MathsICSE Class 10 · Chapter 5

Quadratic Equations in One Variable ICSE Class 10: Methods, Discriminant and Problems

Updated 14 October 2026 · Fact-checked

A quadratic equation in one variable has the form ax² + bx + c = 0 with a ≠ 0. ICSE Class 10 asks you to solve it by factorisation or by the formula, give answers correct to three significant figures when told, judge the nature of roots from b² − 4ac, and solve word problems.

What Quadratic Equations in One Variable is about

A quadratic equation in $x$ is an equation that can be written as $a x^{2} + b x + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a \neq = 0$ . It has at most two roots.

The CISCE 2027 syllabus has three parts: (a) the nature of roots from the discriminant $b^{2} - 4 a c$ ; (b) solving by factorisation and by the quadratic formula; (c) simple problems that lead to a quadratic equation.

ICSE papers often say 'give your answer correct to three significant figures' or 'correct to two decimal places', and 'use mathematical tables if necessary'. That wording is a signal: the equation does not factorise, so use the formula, and find the square root from tables. Examiners' reports for 2024 and 2025 show marks lost on exactly these steps: wrong substitution of signs into the formula, inaccurate square roots, and rounding wrongly or too early. In word problems, especially speed and time, candidates lost marks for not deciding which time is larger before forming the equation.

This chapter is used again in Banking (finding the time in a recurring deposit), in Arithmetic Progressions (finding n), and in mensuration and coordinate problems.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Quadratic Equations in One Variable is in the 2. Algebra (ii) Quadratic Equations in one variable part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Standard form

Before using any method, write the equation as $a x^{2} + b x + c = 0$ : expand brackets, clear fractions, bring every term to one side, and collect like terms. Only then read off $a$ , $b$ and $c$ , with their signs.

An equation is quadratic only if, after simplifying, the $x^{2}$ term remains and no higher power is left. Some equations that look quadratic turn out linear once the $x^{2}$ terms cancel, and some that look cubic turn out quadratic.

Worked example

Write $(x + 2) (x - 3) = 2 x - 1$ in standard form and state $a$ , $b$ , $c$ .

Try it first, then show the solutionSolution

$x^{2} - x - 6 = 2 x - 1$ .
$x^{2} - 3 x - 5 = 0$ .
$a = 1$ , $b = - 3$ , $c = - 5$ .

Answer: $x^{2} - 3 x - 5 = 0$ ; a = 1, b = −3, c = −5

Roots, and finding an unknown coefficient

A number $α$ is a root of $a x^{2} + b x + c = 0$ if $a α^{2} + b α + c = 0$ . So to check a root, substitute it.

If a question says that a number is a root but one coefficient is unknown, substitute the root to get a linear equation in the unknown. Then solve the completed quadratic to find the other root.

Worked example

If $x = 3$ is a root of $2 x^{2} + k x - 12 = 0$ , find $k$ and the other root.

Try it first, then show the solutionSolution

$2 (9) + 3 k - 12 = 0$ , so $3 k = - 6$ and $k = - 2$ .
The equation is $2 x^{2} - 2 x - 12 = 0$ , i.e. $x^{2} - x - 6 = 0$ .
$(x - 3) (x + 2) = 0$ , so the other root is $- 2$ .

Answer: k = −2; other root −2

Solving by factorisation (splitting the middle term)

For $a x^{2} + b x + c = 0$ , find two numbers whose **product is $a c$ and whose sum is $b$ **. Split $b x$ into two terms using them, group in pairs, and take out common factors to get two linear factors. Then use the zero-product rule: if $P Q = 0$ , then $P = 0$ or $Q = 0$ .

Sign guide: if $a c > 0$ , both numbers have the sign of $b$ ; if $a c < 0$ , they have opposite signs and the larger one takes the sign of $b$ .

Worked example

Solve $6 x^{2} - x - 15 = 0$ by factorisation.

Try it first, then show the solutionSolution

$a c = - 90$ ; numbers with product $- 90$ and sum $- 1$ : $9$ and $- 10$ .
$6 x^{2} + 9 x - 10 x - 15 = 0 \Rightarrow 3 x (2 x + 3) - 5 (2 x + 3) = 0$ .
$(2 x + 3) (3 x - 5) = 0$ , so $x = - \frac{3}{2}$ or $x = \frac{5}{3}$ .

Answer: $x = - \frac{3}{2}$ or $\frac{5}{3}$

Solving by the quadratic formula, with rounding

For $a x^{2} + b x + c = 0$ ,
$x = \frac{- b \pm b ^{2} - 4 a c}{2 a} .$

Steps that protect your marks:
1. Write $a$ , $b$ , $c$ with signs on one line.
2. Work out $b^{2} - 4 a c$ separately. Remember $(- q)^{2} = q^{2}$ .
3. Find the square root from the tables (or accurately) to at least four significant figures.
4. Work out both roots, and round only at the end to what the question asks (three significant figures or two decimal places).

Three significant figures: 2.1805 → 2.18; 0.15286 → 0.153 (leading zeros do not count).

Worked example

Solve $3 x^{2} - 7 x + 1 = 0$ , giving your answers correct to three significant figures.

Try it first, then show the solutionSolution

$a = 3$ , $b = - 7$ , $c = 1$ ; $b^{2} - 4 a c = 49 - 12 = 37$ .
$37 = 6.083$ .
$x = \frac{7 \pm 6.083}{6}$ , so $x = \frac{13.083}{6} = 2.1805$ or $x = \frac{0.917}{6} = 0.1528$ .
Correct to 3 s.f.: $x = 2.18$ or $x = 0.153$ .

Answer: x = 2.18 or 0.153

Nature of roots: the discriminant

The discriminant is $D = b^{2} - 4 a c$ . Without solving:

$D > 0$ : two distinct real roots;
$D = 0$ : two equal real roots, each $- \frac{b}{2 a}$ ;
$D < 0$ : no real roots.

If $D$ is a perfect square (and $a$ , $b$ , $c$ are rational), the roots are rational and the equation factorises; otherwise, use the formula.

When $b$ contains a surd, square it carefully: for $b = 7$ , $b^{2} = 7$ , not 49.

Worked example

Find the nature of the roots of $4 x^{2} - 12 x + 9 = 0$ , and find them if they are real.

Try it first, then show the solutionSolution

$D = 144 - 144 = 0$ : two equal real roots.
Each root $= - \frac{b}{2 a} = \frac{12}{8} = \frac{3}{2}$ .

Answer: Equal roots, each 3/2

Finding k from the nature of roots

If a coefficient contains an unknown $k$ and the question states the nature of the roots, set up the matching condition on $D$ :

equal roots: $D = 0$ ;
real roots: $D \geq 0$ ;
distinct real roots: $D > 0$ ;
no real roots: $D < 0$ .

Also keep $a \neq = 0$ : a value of $k$ that makes the $x^{2}$ coefficient zero must be rejected, because the equation would no longer be quadratic.

Worked example

Find $k$ if $k x^{2} - 6 x + 1 = 0$ has two equal roots.

Try it first, then show the solutionSolution

$D = 36 - 4 k = 0$ , so $k = 9$ .
$k = 9 \neq = 0$ , so it is valid. (The equation is $9 x^{2} - 6 x + 1 = 0$ , with root $\frac{1}{3}$ .)

Answer: k = 9

Equations with fractions that reduce to quadratics

First write the values that make a denominator zero; these cannot be roots. Multiply both sides by the LCM of the denominators, simplify to standard form and solve. Finally check that neither root is a restricted value.

Worked example

Solve $\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$ , $x \neq = 0, 1, 2$ .

Try it first, then show the solutionSolution

Multiply by $x (x - 1) (x - 2)$ : $x (x - 1) + 2 x (x - 2) = 6 (x - 1) (x - 2)$ .
$3 x^{2} - 5 x = 6 x^{2} - 18 x + 12$ , so $3 x^{2} - 13 x + 12 = 0$ .
$(3 x - 4) (x - 3) = 0$ , so $x = \frac{4}{3}$ or $x = 3$ . Neither is restricted.

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

Word problems: forming the equation

1. Choose the unknown and write its unit ('Let the width be $x$ m').
2. Express the other quantities in terms of $x$ .
3. Translate the condition into an equation and simplify to standard form.
4. Solve, then reject roots that make no sense (negative lengths, speeds or numbers of people).
5. Answer exactly what is asked, with units.

For speed problems, use time $= \frac{distance}{speed}$ and think first: a lower speed means a longer time, so (time at lower speed) − (time at higher speed) = the extra time.

Worked example

The length of a rectangle is 5 m more than its width and its area is 84 m². Find its dimensions.

Try it first, then show the solutionSolution

Let the width be $x$ m; length $= (x + 5)$ m.
$x (x + 5) = 84 \Rightarrow x^{2} + 5 x - 84 = 0$ .
$(x + 12) (x - 7) = 0$ ; width cannot be negative, so $x = 7$ .

Answer: 7 m by 12 m

Maths in real life: Quadratic Equations in One Variable

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.

Area problem with a uniform border

Suppose you frame a 20 cm by 15 cm photograph with a border of the same width all round, and the framed picture covers 500 cm².

How wide is the border?

Try it first, then show the solutionSolution

Let the width be $x$ cm. Outer size: $(20 + 2 x)$ by $(15 + 2 x)$ .
$(20 + 2 x) (15 + 2 x) = 500 \Rightarrow 4 x^{2} + 70 x - 200 = 0 \Rightarrow 2 x^{2} + 35 x - 100 = 0$ .
$D = 1225 + 800 = 2025$ , $D = 45$ ; $x = \frac{- 35 \pm 45}{4} = 2.5$ or $- 20$ .
Width cannot be negative, so $x = 2.5$ .

Answer: 2.5 cm

Memory hook: Border on both sides: add 2x to each dimension.

Sharing equally: a fractions equation

Imagine your class decides to share the ₹1,200 cost of a farewell gift equally. On the day, 4 students are absent, so each of the others pays ₹10 more.

How many students were in the class, and what did each finally pay?

Try it first, then show the solutionSolution

Let there be $x$ students. $\frac{1200}{x - 4} - \frac{1200}{x} = 10$ .
$\frac{4800}{x ( x - 4 )} = 10 \Rightarrow x^{2} - 4 x - 480 = 0$ .
$(x - 24) (x + 20) = 0 \Rightarrow x = 24$ .
Each paid $\frac{1200}{20} = ₹60$ (instead of ₹50).

Answer: 24 students; ₹60 each

Memory hook: Fewer people sharing → bigger share: new share − old share = the increase.

Two meaningful roots

Suppose you throw a ball straight up, and its height after $t$ seconds is $h = 20 t - 5 t^{2}$ metres.

When is the ball 15 m above the ground?

Try it first, then show the solutionSolution

$20 t - 5 t^{2} = 15 \Rightarrow 5 t^{2} - 20 t + 15 = 0 \Rightarrow t^{2} - 4 t + 3 = 0$ .
$(t - 1) (t - 3) = 0$ , so $t = 1$ or $t = 3$ .
Both make sense: at 1 s on the way up, at 3 s on the way down.

Answer: At 1 second and at 3 seconds.

Memory hook: Not every second root is rubbish: check what it means before rejecting it.

Topics in this chapter

Quadratic Equations in One Variable formula sheet

Standard form: $a x^{2} + b x + c = 0, a \neq = 0$; Read a, b, c with signs only from this form.
Quadratic formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$; Use when told to give answers to 3 s.f. or 2 d.p.
Discriminant: $D = b^{2} - 4 a c$; D > 0 distinct real; D = 0 equal real; D < 0 no real roots.
Equal roots: $x = - \frac{b}{2 a}$; When D = 0.
Splitting the middle term: $pq = a c, p + q = b$; Then ax² + px + qx + c factorises by grouping.
Speed, distance, time: $t = \frac{d}{s}$; Lower speed, longer time.

How the board asks Quadratic Equations in One Variable

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

Question type	How it is asked	Method
Nature of roots MCQ	Section A, Question 1: decide the nature of roots of a given equation, or the condition for equal roots of $p x^{2} - q x + r = 0$ .	Compute $b^{2} - 4 a c$ carefully: $(- q)^{2} = q^{2}$ and $(7)^{2} = 7$ .
Solve to 3 significant figures (3-4 marks)	'Solve … correct to three significant figures. Use mathematical tables if necessary.'	Formula; D on its own line; square root from tables to 4 figures; round both roots only at the end.
Solve by factorisation	An equation with integer, fraction or surd coefficients that factorises.	Standard form, split the middle term, zero-product rule.
Find k for equal or real roots	A coefficient contains k; the roots are equal or real.	D = 0 (or ≥ 0); solve for k; reject k making a = 0; find the roots if asked.
Equations with fractions	Algebraic fractions on both sides.	State restrictions, multiply by the LCM, solve, check against restrictions.
Word problems (4 marks)	Speed and time, numbers and digits, ages, areas, sharing money equally.	Define x with units, form the equation, solve, reject meaningless roots, answer every part asked.

Tricks to remember Quadratic Equations in One Variable

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

Minus b, plus or minus root, all over 2a
Say the formula as a sentence: 'x equals minus b, plus or minus the root of b squared minus four a c, all over two a'.
Use it when: Writing the formula correctly under exam pressure.
Product ac, sum b
For splitting: multiply the ends (ac), add to the middle (b).
Use it when: Factorisation.
Plus two, zero twins, minus none
D positive: two different roots. D zero: twin roots. D negative: no real roots.
Use it when: Nature-of-roots questions.
Round at the end, not on the way
Keep 4 figures until the last line, then round once.
Use it when: 3 s.f. and 2 d.p. answers.
Slow takes long
In speed problems: slower speed → longer time → its time is the bigger fraction.
Use it when: Forming the time equation the right way round.

Quadratic Equations in One Variable: board-level solved questions

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

Example 1: Nature of roots with a surd · 1 mark · MCQBoard level

The equation $2 x^{2} - 26 x + 3 = 0$ has:
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than two real roots

Try it first, then show the solutionSolution

$b^{2} = (26)^{2} = 24$ ; $4 a c = 24$ .
$D = 0$ .

Answer: (b)

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

Assertion (A): The equation $3 x^{2} + p x - 5 = 0$ has two distinct real roots for every real value of $p$ .
Reason (R): If $a$ and $c$ have opposite signs, then $b^{2} - 4 a c > 0$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

$a c < 0$ gives $- 4 a c > 0$ , and $b^{2} \geq 0$ , so $D > 0$ : (R) is true.
Here $D = p^{2} + 60 > 0$ for all $p$ : (A) is true, and it follows from (R).

Answer: (c)

Example 3: Formula and three significant figures · 4 marks · Short answerHarder board level

Solve $3 x^{2} - 8 x + 2 = 0$ and give your answers correct to three significant figures.

Try it first, then show the solutionSolution

$a = 3$ , $b = - 8$ , $c = 2$ ; $D = 64 - 24 = 40$ .
$40 = 6.325$ .
$x = \frac{8 + 6.325}{6} = 2.3875$ or $x = \frac{8 - 6.325}{6} = 0.2792$ .
To 3 s.f.: $x = 2.39$ or $x = 0.279$ .

Answer: x = 2.39 or 0.279

Example 4: Equal roots with k in two coefficients · 4 marks · Short answerHarder board level

Find the values of $k$ for which $(k + 4) x^{2} + (k + 1) x + 1 = 0$ has equal roots. For each value, find the root.

Try it first, then show the solutionSolution

$D = (k + 1)^{2} - 4 (k + 4) = k^{2} - 2 k - 15$ .
$k^{2} - 2 k - 15 = 0 \Rightarrow (k - 5) (k + 3) = 0$ , so $k = 5$ or $- 3$ (neither makes $k + 4 = 0$ ).
$k = 5$ : $9 x^{2} + 6 x + 1 = 0$ , root $- \frac{6}{18} = - \frac{1}{3}$ .
$k = - 3$ : $x^{2} - 2 x + 1 = 0$ , root $1$ .

Answer: k = 5 (root −1/3) or k = −3 (root 1)

Example 5: A fractions equation · 4 marks · Short answerHarder board level

Solve $\frac{x + 1}{x - 1} - \frac{x - 1}{x + 1} = \frac{5}{6}$ , $x \neq = \pm 1$ .

Try it first, then show the solutionSolution

LHS $= \frac{( x + 1 ) ^{2} - ( x - 1 ) ^{2}}{x ^{2} - 1} = \frac{4 x}{x ^{2} - 1}$ .
$24 x = 5 x^{2} - 5 \Rightarrow 5 x^{2} - 24 x - 5 = 0$ .
$(5 x + 1) (x - 5) = 0$ , so $x = 5$ or $- \frac{1}{5}$ ; neither is restricted.

Answer: x = 5 or −1/5

Example 6: Factorisation · 3 marks · Long answerBoard level

Solve by factorisation: $4 x^{2} - 4 x - 15 = 0$ .

Try it first, then show the solutionSolution

$a c = - 60$ ; numbers $- 10$ and $6$ .
$4 x^{2} - 10 x + 6 x - 15 = 2 x (2 x - 5) + 3 (2 x - 5) = (2 x - 5) (2 x + 3)$ .
$x = \frac{5}{2}$ or $- \frac{3}{2}$ .

Answer: x = 5/2 or −3/2

Example 7: Speed and time: how much faster to save time? · 4 marks · Long answerHOTS / competency

A cyclist rides 48 km at a steady speed. (a) If her speed were 4 km/h more, she would take 1 hour less. Find her speed. (b) What original speed would mean that 4 km/h more saves 2 hours? (c) Explain, using your answers, why the time saved is greater when the original speed is lower.

Try it first, then show the solutionSolution

(a) Slower speed takes longer: $\frac{48}{x} - \frac{48}{x + 4} = 1 \Rightarrow \frac{192}{x ( x + 4 )} = 1$ , so $x^{2} + 4 x - 192 = 0$ , $(x + 16) (x - 12) = 0$ , $x = 12$ km/h.
(b) $\frac{192}{x ( x + 4 )} = 2 \Rightarrow x^{2} + 4 x - 96 = 0 \Rightarrow (x + 12) (x - 8) = 0$ , $x = 8$ km/h.
Check: $\frac{48}{8} - \frac{48}{12} = 6 - 4 = 2$ h.
(c) The time saved is $\frac{192}{x ( x + 4 )}$ . A smaller $x$ makes the denominator smaller, so the time saved is larger: the same 4 km/h is a bigger relative increase for a slow rider.

Answer: (a) 12 km/h (b) 8 km/h (c) Time saved = 192/(x(x + 4)) is larger for smaller x.

Example 8: Ages · 4 marks · Long answerHarder board level

The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 124. Find their present ages.

Try it first, then show the solutionSolution

Let the father be $x$ years; son $(45 - x)$ .
Five years ago: $(x - 5) (40 - x) = 124 \Rightarrow - x^{2} + 45 x - 200 = 124$ .
$x^{2} - 45 x + 324 = 0 \Rightarrow (x - 36) (x - 9) = 0$ .
$x = 9$ would make the son older than the father, so $x = 36$ ; son $= 9$ . Check: $31 \times 4 = 124$ .

Answer: Father 36 years, son 9 years

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Laying a path round a school lawn

A school in Dehradun has a rectangular lawn 30 m long and 20 m wide. The gardener plans a path of uniform width $x$ m inside the lawn, along all four edges. The grass left in the middle is to be 336 m². Paving the path costs ₹150 per m².

(i) · 3 marks
Show that $x^{2} - 25 x + 66 = 0$ .
Answer
Answer: $(30 - 2 x) (20 - 2 x) = 336$ simplifies to $x^{2} - 25 x + 66 = 0$ .
Inner rectangle: $(30 - 2 x)$ by $(20 - 2 x)$ . $(30 - 2 x) (20 - 2 x) = 336 \Rightarrow 600 - 100 x + 4 x^{2} = 336 \Rightarrow 4 x^{2} - 100 x + 264 = 0 \Rightarrow x^{2} - 25 x + 66 = 0$ .
(ii) · 3 marks
Solve the equation, explain which root you reject, and find the cost of paving the path.
Answer
Answer: x = 3 m; cost ₹39,600
$(x - 3) (x - 22) = 0$ . $x = 22$ is impossible (twice 22 is more than the width 20 m), so $x = 3$ . Path area $= 600 - 336 = 264$ m²; cost $= 264 \times 150 = ₹39, 600$ .
(iii) · 4 marks
The principal asks for a wider path, leaving only 150 m² of grass. Find the width of the path correct to three significant figures.
Answer
Answer: 5.89 m
$(30 - 2 x) (20 - 2 x) = 150 \Rightarrow 4 x^{2} - 100 x + 450 = 0 \Rightarrow 2 x^{2} - 50 x + 225 = 0$ . $D = 2500 - 1800 = 700$ , $700 = 26.46$ . $x = \frac{50 - 26.46}{4} = 5.886$ (the other root, 19.1, is too wide). Width = 5.89 m.

Prize money shared equally

A quiz club in Pune wins ₹6,000, to be shared equally among the members who took part. Five members later say they do not want a share, and as a result each of the others gets ₹100 more.

(i) · 3 marks
Taking $x$ as the original number of members, form a quadratic equation.
Answer
Answer: $x^{2} - 5 x - 300 = 0$
$\frac{6000}{x - 5} - \frac{6000}{x} = 100 \Rightarrow \frac{30000}{x ( x - 5 )} = 100 \Rightarrow x^{2} - 5 x - 300 = 0$ .
(ii) · 3 marks
Solve it and find the number of members.
Answer
Answer: 20 members
$(x - 20) (x + 15) = 0$ ; members cannot be negative, so $x = 20$ .
(iii) · 4 marks
Find each member's share before and after. Use the discriminant to show that an equation of the form $x^{2} - 5 x - c = 0$ with $c > 0$ always has two real roots, one positive and one negative.
Answer
Answer: ₹300 before, ₹400 after.
Before: $\frac{6000}{20} = ₹300$ ; after: $\frac{6000}{15} = ₹400$ . $D = 25 + 4 c > 0$ , so the roots are real and distinct. Their product is $- c < 0$ (from $(x - α) (x - β) = x^{2} - 5 x - c$ , so $α β = - c$ ), so one root is positive and one negative, which is why one root is always rejected.

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.

Solve by factorisation: $3 x^{2} - 5 x - 2 = 0$ . (3 marks)

Step	Marks
Split: $3 x^{2} - 6 x + x - 2 = 0$	1
$(x - 2) (3 x + 1) = 0$	1
$x = 2$ or $x = - \frac{1}{3}$	1

Solve $x^{2} - 5 x + 2 = 0$ , giving your answers correct to three significant figures. (4 marks)

Step	Marks
Correct formula with $a = 1$ , $b = - 5$ , $c = 2$	1
$D = 25 - 8 = 17$ , $17 = 4.123$	1
$x = \frac{5 + 4.123}{2} = 4.5615$ and $x = \frac{5 - 4.123}{2} = 0.4385$	1
$x = 4.56$ or $0.438$ (3 s.f.)	1

A car travels 400 km. If its speed were 10 km/h more, it would take 2 hours less. Find (a) its speed, (b) the time taken at that speed. (5 marks)

Step	Marks
Let the speed be $x$ km/h: $\frac{400}{x} - \frac{400}{x + 10} = 2$	1
$\frac{4000}{x ( x + 10 )} = 2 \Rightarrow x^{2} + 10 x - 2000 = 0$	1
$(x + 50) (x - 40) = 0$	1
$x = 40$ km/h (speed cannot be negative)	1
Time $= \frac{400}{40} = 10$ hours	1

Where students lose marks in Quadratic Equations in One Variable

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

Mistake: Squaring a surd coefficient wrongly
CISCE examiners' comments, 2025: for $3 x^{2} + 7 x + 2 = 0$ , candidates took $b^{2}$ as 49 instead of 7 and chose 'two distinct real roots'.
Wrong working
$D = 49 - 24 = 25 > 0$
Right working
$D = (7)^{2} - 24 = 7 - 24 = - 17 < 0$ : no real roots
Do this: Write $b^{2} = (\dots)^{2} = \dots$ on its own before finding D.
Usually costs: 1 mark.
Mistake: Treating $(- q)^{2}$ as $- q^{2}$
CISCE examiners' comments, 2024: for the equal-roots condition of $p x^{2} - q x + r = 0$ , candidates wrote $- q^{2} = 4 p r$ , or used $p^{2} = 4 q r$ .
Do this: The square of a negative number is positive: $(- q)^{2} = q^{2}$ . Equal roots: $q^{2} = 4 p r$ .
Usually costs: 1 mark.
Mistake: Inaccurate square roots and early rounding
CISCE examiners' comments, 2024 and 2025: candidates took $60$ as 7.75 instead of 7.746, found square roots by long division with errors instead of using tables, and rounded wrongly to three significant figures.
Do this: Use tables to four figures; keep them until the last line; then round each root to 3 s.f.
Usually costs: 1 to 2 marks of a 3- or 4-mark question.
Mistake: Sign errors in the formula
CISCE examiners' comments, 2024 and 2025: wrong formula or sign errors when substituting, e.g. using $b$ instead of $- b$ .
Wrong working
$2 x^{2} - 5 x - 4 = 0$ : $x = \frac{- 5 \pm 25 - 32}{4}$
Right working
$x = \frac{5 \pm 25 + 32}{4} = \frac{5 \pm 57}{4}$
Do this: List $a$ , $b$ , $c$ with signs; write $- b = \dots$ and $- 4 a c = \dots$ separately.
Usually costs: Most of the question.
Mistake: Forming a speed-time equation the wrong way round
CISCE examiners' comments, 2025: candidates could not decide which of $\frac{350}{x}$ and $\frac{350}{x - 20}$ is larger, so they formed a wrong equation; others stopped after finding the speed and did not find the time asked for.
Do this: Ask: which journey is slower? Its time is larger. Larger time − smaller time = difference. Then answer every sub-part.
Usually costs: The whole question if the equation is wrong.

Practice questions on Quadratic Equations in One Variable

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
Which of these is a quadratic equation?
(a) $x (x + 1) = x^{2} + 5$
(b) $(x + 2)^{3} = x^{3} + 8$
(c) $(x - 1)^{2} = x^{2} - 3$
(d) $x^{3} - x = x^{3} + 1$
Hint
Expand and simplify each.
Answer
(b): $x^{3} + 6 x^{2} + 12 x + 8 = x^{3} + 8$ gives $6 x^{2} + 12 x = 0$ . The others reduce to linear equations: (a) $x = 5$ , (c) $- 2 x + 1 = - 3$ , (d) $- x = 1$ .
Q2 · 1 mark · MCQBoard level
If $x = 2$ is a root of $x^{2} - 7 x + k = 0$ , then $k$ is:
(a) $- 10$
(b) $10$
(c) $5$
(d) $14$
Hint
Substitute x = 2.
Answer
(b) 10, since $4 - 14 + k = 0$ .
Q3 · 1 mark · MCQHarder board level
The roots of $a x^{2} - 2 b x + c = 0$ are real and equal if:
(a) $b^{2} = 4 a c$
(b) $b^{2} = a c$
(c) $4 b^{2} = a c$
(d) $b = 2 a c$
Hint
$(- 2 b)^{2} = 4 b^{2}$ .
Answer
(b): $4 b^{2} - 4 a c = 0 \Rightarrow b^{2} = a c$ .
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): The equation $2 x^{2} - 3 x + 5 = 0$ has no real roots.
Reason (R): A quadratic equation has no real roots when $b^{2} - 4 a c < 0$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Find D.
Answer
(c): $D = 9 - 40 = - 31 < 0$ .
Q5 · 4 marks · Short answerHarder board level
Solve $x^{2} - 6 x + 4 = 0$ , giving your answers correct to two decimal places.
Hint
$20 = 4.472$ .
Answer
$D = 36 - 16 = 20$ ; $x = \frac{6 \pm 4.472}{2} = 5.236$ or $0.764$ ; $x = 5.24$ or $0.76$ .
Q6 · 4 marks · Short answerHarder board level
Find the values of $p$ for which $4 x^{2} + p x + 9 = 0$ has equal roots, and find the root in each case.
Hint
$p^{2} - 144 = 0$ .
Answer
$p = \pm 12$ . For $p = 12$ : root $- \frac{12}{8} = - \frac{3}{2}$ . For $p = - 12$ : root $\frac{3}{2}$ .
Q7 · 3 marks · Long answerBoard level
Solve: $x (2 x + 5) = 25$ .
Hint
Standard form first.
Answer
$2 x^{2} + 5 x - 25 = 0 \Rightarrow (2 x - 5) (x + 5) = 0 \Rightarrow x = \frac{5}{2}$ or $- 5$ .
Q8 · 4 marks · Long answerHarder board level
Solve: $\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$ , $x \neq = - 4, 7$ .
Hint
The left side simplifies to $\frac{- 11}{( x + 4 ) ( x - 7 )}$ .
Answer
$\frac{- 11}{( x + 4 ) ( x - 7 )} = \frac{11}{30} \Rightarrow (x + 4) (x - 7) = - 30 \Rightarrow x^{2} - 3 x + 2 = 0 \Rightarrow x = 1$ or $2$ .
Q9 · 4 marks · Long answerHOTS / competency
(a) Show that $x^{2} + p x - 1 = 0$ has two distinct real roots for every real value of $p$ . (b) If one root is $\frac{1}{2}$ , find $p$ and the other root.
Hint
D = p² + 4.
Answer
(a) $D = p^{2} + 4 > 0$ for all real $p$ . (b) $\frac{1}{4} + \frac{p}{2} - 1 = 0 \Rightarrow p = \frac{3}{2}$ . Equation: $2 x^{2} + 3 x - 2 = 0 \Rightarrow (2 x - 1) (x + 2) = 0$ ; other root $- 2$ .
Q10 · 4 marks · Long answerHarder board level
The product of the digits of a two-digit number is 18. When 27 is subtracted from the number, the digits are reversed. Find the number.
Hint
Let the digits be a (tens) and b (units): $a - b = 3$ .
Answer
$10 a + b - 27 = 10 b + a \Rightarrow a - b = 3$ . With $ab = 18$ : $b (b + 3) = 18 \Rightarrow b^{2} + 3 b - 18 = 0 \Rightarrow b = 3$ (not $- 6$ ), $a = 6$ . Number 63; check $63 - 27 = 36$ .
Q11 · 4 marks · Long answerHarder board level
A motorboat whose speed in still water is 15 km/h goes 30 km upstream and comes back in a total of 4 hours 30 minutes. Find the speed of the stream.
Hint
Upstream speed 15 − s, downstream 15 + s.
Answer
$\frac{30}{15 - s} + \frac{30}{15 + s} = 4.5 \Rightarrow \frac{900}{225 - s ^{2}} = 4.5 \Rightarrow 225 - s^{2} = 200 \Rightarrow s = 5$ km/h.
Q12 · 3 marks · Short answerHOTS / competency
Show that a rectangle with perimeter 40 cm cannot have an area of 110 cm². What is the greatest possible area?
Hint
Sides x and 20 − x; use the discriminant.
Answer
$x (20 - x) = 110 \Rightarrow x^{2} - 20 x + 110 = 0$ , $D = 400 - 440 = - 40 < 0$ : no real $x$ . For area $A$ , $x^{2} - 20 x + A = 0$ needs $400 - 4 A \geq 0$ , so $A \leq 100$ : the greatest area is 100 cm² (a 10 cm square).

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.Write the quadratic formula.Show answer
$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
2.What does D = 0 tell you?Show answer
Two equal real roots, each −b/(2a).
3.What does 'use mathematical tables' usually signal?Show answer
The equation does not factorise; use the formula and take the square root from tables.
4.Round 0.15286 to three significant figures.Show answer
0.153
5.Condition for no real roots?Show answer
b² − 4ac < 0
6.When splitting the middle term, what product and sum do you need?Show answer
Product ac, sum b.
7.Why must you check fractional equations at the end?Show answer
A root might make a denominator zero, so it would have to be rejected.
8.If a and c have opposite signs, what is the nature of the roots?Show answer
Real and distinct (D > 0).

Best order to study it

Practise writing equations in standard form and checking roots.
Master factorisation by splitting the middle term.
Learn the formula and practise answers to 3 s.f. and 2 d.p. with tables.
Learn the discriminant and the 'find k' questions.
Practise fraction equations, then word problems (speed, ages, area, sharing).

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 signs?
Can you factorise by splitting the middle term quickly?
Can you use the formula and give roots to 3 s.f. correctly?
Can you find the nature of roots, including surd coefficients?
Can you find k for equal or real roots and reject k that makes a = 0?
Can you solve equations with fractions and check restrictions?
Can you form speed-time equations the right way round and reject meaningless roots?

Quadratic Equations in One Variable: frequently asked questions

Is completing the square in the ICSE 2027 syllabus?

The syllabus names two methods for solving: factorisation and using the formula. Use these in the exam.

How do I know whether to factorise or use the formula?

If the question asks for answers correct to three significant figures or two decimal places, or mentions tables, use the formula. Otherwise, check whether b² − 4ac is a perfect square; if it is, factorisation is quicker.

What does 'correct to three significant figures' mean?

Keep three digits counting from the first non-zero digit: 2.1805 becomes 2.18 and 0.27924 becomes 0.279.

Should I reject the negative root in every word problem?

Only when it makes no sense in the situation, such as a negative length, speed or number of people. In some problems, like a ball's height at two times, both roots are meaningful.

Do I need to show working for nature-of-roots MCQs?

No. In Question 1 you write only the correct option, but work out the discriminant carefully in rough.

Based on: CISCE ICSE 2027 Mathematics syllabus; CISCE ICSE 2027 Mathematics specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2025; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2024