MathsCBSE Class 10 · Quadratic Equations

Solving Quadratic Equations by Factorisation (Class 10)

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

Use factorisation when the coefficients are manageable and you can quickly find two numbers with the right product and sum. This happens when the roots are rational. If the question says 'by factorisation', you must use it. Otherwise, switch to the formula if no pair works after a fair try.

Q: What if I cannot find the two numbers for splitting the middle term?

First recheck that the equation is correctly in standard form, because expansion mistakes are the usual cause. If it is correct and no pair exists, the roots are irrational or not real, and factorisation with simple numbers will not work. In that case use the quadratic formula, unless the question insists on factorisation.

Q: Is solving by completing the square in the 2026-27 syllabus?

No. Completing the square is not in the 2026-27 syllabus. You solve by factorisation and by the quadratic formula only.

Q: What kind of questions are in NCERT Exercise 4.2?

Exercise 4.2 asks you to find roots of quadratic equations by factorisation, including ones with surds and fractions. It also asks you to form quadratic equations from everyday situations and solve them. Practise both kinds, and always reject roots that make no sense in the situation.

Q: Do I have to verify the roots in the board exam?

Only when the question asks for it. Even when it does not, a quick check with the sum and product of roots, or by substituting one root, catches errors before you move on. Examiners give marks for correct roots, so the check protects your marks.

Updated 13 October 2026 · Fact-checked

Factorisation solves a quadratic equation by writing it as a product of two linear factors and setting each factor equal to zero. You split the middle term into two parts whose product is a times c and whose sum is b. Board questions ask you to solve equations with integers, surds or fractions this way, often inside word problems.

Find two numbers with product ac and sum b, split the middle term, group, then set each factor equal to zero.

Illustration drawn for QuestionBuddy.

Understand Solving Quadratic Equations by Factorisation

A quadratic equation in standard form $a x^{2} + b x + c = 0$ , with $a \neq = 0$ , has at most two real roots. Factorisation finds them by rewriting the left side as a product of two linear factors, such as $(2 x + 3) (3 x - 5)$ . Once the equation looks like this, you can read the roots off almost at once.

Why does this work? It rests on the zero product rule: if $pq = 0$ , then $p = 0$ or $q = 0$ . So $(2 x + 3) (3 x - 5) = 0$ means $2 x + 3 = 0$ or $3 x - 5 = 0$ , giving $x = - \frac{3}{2}$ or $x = \frac{5}{3}$ . The rule needs the right side to be zero. If $(x - 3) (x + 2) = 14$ , you cannot say either bracket equals 14. You must first expand and bring everything to one side.

To find the factors, split the middle term. Multiply $a$ and $c$ . Then find two numbers whose product is $a c$ and whose sum is $b$ . Here is why the product must be $a c$ and not just $c$ . When you expand $(p x + q) (r x + s)$ , you get $p r x^{2} + (p s + q r) x + q s$ . The two middle pieces $p s$ and $q r$ add up to $b$ , and their product is $p r \times q s$ , which is $a \times c$ . Write $b x$ as the sum of these two pieces, group the four terms in pairs, and take out the common factor. The two brackets left must match exactly.

This method links the whole chapter together. The roots you find are the zeroes of the polynomial $a x^{2} + b x + c$ from the Polynomials chapter. So you can check them quickly: their sum is $- \frac{b}{a}$ and their product is $\frac{c}{a}$ . Equations with surds, with letters as coefficients, or with $x$ in the denominators all come back to this same step once you reach standard form. Most word problems also end with a factorisation.

When does it work neatly? For integer coefficients, the two numbers exist exactly when $b^{2} - 4 a c$ is a perfect square, that is, when the roots are rational. If you have checked the factor pairs of $a c$ and none adds up to $b$ , switch to the quadratic formula. In this syllabus you deal only with real roots. Remember one rule above all: never divide both sides by $x$ , because that throws away the root $x = 0$ . Factor $x$ out instead.

Solving Quadratic Equations by Factorisation: formulas to remember

Zero product rule: $pq = 0 \Rightarrow p = 0 or q = 0$; The reason factorisation works. The right side must be 0 before you use it.
Splitting the middle term: $a x^{2} + b x + c = a x^{2} + p x + q x + c, p + q = b, pq = a c$; Find p and q first. Then group the four terms in pairs.
Sign guide for p and q: $a c > 0 \Rightarrow p, q same sign as b; a c < 0 \Rightarrow opposite signs, larger has sign of b$; This cuts down the pairs you need to try.
Factor form of a quadratic: $a x^{2} + b x + c = a (x - α) (x - β)$; Here α and β are the roots. It is useful for checking and for forming an equation from given roots.
Root from a linear factor: $p x + q = 0 \Rightarrow x = - \frac{q}{p}$; Change the sign of the constant and divide by the coefficient of x. So 3x - 5 gives 5/3, not 3/5.
Perfect square case (equal roots): $(p x + q)^{2} = 0 \Rightarrow x = - \frac{q}{p}, - \frac{q}{p}$; When p and q come out equal, the two factors are the same. Write the root twice or say 'equal roots'.
Taking out a common x: $a x^{2} + b x = 0 \Rightarrow x (a x + b) = 0 \Rightarrow x = 0 or x = - \frac{b}{a}$; Never divide by x. You would lose the root 0.
Sum and product check: $α + β = - \frac{b}{a}, α β = \frac{c}{a}$; A 20-second check on your two roots after factorising.
Useful identities: $a^{2} - b^{2} = (a - b) (a + b), a^{2} \pm 2 ab + b^{2} = (a \pm b)^{2}$; These spot factors instantly when the quadratic is a difference of squares or a perfect square.

How to solve Solving Quadratic Equations by Factorisation questions

This method works for every factorisation question in the board paper: plain integers, surds, letter coefficients, or fractions that reduce to a quadratic.

1If x appears in a denominator, first write the restrictions (the values that make any denominator zero). Then combine the fractions and cross-multiply.
2Expand all brackets and bring every term to one side to get $a x^{2} + b x + c = 0$ . Divide by any common numerical factor to keep the numbers small.
3Multiply $a$ and $c$ . Use the sign guide to decide the signs, then list factor pairs of $a c$ until you find the pair that adds up to $b$ .
4Rewrite $b x$ as the sum of the two terms. Group the four terms in pairs and take out the common factor from each pair.
5Check that the two brackets are identical, then write the product of the two factors equal to 0.
6Set each factor equal to 0 and solve. Rationalise any surd in a denominator.
7Reject any value that breaks a restriction, or that makes no sense in a word problem, and give the reason in one line.
8Check your roots using the sum $- \frac{b}{a}$ and product $\frac{c}{a}$ , or by substituting one root.

Solving Quadratic Equations by Factorisation 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.

Solving by factorisation (splitting the middle term)

Suppose your family is retiling the floor of a rectangular hall at home. The length of the hall is 3 m more than its breadth, and the floor area is 108 square metres. The square tiles you like have a side of 50 cm, and suppose each tile costs Rs 40.

Find the length and breadth of the hall. Then find how many tiles are needed to cover the floor exactly, and the total cost of the tiles.

Try it first, then show the solutionSolution

Let the breadth be $x$ m, so the length is $(x + 3)$ m.
Area: $x (x + 3) = 108 \Rightarrow x^{2} + 3 x - 108 = 0$ .
We need two numbers with product $- 108$ and sum $3$ . These are $12$ and $- 9$ .
$x^{2} + 12 x - 9 x - 108 = 0 \Rightarrow x (x + 12) - 9 (x + 12) = 0$ .
$(x + 12) (x - 9) = 0 \Rightarrow x = 9$ or $x = - 12$ .
A breadth cannot be negative, so $x = 9$ . The breadth is 9 m and the length is 12 m. Check: $9 \times 12 = 108$ .
Area of one tile $= 0.5 \times 0.5 = 0.25$ square metres.
Number of tiles $= \frac{108}{0.25} = 432$ .
Cost $= 432 \times 40 =$ Rs 17280.

Answer: The hall is 12 m long and 9 m wide. It needs 432 tiles, costing Rs 17280.

Memory hook: Floor area gives the product, the 3 m gap gives the middle term: split it and the hall's size falls out.

Solving by factorisation (splitting the middle term)

Imagine that in a school cricket tournament, your house team scores 240 runs in its innings at a steady run rate. The coach points out that if the batters had scored 2 runs per over more, the same 240 runs would have come in 10 fewer overs.

Find the number of overs the team actually batted and its run rate.

Try it first, then show the solutionSolution

Let the number of overs be $x$ . The run rate is $\frac{240}{x}$ runs per over.
With 10 fewer overs, the run rate would be $\frac{240}{x - 10}$ .
$\frac{240}{x - 10} - \frac{240}{x} = 2$ .
$\frac{240 x - 240 ( x - 10 )}{x ( x - 10 )} = 2 \Rightarrow \frac{2400}{x ( x - 10 )} = 2$ .
$2 x (x - 10) = 2400 \Rightarrow x^{2} - 10 x - 1200 = 0$ .
We need two numbers with product $- 1200$ and sum $- 10$ . These are $- 40$ and $30$ .
$x^{2} - 40 x + 30 x - 1200 = 0 \Rightarrow x (x - 40) + 30 (x - 40) = 0 \Rightarrow (x - 40) (x + 30) = 0$ .
$x = 40$ or $x = - 30$ . The number of overs cannot be negative, so $x = 40$ .
Run rate $= \frac{240}{40} = 6$ runs per over. Check: $\frac{240}{30} = 8$ , which is 2 more.

Answer: The team batted 40 overs at a run rate of 6 runs per over.

Memory hook: Same runs, faster rate, fewer overs: the gap in run rates becomes a quadratic you split.

Tricks to remember Solving Quadratic Equations by Factorisation

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

Multiply the ends, add to the middle
Multiply the first and last coefficients ( $a \times c$ ). The two numbers you want multiply to that and add to the middle coefficient $b$ .
Use it when: Every time you split the middle term, including surds such as $23 \times 3 = 6$ .
Plus product: same sign. Minus product: fight, and the bigger wins
If $a c$ is positive, both numbers take the sign of $b$ . If $a c$ is negative, they have opposite signs and the bigger one takes the sign of $b$ .
Use it when: Choosing the signs before you try factor pairs, so you do not end up with a sum of $+ 4$ when you need $- 4$ .
Twin brackets
After grouping, the two brackets must be twins. If they are not identical, your pair or a sign is wrong. Fix it now, not after solving.
Use it when: A built-in check in the middle of every factorisation.
Never divide by x
Seeing $x$ on both sides? Factor it out, never cancel it. Cancelling $x$ cancels a root.
Use it when: Equations like $5 x^{2} = 20 x$ , or fraction equations that reduce to $x^{2} - 2 x = 0$ .

Solving Quadratic Equations by Factorisation: solved examples

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

Example 1: Integer coefficients with a negative product · 2 marks · Very short answerBoard level

Solve by factorisation and verify both roots: $6 x^{2} - x - 15 = 0$ .

Try it first, then show the solutionSolution

$a c = 6 \times (- 15) = - 90$ and $b = - 1$ . We need two numbers with product $- 90$ and sum $- 1$ . These are $- 10$ and $9$ . (½)
$6 x^{2} - 10 x + 9 x - 15 = 0 \Rightarrow 2 x (3 x - 5) + 3 (3 x - 5) = 0$ (½)
$(3 x - 5) (2 x + 3) = 0$ (½)
$x = \frac{5}{3}$ or $x = - \frac{3}{2}$ (½)
Verify $x = \frac{5}{3}$ : $6 \times \frac{25}{9} - \frac{5}{3} - 15 = \frac{50}{3} - \frac{5}{3} - \frac{45}{3} = 0$ .
Verify $x = - \frac{3}{2}$ : $6 \times \frac{9}{4} + \frac{3}{2} - 15 = \frac{27}{2} + \frac{3}{2} - 15 = 15 - 15 = 0$ .

Answer: $x = \frac{5}{3}$ or $x = - \frac{3}{2}$ . Both roots satisfy the equation.

Example 2: Factorising with surd coefficients · 3 marks · Short answerHarder board level

Solve by factorisation: $23 x^{2} - 5 x + 3 = 0$ . Verify that the sum of the roots equals $- \frac{b}{a}$ .

Try it first, then show the solutionSolution

$a c = 23 \times 3 = 6$ and $b = - 5$ . Since $a c > 0$ and $b < 0$ , both numbers are negative: $- 3$ and $- 2$ . (½)
$23 x^{2} - 3 x - 2 x + 3 = 0$ (½)
Write $- 3 x = 3 x \times (- 3)$ : $3 x (2 x - 3) - 1 (2 x - 3) = 0$ (½)
$(2 x - 3) (3 x - 1) = 0$ (½)
$x = \frac{3}{2}$ or $x = \frac{1}{3} = \frac{3}{3}$ (½)
Sum $= \frac{3}{2} + \frac{3}{3} = \frac{5 3}{6}$ , and $- \frac{b}{a} = \frac{5}{2 3} = \frac{5 3}{6}$ . The two are equal, so the roots are verified. (½)

Answer: $x = \frac{3}{2}$ or $x = \frac{3}{3}$ . Their sum is $\frac{5 3}{6} = - \frac{b}{a}$ .

Example 3: A fraction equation that hides a root at zero · 3 marks · Short answerHarder board level

Solve for $x$ : $\frac{1}{x + 3} - \frac{1}{x - 5} = \frac{8}{15}$ , where $x \neq = - 3, 5$ .

Try it first, then show the solutionSolution

Restrictions: $x \neq = - 3, 5$ . Combine the fractions: $\frac{( x - 5 ) - ( x + 3 )}{( x + 3 ) ( x - 5 )} = \frac{8}{15}$
The numerator is $- 8$ , so $\frac{- 8}{( x + 3 ) ( x - 5 )} = \frac{8}{15}$ (1)
Cross-multiply: $- 8 \times 15 = 8 (x + 3) (x - 5) \Rightarrow (x + 3) (x - 5) = - 15$
$x^{2} - 2 x - 15 = - 15 \Rightarrow x^{2} - 2 x = 0$ (1)
Take $x$ out as a factor (do not divide by it): $x (x - 2) = 0 \Rightarrow x = 0$ or $x = 2$ (½)
Neither $0$ nor $2$ is a restricted value. Check $x = 0$ : $\frac{1}{3} + \frac{1}{5} = \frac{8}{15}$ . Both roots are accepted. (½)

Answer: $x = 0$ or $x = 2$ .

Example 4: Letter coefficients and a reciprocal fraction equation · 5 marks · Long answerHarder board level

(i) Solve for $x$ by factorisation: $6 x^{2} - (9 a - 2 b) x - 3 ab = 0$ .
(ii) Solve for $x$ : $\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15}$ , where $x \neq = 0, - 1$ .

Try it first, then show the solutionSolution

(i) $a c = 6 \times (- 3 ab) = - 18 ab$ , and the middle coefficient is $- (9 a - 2 b) = - 9 a + 2 b$ .
The two terms are $- 9 a$ and $2 b$ , since $(- 9 a) (2 b) = - 18 ab$ and their sum is $- 9 a + 2 b$ . (½)
$6 x^{2} - 9 a x + 2 b x - 3 ab = 0 \Rightarrow 3 x (2 x - 3 a) + b (2 x - 3 a) = 0$ (½)
$(2 x - 3 a) (3 x + b) = 0$ (½)
$x = \frac{3 a}{2}$ or $x = - \frac{b}{3}$ (½)
(ii) Take the LCM: $\frac{x ^{2} + ( x + 1 ) ^{2}}{x ( x + 1 )} = \frac{34}{15}$ (½)
$15 (2 x^{2} + 2 x + 1) = 34 (x^{2} + x) \Rightarrow 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x$ (½)
$4 x^{2} + 4 x - 15 = 0$ (½)
$a c = - 60$ and $b = 4$ , so the pair is $10$ and $- 6$ : $4 x^{2} + 10 x - 6 x - 15 = 0 \Rightarrow 2 x (2 x + 5) - 3 (2 x + 5) = 0$ (½)
$(2 x + 5) (2 x - 3) = 0 \Rightarrow x = \frac{3}{2}$ or $x = - \frac{5}{2}$ (½)
Neither value is 0 or $- 1$ . Check $x = \frac{3}{2}$ : $\frac{3}{5} + \frac{5}{3} = \frac{34}{15}$ . Both are accepted. (½)

Answer: (i) $x = \frac{3 a}{2}$ or $x = - \frac{b}{3}$ . (ii) $x = \frac{3}{2}$ or $x = - \frac{5}{2}$ .

Example 5: Case study: a path inside a school garden · 4 marks · Case-basedHOTS / competency

Suppose your school has a rectangular garden 30 m long and 20 m wide. A path of uniform width $x$ m is to be laid inside the garden along all four sides. The remaining inner rectangle is a lawn of area 336 square metres.
(i) Show that $x$ satisfies $x^{2} - 25 x + 66 = 0$ . (1 mark)
(ii) Find the area of the path. (1 mark)
(iii) Find the width of the path by factorisation, explaining why one root is rejected. (2 marks)

Try it first, then show the solutionSolution

(i) The lawn has length $(30 - 2 x)$ m and breadth $(20 - 2 x)$ m, so $(30 - 2 x) (20 - 2 x) = 336$ .
$600 - 100 x + 4 x^{2} = 336 \Rightarrow 4 x^{2} - 100 x + 264 = 0$ . Dividing by 4 gives $x^{2} - 25 x + 66 = 0$ . (1)
(ii) Area of the path $=$ garden area $-$ lawn area $= 600 - 336 = 264$ square metres. (1)
(iii) We need two numbers with product $66$ and sum $- 25$ . These are $- 22$ and $- 3$ : $x^{2} - 22 x - 3 x + 66 = 0 \Rightarrow x (x - 22) - 3 (x - 22) = 0$
$(x - 22) (x - 3) = 0 \Rightarrow x = 22$ or $x = 3$ (1)
$x = 22$ would make the lawn breadth $20 - 44 = - 24$ m, which is impossible, so it is rejected. Hence $x = 3$ . Check: $24 \times 14 = 336$ . (1)

Answer: (ii) The path covers 264 square metres. (iii) The path is 3 m wide; $x = 22$ m is rejected because it gives a negative lawn breadth.

Where students lose marks in Solving Quadratic Equations by Factorisation

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

Mistake: Dividing both sides by x
It feels like simplifying, but dividing by $x$ assumes $x \neq = 0$ . That silently deletes the root $x = 0$ .
Wrong working
$5 x^{2} = 20 x \Rightarrow 5 x = 20 \Rightarrow x = 4$
Right working
$5 x^{2} - 20 x = 0 \Rightarrow 5 x (x - 4) = 0 \Rightarrow x = 0$ or $x = 4$
Do this: Bring everything to one side and take $x$ out as a common factor. Never cancel it.
Usually costs: The answer mark for the missing root, often ½ to 1 mark.
Mistake: Using the zero product rule when the right side is not zero
Students see two brackets and jump straight to setting each one equal to the number on the right.
Wrong working
$(x - 2) (x + 1) = 10 \Rightarrow x - 2 = 10$ or $x + 1 = 10 \Rightarrow x = 12$ or $x = 9$
Right working
$x^{2} - x - 2 = 10 \Rightarrow x^{2} - x - 12 = 0 \Rightarrow (x - 4) (x + 3) = 0 \Rightarrow x = 4$ or $x = - 3$
Do this: Expand and get 0 on the right side first. Only then factorise.
Usually costs: Almost the whole question, because every later step is built on a wrong equation.
Mistake: Choosing the pair with the wrong signs
The product $a c$ is right, but the signs are flipped, so the pair adds to $+ 4$ instead of $- 4$ . The grouping still works, so the error goes unnoticed.
Wrong working
$3 x^{2} - 4 x - 4 = 0$ : $3 x^{2} + 6 x - 2 x - 4 = (3 x - 2) (x + 2) = 0 \Rightarrow x = \frac{2}{3}, - 2$
Right working
The pair needs sum $- 4$ , so use $- 6$ and $2$ : $3 x^{2} - 6 x + 2 x - 4 = 3 x (x - 2) + 2 (x - 2) = (3 x + 2) (x - 2) = 0 \Rightarrow x = 2, - \frac{2}{3}$
Do this: After writing the split, add your two middle terms and compare with the original $b x$ before you group.
Usually costs: 1 to 2 marks: the factorisation and both roots are wrong.
Mistake: Reading the root wrongly from a factor
Students flip the fraction or forget to change the sign when solving $p x + q = 0$ .
Wrong working
$(3 x + 2) (x - 2) = 0 \Rightarrow x = - \frac{3}{2}$ or $x = 2$
Right working
$3 x + 2 = 0 \Rightarrow x = - \frac{2}{3}$ , and $x - 2 = 0 \Rightarrow x = 2$
Do this: Write each factor equal to 0 on its own line and solve it like a linear equation. Then check using the product of roots, $\frac{c}{a}$ .
Usually costs: ½ mark per wrong root.

Exam tips

Show the split line, such as $6 x^{2} - 10 x + 9 x - 15$ , and the grouped line. These carry step marks even if a later sign slips.
For fraction equations, write the restrictions at the start and say at the end whether each root is accepted. A restricted value left in the answer loses the final mark.
When surds appear, write out the product $a c$ , for example $23 \times 3 = 6$ , so the examiner can see where your pair came from.
If a question says 'by factorisation', do not use the quadratic formula; the method itself is being tested. If no pair works after a fair try, recheck your standard form, because an expansion error is the usual cause.
Before moving on, check your roots with the sum and product of roots. It takes seconds and catches most sign errors.

Practice questions on Solving Quadratic Equations by Factorisation

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 for $x$ by factorisation: $10 x - \frac{1}{x} = 3$ , where $x \neq = 0$ .
Hint
Multiply through by $x$ to reach standard form. Then look for two numbers with product $- 10$ and sum $- 3$ .
Answer
$10 x^{2} - 3 x - 1 = 0 \Rightarrow 10 x^{2} - 5 x + 2 x - 1 = 0 \Rightarrow 5 x (2 x - 1) + 1 (2 x - 1) = 0 \Rightarrow (2 x - 1) (5 x + 1) = 0$ . So $x = \frac{1}{2}$ or $x = - \frac{1}{5}$ . Check $x = - \frac{1}{5}$ : $- 2 + 5 = 3$ .
Q2 · 3 marks · Short answerHarder board level
Solve for $x$ : $\frac{5}{x} - 3 = \frac{4}{x + 1}$ , where $x \neq = 0, - 1$ .
Hint
Write the left side as one fraction $\frac{5 - 3 x}{x}$ , cross-multiply, and collect all terms on one side.
Answer
$(5 - 3 x) (x + 1) = 4 x \Rightarrow 5 x + 5 - 3 x^{2} - 3 x = 4 x \Rightarrow 3 x^{2} + 2 x - 5 = 0$ . Split using $5 x$ and $- 3 x$ : $3 x^{2} + 5 x - 3 x - 5 = x (3 x + 5) - 1 (3 x + 5) = (3 x + 5) (x - 1) = 0$ . So $x = 1$ or $x = - \frac{5}{3}$ . Neither is restricted. Check $x = 1$ : $5 - 3 = 2$ and $\frac{4}{2} = 2$ .
Q3 · 3 marks · Short answerHarder board level
Solve for $x$ by factorisation: $12 ab x^{2} - (9 a^{2} - 8 b^{2}) x - 6 ab = 0$ , where $a \neq = 0, b \neq = 0$ .
Hint
The product $a c$ is $- 72 a^{2} b^{2}$ . Split the middle coefficient $- 9 a^{2} + 8 b^{2}$ into two terms with exactly this product.
Answer
Split as $- 9 a^{2} x + 8 b^{2} x$ : $12 ab x^{2} - 9 a^{2} x + 8 b^{2} x - 6 ab = 0 \Rightarrow 3 a x (4 b x - 3 a) + 2 b (4 b x - 3 a) = 0 \Rightarrow (4 b x - 3 a) (3 a x + 2 b) = 0$ . So $x = \frac{3 a}{4 b}$ or $x = - \frac{2 b}{3 a}$ .
Q4 · 5 marks · Long answerHarder board level
(i) Solve by factorisation: $5 x^{2} - 210 x + 2 = 0$ .
(ii) Solve for $x$ : $\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}$ , where $x \neq = 0, 2, 3$ .
Hint
(i) Here $a c = 10$ , and you need two numbers that add to $- 210$ : try $- 10$ twice. (ii) Combine the left side into $\frac{3 x - 8}{( x - 3 ) ( x - 2 )}$ before cross-multiplying.
Answer
(i) $5 x^{2} - 10 x - 10 x + 2 = 5 x (5 x - 2) - 2 (5 x - 2) = (5 x - 2)^{2} = 0$ . So $x = \frac{2}{5} = \frac{10}{5}$ , and the two roots are equal. (2 marks)
(ii) $x (3 x - 8) = 8 (x^{2} - 5 x + 6) \Rightarrow 3 x^{2} - 8 x = 8 x^{2} - 40 x + 48 \Rightarrow 5 x^{2} - 32 x + 48 = 0$ . Split using $- 20 x$ and $- 12 x$ : $5 x (x - 4) - 12 (x - 4) = (5 x - 12) (x - 4) = 0$ . So $x = 4$ or $x = \frac{12}{5}$ , and both are allowed. Check $x = 4$ : $1 + 1 = 2 = \frac{8}{4}$ . (3 marks)
Q5 · 3 marks · Short answerHOTS / competency
Aman wrote these two solutions.
Solution 1: $(x - 3) (x + 2) = 14$ , so $x - 3 = 14$ or $x + 2 = 14$ , giving $x = 17$ or $x = 12$ .
Solution 2: $3 x^{2} = 12 x$ . Dividing both sides by $x$ gives $3 x = 12$ , so $x = 4$ .
Find the error in each solution and give the correct roots, with a check.
Hint
The zero product rule only works when the right side is 0. Also ask yourself what happens to $x = 0$ when you divide by $x$ .
Answer
Solution 1 is wrong because a product equal to 14 does not force either factor to be 14. Correct working: $x^{2} - x - 6 = 14 \Rightarrow x^{2} - x - 20 = 0 \Rightarrow (x - 5) (x + 4) = 0$ , so $x = 5$ or $x = - 4$ . Check: $2 \times 7 = 14$ and $(- 7) (- 2) = 14$ .
Solution 2 is wrong because dividing by $x$ loses the root $x = 0$ . Correct working: $3 x^{2} - 12 x = 0 \Rightarrow 3 x (x - 4) = 0$ , so $x = 0$ or $x = 4$ .

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 must the right side be before you use the zero product rule?Show answer
Zero. Expand and bring every term to one side first.
2.In splitting the middle term of $a x^{2} + b x + c$ , the two numbers must have what product and what sum?Show answer
Product $a c$ and sum $b$ .
3.Which root does the factor $5 x - 2$ give?Show answer
$x = \frac{2}{5}$
4.If $a c$ is negative, what signs do the two numbers have?Show answer
Opposite signs. The larger one takes the sign of $b$ .
5.What are the roots of $x (x - 7) = 0$ ?Show answer
$x = 0$ and $x = 7$ .
6.What quick check confirms two roots $α$ and $β$ ?Show answer
$α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ .

Solving Quadratic Equations by Factorisation: frequently asked questions

When should I use factorisation instead of the quadratic formula?

Use factorisation when the coefficients are manageable and you can quickly find two numbers with the right product and sum. This happens when the roots are rational. If the question says 'by factorisation', you must use it. Otherwise, switch to the formula if no pair works after a fair try.

What if I cannot find the two numbers for splitting the middle term?

First recheck that the equation is correctly in standard form, because expansion mistakes are the usual cause. If it is correct and no pair exists, the roots are irrational or not real, and factorisation with simple numbers will not work. In that case use the quadratic formula, unless the question insists on factorisation.

Is solving by completing the square in the 2026-27 syllabus?

No. Completing the square is not in the 2026-27 syllabus. You solve by factorisation and by the quadratic formula only.

What kind of questions are in NCERT Exercise 4.2?

Exercise 4.2 asks you to find roots of quadratic equations by factorisation, including ones with surds and fractions. It also asks you to form quadratic equations from everyday situations and solve them. Practise both kinds, and always reject roots that make no sense in the situation.

Do I have to verify the roots in the board exam?

Only when the question asks for it. Even when it does not, a quick check with the sum and product of roots, or by substituting one root, catches errors before you move on. Examiners give marks for correct roots, so the check protects your marks.