $x = \frac{35 \pm 23}{4} = 14.5$ or $3$ .
(ii) $x = 14.5$ would make the width $15 - 29 = - 14$ cm, which is impossible ( $x$ must be less than 7.5). So $x = 3$ cm, the base is $14 \times 9 = 126$ square cm, and the volume is $126 \times 3 = 378$ cubic cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What condition makes $a x^{2} + b x + c = 0$ a quadratic equation?Show answer
$a \neq = 0$ ; otherwise it is linear.
2.How many roots can a quadratic equation have at most?Show answer
Two.
3.Write the discriminant of $a x^{2} + b x + c = 0$ .Show answer
$D = b^{2} - 4 a c$
4.If $D = 0$ , what are the roots?Show answer
Two equal real roots, each $- \frac{b}{2 a}$ .
5.What does $D < 0$ tell you?Show answer
The equation has no real roots.
6.Condition for real roots (equal or distinct)?Show answer
$D \geq 0$
7.Is $x = 3$ a root of $x^{2} - 5 x + 6 = 0$ ?Show answer
Yes: $9 - 15 + 6 = 0$ .
8.Nature of the roots of $x^{2} + x + 1 = 0$ ?Show answer
$D = 1 - 4 = - 3 < 0$ , so no real roots.

Best order to study it

Learn the standard form and practise expanding equations to decide whether they are quadratic, and testing whether a number is a root.
Master splitting the middle term with integers first, then with surds and with fractions that need restrictions.
Learn the quadratic formula and always compute D on a separate line; practise equations with irrational roots and with letter coefficients.
Study the discriminant: the three cases, 'find k' questions, ranges with inequalities and short proofs.
Finish with word problems of every type (speed, boats, work, area, numbers, ages), always rejecting impossible roots, then attempt case-study style questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write any equation in standard form and read a, b, c with correct signs?
Can you show whether an equation like $(x + 1)^{3} = x^{3} + 2$ is quadratic by expanding it?
Can you find k when one root is given, and then find the other root?
Can you split the middle term when the coefficients contain $2$ or $3$ ?
Can you solve an equation with x in two denominators and state which values are not allowed?
Can you quote the quadratic formula and solve an equation with letters a and b in it?
Can you state the three discriminant cases and the root formula when $D = 0$ ?
Can you find the range of k for real roots, using $\geq$ correctly?
Can you form and solve a speed-time or work-rate equation and reject the negative root with a reason?

Quadratic Equations: frequently asked questions

Is the method of completing the square in the 2026-27 syllabus?

No. Solving by completing the square is not in the 2026-27 syllabus. You solve quadratics by factorisation or by quoting and using the quadratic formula.

Should I use factorisation or the quadratic formula in the exam?

Use the method the question names, if it names one. If it does not, factorisation is quicker when the numbers split easily; otherwise use the formula, which always works when real roots exist. Checking whether the discriminant is a perfect square helps you decide fast.

What should I write if the discriminant is negative?

State clearly that the discriminant is negative, so the equation has no real roots. Only real roots are in your syllabus, so you stop there and do not try to find square roots of negative numbers.

Do I really need to write why I rejected a root?

Yes. A one-line reason, such as speed cannot be negative or this value makes a denominator zero, is part of a complete answer. Leaving an impossible value in your final answer can cost you the final mark.

Is the Quadratic Equations chapter the same for Mathematics Standard and Basic?

Yes. Both papers follow the same syllabus for this chapter, so the concepts and methods are identical. The difference is only in the difficulty of the questions set in the paper.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)