MathsCBSE Class 10 · Quadratic Equations

Quadratic Formula for Class 10: How to Use It

Q: Is the quadratic formula the same as Sridharacharya's formula?

Yes. In India the quadratic formula is often called Sridharacharya's formula. It is the same result, and in the board exam you can call it either name.

Q: Do I need to derive the quadratic formula for the 2026-27 board exam?

No. Solving by completing the square, the method used to derive the formula, is not in the 2026-27 syllabus. You only need to quote the formula and use it correctly.

Q: When should I use the formula instead of factorisation?

Use the formula when the question asks for it, when splitting the middle term is not quick, or when the roots look irrational. If D turns out to be a perfect square, factorisation would also have worked, but the formula still gives the right answer.

Q: What should I write if D comes out negative?

Write that since D is less than zero, the equation has no real roots, and stop. Your syllabus covers only real roots, so do not try to take the square root of a negative number.

Q: Can I leave my answer with a square root in it?

Yes. Roots such as 2 plus root 3 are exact answers and are expected in surd form. Simplify the surd, cancel any common factor, and do not convert to decimals unless the question asks.

Updated 13 October 2026 · Fact-checked

The quadratic formula gives the roots of ax² + bx + c = 0 as x = (−b ± √(b² − 4ac)) / 2a, provided b² − 4ac is not negative. In the board exam you use it to solve equations that do not factorise easily, including ones with surds, fractions or letter coefficients, and in word problems.

Understand Quadratic Formula

A quadratic equation $a x^{2} + b x + c = 0$ (with $a \neq = 0$ ) has at most two roots. Factorisation finds them only when you can spot two numbers with product $a c$ and sum $b$ . That fails when the roots are irrational, as in $x^{2} - 4 x + 1 = 0$ , and is slow when the numbers are large, as in $x^{2} + 20 x - 4800 = 0$ . The quadratic formula removes the guesswork. It gives both roots directly from $a$ , $b$ and $c$ :

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

The $\pm$ sign carries two answers. Take $+$ for one root and $-$ for the other. The expression under the root, $D = b^{2} - 4 a c$ , is the discriminant. It must not be negative, because a negative number has no real square root. Your syllabus deals with real roots only, so if $D < 0$ you simply write that the equation has no real roots.

Why can you trust the formula? Call the two values $α$ and $β$ . Adding them, the $D$ parts cancel and you get $\frac{- 2 b}{2 a} = - \frac{b}{a}$ . Multiplying them gives $\frac{b ^{2} - D}{4 a ^{2}} = \frac{4 a c}{4 a ^{2}} = \frac{c}{a}$ . These are exactly the sum and product of zeroes from the Polynomials chapter. So the formula always produces the right pair. You may quote and use it directly. You are not asked to derive it.

The formula connects the whole chapter. If $D$ is a perfect square such as $169$ , the roots are rational and the equation could also have been factorised. If $D$ is not a perfect square, the roots are surds such as $2 \pm 3$ . Only the formula finds those cleanly. The sign of $D$ also tells you the nature of the roots, which has its own page. Word problems often end in an equation with awkward numbers, and the formula is the safe way to finish them.

The formula works under three conditions. The equation must be in standard form, with everything on one side and zero on the other. You must read $a$ , $b$ , $c$ with their signs, so for $x^{2} - 3 x - 1 = 0$ , $b = - 3$ and $- b = 3$ . And the whole numerator, $- b \pm D$ , is divided by $2 a$ , not just the square root.

Quadratic Formula: formulas to remember

Quadratic formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}, a \neq = 0, b^{2} - 4 a c \geq 0$; Works for every quadratic with real roots. Quote it in your answer, then substitute.
Discriminant: $D = b^{2} - 4 a c$; Compute it on its own line first. It decides whether real roots exist and keeps the square root simple.
The two roots written separately: $α = \frac{- b + D}{2 a}, β = \frac{- b - D}{2 a}$; Simplify each root on its own line, especially when D is a perfect square.
Equal roots case: $D = 0 \Rightarrow x = - \frac{b}{2 a}$; Both values from the formula coincide.
No real roots case: $D < 0$; Stop and write 'no real roots'. Never take the square root of a negative number.
Checking the roots: $α + β = - \frac{b}{a}, α β = \frac{c}{a}$; A quick check after solving, especially with surds or letter coefficients.
Simplifying the surd: $k^{2} m = k m, e.g. 12 = 23, 8 = 22$; Take out square factors before dividing, then cancel a common factor from every term of the numerator and the denominator.

How to solve Quadratic Formula questions

Use this same routine for every quadratic formula question, from a 1-mark MCQ to a 5-mark word problem. Examiners give marks for the standard form, the value of D, the substitution and the final simplified roots.

1Write any restrictions first if x appears in a denominator, for example $x \neq = - 1, - 3$ .
2Bring the equation to standard form $a x^{2} + b x + c = 0$ . Expand brackets, clear fractions and move every term to one side.
3If all coefficients share a common factor, divide it out. Then write $a$ , $b$ , $c$ clearly with their signs. If the question already uses letters $a$ , $b$ , call your coefficients $A$ , $B$ , $C$ .
4Compute $D = b^{2} - 4 a c$ on its own line. If $D < 0$ , state that there are no real roots and stop.
5Simplify $D$ : either a whole number, a surd such as $22$ , or an expression such as $3 b$ when the coefficients contain letters.
6Substitute into $x = \frac{- b \pm D}{2 a}$ and write the two roots separately, simplifying each one fully.
7Check the roots against the restrictions or the situation. Reject values that make a denominator zero, or negative lengths, speeds or counts, and give a reason.
8Verify quickly using the sum $- \frac{b}{a}$ and product $\frac{c}{a}$ , or by substituting one root. Then write the final answer with units.

Quadratic Formula in real life

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.

The quadratic formula

Imagine your family wants to buy a new flat-screen TV for the living room. The screen's diagonal is 100 cm, and the screen is 20 cm wider than it is tall. The space on the wall between two shelves is 85 cm wide.

Find the height and width of the screen, and its area. Decide whether the TV will fit in the 85 cm wide space.

Try it first, then show the solutionSolution

Let the height be $h$ cm, so the width is $(h + 20)$ cm.
The height, width and diagonal form a right triangle, so $h^{2} + (h + 20)^{2} = 10 0^{2}$ .
Expand: $2 h^{2} + 40 h + 400 = 10000$ , so $2 h^{2} + 40 h - 9600 = 0$ .
Divide by 2: $h^{2} + 20 h - 4800 = 0$ , so $a = 1$ , $b = 20$ , $c = - 4800$ .
$D = 2 0^{2} - 4 (1) (- 4800) = 400 + 19200 = 19600$ , so $D = 140$ .
$h = \frac{- 20 \pm 140}{2}$ , giving $h = 60$ or $h = - 80$ .
A height cannot be negative, so $h = 60$ cm and the width is $60 + 20 = 80$ cm.
Check: $6 0^{2} + 8 0^{2} = 3600 + 6400 = 10000 = 10 0^{2}$ .
Area $= 60 \times 80 = 4800$ square cm. The width of 80 cm is less than 85 cm, so the TV fits.

Answer: Height 60 cm, width 80 cm, area 4800 square cm. The TV fits, with 5 cm to spare.

Memory hook: The TV trick: a diagonal means Pythagoras, and big numbers mean the formula.

Using the formula inside a word problem

Suppose you decide to save for a new bicycle that costs Rs 2000. In week 1 you put Rs 10 in a box, in week 2 Rs 20, in week 3 Rs 30, and so on, adding Rs 10 more each week. You add money only once, at the end of each week.

Write an expression for your total savings after $n$ weeks and form a quadratic equation for the total reaching Rs 2000. Use the quadratic formula to find after how many complete weeks you first have enough, and how much money is left over after you buy the bicycle.

Try it first, then show the solutionSolution

Total after $n$ weeks $= 10 + 20 + \dots + 10 n = 10 (1 + 2 + \dots + n) = 10 \times \frac{n ( n + 1 )}{2} = 5 n (n + 1)$ .
Set the total equal to the price: $5 n (n + 1) = 2000$ , so $n^{2} + n - 400 = 0$ .
$a = 1$ , $b = 1$ , $c = - 400$ : $D = 1 + 1600 = 1601$ .
$n = \frac{- 1 \pm 1601}{2}$ . The number of weeks cannot be negative, so $n = \frac{- 1 + 1601}{2}$ .
Since $4 0^{2} = 1600$ and $4 1^{2} = 1681$ , $1601$ is just above 40, so $n$ is just above $\frac{39}{2} = 19.5$ . This is not a whole number, so the total never equals exactly 2000.
Check the whole weeks on either side: after 19 weeks, $5 \times 19 \times 20 = 1900 < 2000$ ; after 20 weeks, $5 \times 20 \times 21 = 2100 > 2000$ .
So you first have enough after 20 weeks, with $2100 - 2000 = 100$ left over.

Answer: After 20 complete weeks you have Rs 2100, so Rs 100 is left over after buying the bicycle.

Memory hook: Piggy bank rule: if the root falls between two whole weeks, round up to the next full week.

Tricks to remember Quadratic Formula

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

The chant
Say it as one sentence: 'minus b, plus or minus root of b squared minus four a c, all over two a'. The words 'all over' remind you that $2 a$ divides the whole numerator.
Use it when: Writing the formula from memory at the start of every answer.
D before x
Think of D as the gatekeeper. Find D first: if it is negative, the gate is shut (no real roots); if it is a perfect square, the roots are rational.
Use it when: Saves time and avoids square roots of negative numbers.
Minus b flips the sign
Whatever sign b has, $- b$ has the opposite. So if $b = - 6$ , the numerator starts with $+ 6$ . And $b^{2}$ is always positive, even when b is negative.
Use it when: Avoiding the most common sign errors in the numerator and in D.
Sum and product check
Your two roots must add to $- \frac{b}{a}$ and multiply to $\frac{c}{a}$ . With surds, the root parts cancel in the sum, so the check is quick.
Use it when: Catching an arithmetic slip before you move on.

Quadratic Formula: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Rearrange, then use the formula · 2 marks · Very short answerBoard level

Using the quadratic formula, find the roots of $x (2 x + 3) = 1$ .

Try it first, then show the solutionSolution

Expand and rearrange: $2 x^{2} + 3 x - 1 = 0$ , so $a = 2$ , $b = 3$ , $c = - 1$ . (½ mark)
$D = 3^{2} - 4 (2) (- 1) = 9 + 8 = 17 > 0$ , so real roots exist. (½ mark)
$x = \frac{- 3 \pm 17}{2 \times 2} = \frac{- 3 \pm 17}{4}$ . (1 mark)
Check: the sum of roots is $\frac{- 6}{4} = - \frac{3}{2} = - \frac{b}{a}$ .

Answer: $x = \frac{- 3 + 17}{4}$ or $x = \frac{- 3 - 17}{4}$

Example 2: Letter coefficients where D is a perfect square · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

Compare with $A x^{2} + B x + C = 0$ : $A = 1$ , $B = - (2 a - b)$ , $C = a^{2} - ab - 2 b^{2}$ . (½ mark)
$D = B^{2} - 4 A C = (2 a - b)^{2} - 4 (a^{2} - ab - 2 b^{2})$ .
$D = 4 a^{2} - 4 ab + b^{2} - 4 a^{2} + 4 ab + 8 b^{2} = 9 b^{2}$ , so $D = 3 b$ . (The sign does not matter because of the $\pm$ .) (1 mark)
$x = \frac{( 2 a - b ) \pm 3 b}{2}$ . (½ mark)
Taking $+$ : $x = \frac{2 a + 2 b}{2} = a + b$ . (½ mark)
Taking $-$ : $x = \frac{2 a - 4 b}{2} = a - 2 b$ . (½ mark)
Check: $(a + b) (a - 2 b) = a^{2} - ab - 2 b^{2} = C$ , and $(a + b) + (a - 2 b) = 2 a - b = - B$ .

Answer: $x = a + b$ or $x = a - 2 b$

Example 3: Fractions that lead to irrational roots · 3 marks · Short answerBoard level

Solve for $x$ : $\frac{1}{x + 1} + \frac{1}{x + 3} = 1$ , where $x \neq = - 1, - 3$ .

Try it first, then show the solutionSolution

Combine the fractions: $\frac{( x + 3 ) + ( x + 1 )}{( x + 1 ) ( x + 3 )} = 1$ , so $\frac{2 x + 4}{x ^{2} + 4 x + 3} = 1$ . (½ mark)
Cross-multiply: $2 x + 4 = x^{2} + 4 x + 3$ , so $x^{2} + 2 x - 1 = 0$ . (1 mark)
$a = 1$ , $b = 2$ , $c = - 1$ : $D = 4 + 4 = 8$ , so $D = 22$ . (½ mark)
$x = \frac{- 2 \pm 2 2}{2} = - 1 \pm 2$ . (½ mark)
Neither $- 1 + 2$ nor $- 1 - 2$ equals $- 1$ or $- 3$ , so both roots are valid. (½ mark)
Check: the sum is $- 2 = - \frac{b}{a}$ and the product is $1 - 2 = - 1 = \frac{c}{a}$ .

Answer: $x = - 1 + 2$ or $x = - 1 - 2$

Example 4: A journey problem with large numbers · 5 marks · Long answerBoard level

A car covers a distance of 240 km at a uniform speed. If its speed had been 20 km/h more, it would have taken 1 hour less for the same journey. Using the quadratic formula, find the original speed of the car and the time it took.

Try it first, then show the solutionSolution

Let the original speed be $x$ km/h. The time taken is $\frac{240}{x}$ h, and at the higher speed it is $\frac{240}{x + 20}$ h.
Given: $\frac{240}{x} - \frac{240}{x + 20} = 1$ . (1 mark)
Simplify: $\frac{240 ( x + 20 ) - 240 x}{x ( x + 20 )} = 1$ , so $4800 = x^{2} + 20 x$ , that is $x^{2} + 20 x - 4800 = 0$ . (1 mark)
$a = 1$ , $b = 20$ , $c = - 4800$ : $D = 400 + 19200 = 19600$ , so $D = 140$ . (1 mark)
$x = \frac{- 20 \pm 140}{2}$ , so $x = 60$ or $x = - 80$ . (1 mark)
A speed cannot be negative, so $x = 60$ km/h. The time is $\frac{240}{60} = 4$ h. Check: at 80 km/h the time is 3 h, which is 1 hour less. (1 mark)

Answer: The original speed is 60 km/h and the journey took 4 hours.

Example 5: Case study: the art club banner · 4 marks · Case-basedBoard level

Imagine your school's art club is making a rectangular cloth banner for the annual day. The length of the banner is 3 m more than twice its breadth, and its area must be exactly 20 square metres. The edge of the banner will be lined with golden lace costing Rs 40 per metre.
(i) Taking the breadth as $x$ m, form a quadratic equation in standard form. (1 mark)
(ii) Find the discriminant and state the nature of the roots. (1 mark)
(iii) Using the quadratic formula, find the length and breadth of the banner, and the cost of the lace. (2 marks)

Try it first, then show the solutionSolution

(i) Length $= (2 x + 3)$ m, so $x (2 x + 3) = 20$ , that is $2 x^{2} + 3 x - 20 = 0$ . (1 mark)
(ii) $D = 3^{2} - 4 (2) (- 20) = 9 + 160 = 169 > 0$ , so there are two distinct real roots. Because 169 is a perfect square, they are rational. (1 mark)
(iii) $x = \frac{- 3 \pm 13}{4}$ , so $x = \frac{10}{4} = \frac{5}{2}$ or $x = - 4$ .
A breadth cannot be negative, so the breadth is $2.5$ m and the length is $2 (2.5) + 3 = 8$ m. Check: $2.5 \times 8 = 20$ . (1 mark)
Perimeter $= 2 (8 + 2.5) = 21$ m, so the cost of lace $= 21 \times 40 =$ Rs 840. (1 mark)

Answer: (i) $2 x^{2} + 3 x - 20 = 0$ (ii) $D = 169$ , two distinct real (rational) roots (iii) Breadth 2.5 m, length 8 m, cost of lace Rs 840.

Example 6: Two taps: a positive root that still fails · 5 marks · Long answerHarder board level

Two taps together can fill a water tank in 12 hours. The larger tap alone takes 10 hours less than the smaller tap alone. Using the quadratic formula, find the time each tap takes on its own to fill the tank.

Try it first, then show the solutionSolution

Let the smaller tap take $x$ hours, so the larger tap takes $(x - 10)$ hours. Both times must be positive, so $x > 10$ .
In 1 hour the taps fill $\frac{1}{x}$ and $\frac{1}{x - 10}$ of the tank, so $\frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12}$ . (1 mark)
Combine: $\frac{2 x - 10}{x ( x - 10 )} = \frac{1}{12}$ , so $24 x - 120 = x^{2} - 10 x$ , that is $x^{2} - 34 x + 120 = 0$ . (1 mark)
$a = 1$ , $b = - 34$ , $c = 120$ : $D = 1156 - 480 = 676$ , so $D = 26$ . (1 mark)
$x = \frac{34 \pm 26}{2}$ , so $x = 30$ or $x = 4$ . (1 mark)
$x = 4$ is positive but fails $x > 10$ , because the larger tap would take $4 - 10 = - 6$ hours. So $x = 30$ , and the larger tap takes 20 hours. Check: $\frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{1}{12}$ . (1 mark)

Answer: The smaller tap takes 30 hours and the larger tap takes 20 hours.

Example 7: Boat and stream: upstream and downstream · 5 marks · Long answerHarder board level

A motorboat's speed in still water is 15 km/h. It goes 40 km upstream and then returns 40 km downstream. The upstream trip takes 2 hours more than the downstream trip. Using the quadratic formula, find the speed of the stream and the total time for the round trip.

Try it first, then show the solutionSolution

Let the speed of the stream be $x$ km/h, with $0 < x < 15$ . Upstream speed $= (15 - x)$ km/h and downstream speed $= (15 + x)$ km/h.
Given: $\frac{40}{15 - x} - \frac{40}{15 + x} = 2$ . (1 mark)
Simplify: $\frac{40 ( 15 + x ) - 40 ( 15 - x )}{225 - x ^{2}} = 2$ , so $80 x = 450 - 2 x^{2}$ , that is $x^{2} + 40 x - 225 = 0$ . (1 mark)
$a = 1$ , $b = 40$ , $c = - 225$ : $D = 1600 + 900 = 2500$ , so $D = 50$ . (1 mark)
$x = \frac{- 40 \pm 50}{2}$ , so $x = 5$ or $x = - 45$ . (1 mark)
A speed cannot be negative, so $x = 5$ km/h. Upstream time $= \frac{40}{10} = 4$ h and downstream time $= \frac{40}{20} = 2$ h, which differ by 2 h as given. Total time $= 4 + 2 = 6$ h. (1 mark)

Answer: The stream flows at 5 km/h and the round trip takes 6 hours.

Example 8: Equal roots: find k, then the root · 3 marks · Short answerHOTS / competency

Find the value of $k$ for which the equation $k x (x - 2) + 6 = 0$ has two equal real roots. Hence find the root. Explain why any other value of $k$ you obtain must be rejected.

Try it first, then show the solutionSolution

Standard form: $k x^{2} - 2 k x + 6 = 0$ , so $a = k$ , $b = - 2 k$ , $c = 6$ . (½ mark)
$D = (- 2 k)^{2} - 4 (k) (6) = 4 k^{2} - 24 k = 4 k (k - 6)$ . (½ mark)
Equal roots need $D = 0$ , so $4 k (k - 6) = 0$ , giving $k = 0$ or $k = 6$ . (½ mark)
If $k = 0$ , then $a = 0$ and the equation becomes $6 = 0$ , which is not a quadratic and has no root. So reject $k = 0$ . (½ mark)
For $k = 6$ : $x = - \frac{b}{2 a} = \frac{12}{12} = 1$ . (½ mark)
Check: $6 x^{2} - 12 x + 6 = 6 (x - 1)^{2} = 0$ gives $x = 1$ twice. (½ mark)

Answer: $k = 6$ , and the repeated root is $x = 1$ . The value $k = 0$ is rejected because the equation is then not quadratic.

Where students lose marks in Quadratic Formula

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

Mistake: Dividing only the square root by 2a
The fraction bar is written short, so the student divides $D$ by $2 a$ and leaves $- b$ undivided.
Wrong working
$2 x^{2} + 3 x - 20 = 0$ : $x = - 3 \pm \frac{13}{4}$ , so $x = \frac{1}{4}$ or $x = - \frac{25}{4}$
Right working
$x = \frac{- 3 \pm 13}{4}$ , so $x = \frac{5}{2}$ or $x = - 4$
Do this: Always draw one long fraction bar under the whole of $- b \pm D$ .
Usually costs: 1 to 2 marks: both roots are wrong.
Mistake: Getting the sign of -b wrong
The student copies b with its own sign instead of changing it.
Wrong working
$x^{2} - 3 x - 1 = 0$ : $x = \frac{- 3 \pm 13}{2}$
Right working
$b = - 3$ , so $- b = 3$ : $x = \frac{3 \pm 13}{2}$
Do this: Write the line ' $a = \dots$ , $b = \dots$ , $c = \dots$ ' with signs, then write $- b$ separately before substituting.
Usually costs: 1 mark for the roots, plus any later part that uses them.
Mistake: Reading a, b, c before reaching standard form
The student takes the coefficients from an equation that still has terms on both sides.
Wrong working
$x^{2} = 4 x - 2$ : $a = 1$ , $b = 4$ , $c = - 2$ , so $D = 24$ and $x = - 2 \pm 6$
Right working
$x^{2} - 4 x + 2 = 0$ : $a = 1$ , $b = - 4$ , $c = 2$ , so $D = 8$ and $x = \frac{4 \pm 2 2}{2} = 2 \pm 2$
Do this: First line of every solution: write the equation as ' $\dots = 0$ '.
Usually costs: Usually the full question, because every later step uses the wrong values.
Mistake: Cancelling only part of the numerator
The student cancels the denominator against one term of the numerator.
Wrong working
$3 x^{2} - 6 x + 2 = 0$ : $x = \frac{6 \pm 2 3}{6} = 1 \pm 23$
Right working
$x = \frac{6 \pm 2 3}{6} = \frac{2 ( 3 \pm 3 )}{6} = \frac{3 \pm 3}{3}$
Do this: Take a common factor out of every term of the numerator before cancelling.
Usually costs: ½ to 1 mark for the final roots.
Mistake: Treating b squared as negative
When b is negative, the student squares only the number and keeps the minus sign.
Wrong working
$x^{2} - 5 x + 3 = 0$ : $D = - 25 - 12 = - 37$ , so 'no real roots'
Right working
$D = (- 5)^{2} - 4 (1) (3) = 25 - 12 = 13$ , so $x = \frac{5 \pm 13}{2}$
Do this: Put negative values in brackets when you square them: $(- 5)^{2} = 25$ .
Usually costs: The whole question, because the conclusion changes.

Exam tips

If the question says 'using the quadratic formula', show the standard form, the values of a, b, c, the value of D and the substitution. An answer found only by factorising does not follow the instruction.
Write D on its own line. Many marking schemes give a separate mark for the correct discriminant even if a later step goes wrong.
In letter-coefficient questions, expect D to simplify to a perfect square such as $9 b^{2}$ or $(b^{2} + a c)^{2}$ . If it does not, recheck your expansion.
In word problems, write one line rejecting the negative root with a reason, then state the answer in words with units.
When an MCQ asks for roots, you can save time by checking the options with the sum $- \frac{b}{a}$ and the product $\frac{c}{a}$ instead of solving fully.

Practice questions on Quadratic Formula

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

Q1 · 1 mark · MCQBoard level
The roots of the equation $x^{2} - 4 x + 1 = 0$ are:
(a) $2 \pm 3$
(b) $- 2 \pm 3$
(c) $2 \pm 5$
(d) $4 \pm 3$
Hint
Here $b = - 4$ , so $- b = 4$ . Simplify $12$ before dividing by 2.
Answer
(a) $2 \pm 3$ . $D = 16 - 4 = 12$ , $12 = 23$ , and $x = \frac{4 \pm 2 3}{2} = 2 \pm 3$ .
Q2 · 3 marks · Short answerHarder board level
Using the quadratic formula, solve for $x$ : $ab x^{2} + (b^{2} - a c) x - b c = 0$ , where $a \neq = 0$ , $b \neq = 0$ .
Hint
Expand $D$ fully. The middle terms combine to give $+ 2 a b^{2} c$ , and $D$ becomes a perfect square.
Answer
$D = (b^{2} - a c)^{2} + 4 a b^{2} c = b^{4} + 2 a b^{2} c + a^{2} c^{2} = (b^{2} + a c)^{2}$ . Then $x = \frac{- ( b ^{2} - a c ) \pm ( b ^{2} + a c )}{2 ab}$ . Taking $+$ gives $x = \frac{2 a c}{2 ab} = \frac{c}{b}$ ; taking $-$ gives $x = \frac{- 2 b ^{2}}{2 ab} = - \frac{b}{a}$ . So $x = \frac{c}{b}$ or $x = - \frac{b}{a}$ .
Q3 · 3 marks · Short answerBoard level
Solve for $x$ : $\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$ , where $x \neq = 0, 1, 2$ .
Hint
Combine the left side into one fraction, cross-multiply with $\frac{6}{x}$ , and collect terms into standard form before using the formula.
Answer
The left side is $\frac{3 x - 5}{x ^{2} - 3 x + 2}$ . Cross-multiplying: $3 x^{2} - 5 x = 6 x^{2} - 18 x + 12$ , so $3 x^{2} - 13 x + 12 = 0$ . $D = 169 - 144 = 25$ , so $x = \frac{13 \pm 5}{6}$ , giving $x = 3$ or $x = \frac{4}{3}$ . Neither is restricted. Check $x = 3$ : $1 + 1 = 2 = \frac{6}{3}$ .
Q4 · 5 marks · Long answerHarder board level
Two pipes together can fill a tank in 6 hours. One pipe alone takes 5 hours more than the other pipe alone.
(i) Using the quadratic formula, find the time each pipe takes on its own to fill the tank. (3 marks)
(ii) One day only the faster pipe is opened for the first 4 hours, and then both pipes run together. Find the total time taken to fill the tank. (2 marks)
Hint
Let the faster pipe take $x$ hours. Add the fractions of the tank filled in 1 hour. In (ii), find the fraction still empty after 4 hours and divide it by the joint rate $\frac{1}{6}$ per hour.
Answer
(i) $\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}$ gives $6 (2 x + 5) = x^{2} + 5 x$ , so $x^{2} - 7 x - 30 = 0$ . $D = 49 + 120 = 169$ , so $x = \frac{7 \pm 13}{2} = 10$ or $- 3$ . Reject $- 3$ , as time cannot be negative. The pipes take 10 hours and 15 hours. Check: $\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{1}{6}$ . (ii) In 4 hours the faster pipe fills $\frac{4}{10} = \frac{2}{5}$ of the tank. The remaining $\frac{3}{5}$ at $\frac{1}{6}$ per hour takes $\frac{3}{5} \times 6 = \frac{18}{5} = 3 \frac{3}{5}$ hours. Total time $= 4 + 3 \frac{3}{5} = 7 \frac{3}{5}$ hours, that is 7 hours 36 minutes.
Q5 · 3 marks · Short answerHOTS / competency
Kabir solved $2 x^{2} - 6 x + 5 = 0$ as follows: $x = \frac{6 \pm 36 - 40}{4} = \frac{6 \pm 2}{4}$ , so $x = 2$ or $x = 1$ .
(i) Explain his error and state the correct conclusion. (1 mark)
(ii) Find the greatest integer value of $c$ for which $2 x^{2} - 6 x + c = 0$ has real roots, and solve the equation for that value of $c$ . (2 marks)
Hint
Real roots need $D \geq 0$ . Write $D$ in terms of $c$ and solve the inequality.
Answer
(i) $D = 36 - 40 = - 4 < 0$ . A negative number has no real square root, so $- 4$ is not 2. The equation has no real roots. (ii) $D = 36 - 8 c \geq 0$ gives $c \leq \frac{9}{2}$ , so the greatest integer is $c = 4$ . Then $2 x^{2} - 6 x + 4 = 0$ has $D = 36 - 32 = 4$ , so $x = \frac{6 \pm 2}{4}$ , giving $x = 2$ or $x = 1$ . Kabir's numbers solve this equation, not the one he was given.
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): The equation $x^{2} - 6 x + 9 = 0$ has one repeated root, $x = 3$ .
Reason (R): When $D = 0$ , the quadratic formula gives both roots as $x = - \frac{b}{2 a}$ .
(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
Find $D$ first. If it is zero, compute $- \frac{b}{2 a}$ and compare with A.
Answer
(a) Both A and R are true and R is the correct explanation of A. $D = 36 - 36 = 0$ , so the formula gives $x = \frac{6 \pm 0}{2} = 3$ twice, which is exactly $- \frac{b}{2 a} = \frac{6}{2} = 3$ .
Q7 · 2 marks · Very short answerBoard level
If $x = 2$ is a root of $3 x^{2} + k x - 10 = 0$ , find the value of $k$ . Then use the quadratic formula to find the other root.
Hint
Substitute $x = 2$ to find $k$ first, then solve the complete equation.
Answer
$3 (4) + 2 k - 10 = 0$ , so $k = - 1$ . The equation is $3 x^{2} - x - 10 = 0$ . $D = 1 + 120 = 121$ , so $x = \frac{1 \pm 11}{6}$ , giving $x = 2$ or $x = - \frac{5}{3}$ . The other root is $- \frac{5}{3}$ .
Q8 · 5 marks · Long answerHarder board level
A rectangular plot has a perimeter of 28 m and an area of 44 square metres.
(i) Using the quadratic formula, find its length and breadth in surd form. (3 marks)
(ii) Find the length of its diagonal in simplest surd form. (1 mark)
(iii) Show that no rectangle can have a perimeter of 28 m and an area of 50 square metres. (1 mark)
Hint
Length plus breadth is 14 m, so take the sides as $x$ and $14 - x$ . For the diagonal, add the squares of the two sides before taking the square root.
Answer
(i) $x (14 - x) = 44$ gives $x^{2} - 14 x + 44 = 0$ . $D = 196 - 176 = 20$ and $20 = 25$ , so $x = \frac{14 \pm 2 5}{2} = 7 \pm 5$ . Both values are positive and they are the two sides: length $(7 + 5)$ m, breadth $(7 - 5)$ m. Check: $(7 + 5) (7 - 5) = 49 - 5 = 44$ . (ii) $(7 + 5)^{2} + (7 - 5)^{2} = (54 + 145) + (54 - 145) = 108$ , so the diagonal is $108 = 63$ m. (iii) $x (14 - x) = 50$ gives $x^{2} - 14 x + 50 = 0$ , with $D = 196 - 200 = - 4 < 0$ . There are no real roots, so no such rectangle exists.
Q9 · 5 marks · Long answerHarder board level
A train travels 480 km at a uniform speed. If its speed had been 8 km/h less, it would have taken 2 hours more for the same journey. Using the quadratic formula, find the original speed of the train and the time taken at each speed.
Hint
Let the original speed be $x$ km/h, with $x > 8$ . The slower journey takes longer, so subtract $\frac{480}{x}$ from $\frac{480}{x - 8}$ .
Answer
$\frac{480}{x - 8} - \frac{480}{x} = 2$ gives $480 \times 8 = 2 x (x - 8)$ , so $x^{2} - 8 x - 1920 = 0$ . $D = 64 + 7680 = 7744 = 8 8^{2}$ , so $x = \frac{8 \pm 88}{2} = 48$ or $- 40$ . Reject $- 40$ , as speed cannot be negative. The original speed is 48 km/h and the time is $\frac{480}{48} = 10$ hours. At 40 km/h the time is $\frac{480}{40} = 12$ hours, which is 2 hours more.
Q10 · 3 marks · Short answerHOTS / competency
Find the value of $m$ for which $x^{2} - 2 (m + 1) x + (m^{2} + 7) = 0$ has equal roots, and solve the equation for that value of $m$ . For which values of $m$ does the equation have real roots?
Hint
Write $D$ in terms of $m$ and simplify fully: the $m^{2}$ terms cancel.
Answer
$D = 4 (m + 1)^{2} - 4 (m^{2} + 7) = 4 (m^{2} + 2 m + 1 - m^{2} - 7) = 8 (m - 3)$ . Equal roots need $D = 0$ , so $m = 3$ . The equation becomes $x^{2} - 8 x + 16 = 0$ , and $x = - \frac{b}{2 a} = \frac{8}{2} = 4$ (repeated). Check: $(x - 4)^{2} = 0$ . Real roots need $D \geq 0$ , that is $m \geq 3$ .
Q11 · 4 marks · Case-basedHOTS / competency
Suppose your class plans a trip and hires a bus for Rs 4800, with the cost shared equally by the students going. If 8 more students join, each student pays Rs 20 less. The bus has 50 seats, and 2 teachers will also travel in it.
(i) Taking the original number of students as $n$ , form a quadratic equation in standard form. (1 mark)
(ii) Find the discriminant of this equation. (1 mark)
(iii) Find $n$ and each student's share after the 8 students join. Decide whether everyone, including the teachers, fits in the bus. (2 marks)
Hint
Each share is the total cost divided by the number of students. The difference between the two shares is 20.
Answer
(i) $\frac{4800}{n} - \frac{4800}{n + 8} = 20$ gives $38400 = 20 n (n + 8)$ , so $n^{2} + 8 n - 1920 = 0$ . (ii) $D = 64 + 7680 = 7744$ . (iii) $D = 88$ , so $n = \frac{- 8 \pm 88}{2} = 40$ or $- 48$ ; reject $- 48$ . So 40 students first paid Rs 120 each, and with 48 students each pays Rs 100, which is Rs 20 less. People on the bus $= 48 + 2 = 50$ , equal to the 50 seats, so everyone fits exactly with no seat to spare.

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.State the quadratic formula for $a x^{2} + b x + c = 0$ .Show answer
$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , valid when $b^{2} - 4 a c \geq 0$ .
2.What is the discriminant?Show answer
$D = b^{2} - 4 a c$ , the expression under the square root in the formula.
3.When does the formula give no real roots?Show answer
When $D < 0$ , because a negative number has no real square root.
4.What single root does the formula give when $D = 0$ ?Show answer
$x = - \frac{b}{2 a}$ , repeated twice.
5.What must you do before reading a, b and c?Show answer
Rewrite the equation in standard form $a x^{2} + b x + c = 0$ .
6.If $D$ is a perfect square, what kind of roots do you get?Show answer
Rational roots, so the equation could also have been factorised.

Quadratic Formula: frequently asked questions

Is the quadratic formula the same as Sridharacharya's formula?

Yes. In India the quadratic formula is often called Sridharacharya's formula. It is the same result, and in the board exam you can call it either name.

Do I need to derive the quadratic formula for the 2026-27 board exam?

No. Solving by completing the square, the method used to derive the formula, is not in the 2026-27 syllabus. You only need to quote the formula and use it correctly.

When should I use the formula instead of factorisation?

Use the formula when the question asks for it, when splitting the middle term is not quick, or when the roots look irrational. If D turns out to be a perfect square, factorisation would also have worked, but the formula still gives the right answer.

What should I write if D comes out negative?

Write that since D is less than zero, the equation has no real roots, and stop. Your syllabus covers only real roots, so do not try to take the square root of a negative number.

Can I leave my answer with a square root in it?

Yes. Roots such as 2 plus root 3 are exact answers and are expected in surd form. Simplify the surd, cancel any common factor, and do not convert to decimals unless the question asks.