What is the factorisation method in Quadratic Equations?

The factorisation method involves expressing the quadratic equation in the form (x−p)(x−q)=0 and finding the values of x that satisfy the equation.

Are factorisation questions important in Class 10 Quadratic Equations?

Yes, factorisation-based questions are very important in Class 10 Maths Chapter 4 and frequently appear in CBSE board exams as 1, 2, and 3 mark questions.

Does this page contain CBSE previous year questions on factorisation?

Yes, this page contains CBSE previous year questions (PYQs) specifically focused on solving quadratic equations by factorisation, with complete step-by-step solutions.

Which types of questions are covered using factorisation method?

This page covers MCQs, SA-I, SA-II, radical equations, rational equations, and case-based questions solvable using factorisation.

Solving By Factorisation PYQs

Overview

This page provides comprehensive Class 10 Maths Solving By Factorisation PYQs | Quadratic Equations. Solving By Factorisation previous year questions for Class 10 Maths Quadratic Equations. Practice CBSE board PYQs with step-by-step solutions on SJMaths.

Practice Previous Year Questions Topic-wise

Q1 2023

00:00

The roots of the equation $x^2 + 3x - 10 = 0$ are:

(a)$2, -5$

(b)$-2, 5$

(c)$2, 5$

(d)$-2, -5$

Factorising:

$x^2+5x-2x-10=0$

$(x+5)(x-2)=0$

$x=-5, 2$

Option (a) $2, -5$

Q2 SA-I 2021–22

00:00

Solve for $x$: $x^2 - 2ax - (4b^2 - a^2) = 0$

Rewrite equation:

$x^2-2ax+a^2-4b^2=0$

$(x-a)^2-(2b)^2=0$

$(x-a-2b)(x-a+2b)=0$

$x=a\pm 2b$

Q3 SA-I 2021–22

00:00

Solve for $x$: $2x^2-2\sqrt2x+1=0$

$a=2, b=-2\sqrt2, c=1$

$D=b^2-4ac=8-8=0$

Roots are real and equal

$x=\frac{-b}{2a}=\frac{2\sqrt2}{4}=\frac{\sqrt2}{2}$

$x=\frac{\sqrt2}{2}$

Q4 SA-II 2016

00:00

Solve for $x$: $\sqrt{2x+9}+x=13$

$\sqrt{2x+9}=13-x$

Squaring both sides:

$2x+9=(13-x)^2$

$x^2-28x+160=0$

$(x-8)(x-20)=0$

Check: $x=8$ valid, $x=20$ rejected

$x=8$

Q5 SA-II 2016

00:00

Solve for $x$: $\sqrt{6x+7}-(2x-7)=0$

$\sqrt{6x+7}=2x-7$

Squaring both sides:

$6x+7=4x^2-28x+49$

$4x^2-34x+42=0$

$(2x-7)(2x-6)=0$

Valid solution: $x=\frac{7}{2}$

$x=\frac{7}{2}$

Q6 SA-II 2016

00:00

Solve: $\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}=\dfrac{3}{2}$ ($x\neq1,2,3$)

Taking LCM and simplifying

$2(2x-4)=3(x-1)(x-3)$

$3x^2-16x+17=0$

$x= \frac{17}{3}, 1$

$x=\frac{17}{3}$

Q7 SA-II 2016

00:00

Solve for $x$: $\dfrac{1}{x}+\dfrac{2}{2x-3}=\dfrac{1}{x-2}$

Taking LCM and simplifying

$2x^2-7x+6=0$

$(2x-3)(x-2)=0$

Valid solution: $x=\frac{3}{2}$

$x=\frac{3}{2}$

Q8 SA-II 2019

00:00

Solve: $\dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{7}{x+5}$ ($x\neq-1,-2,-5$)

Taking LCM and simplifying

$x^2+4x+1=0$

$x=-2\pm\sqrt3$

$x=-2\pm\sqrt3$

Q9 SA-II 2017

00:00

Solve for $x$: $\dfrac{1}{x+1}+\dfrac{3}{5x+1}=\dfrac{5}{x+4}$ ($x\neq-1,-\frac15,-4$)

Taking LCM: $(x+1)(5x+1)(x+4)$

$(5x+1)(x+4)+3(x+1)(x+4)=5(x+1)(5x+1)$

$5x^2+21x+4+3x^2+15x+12=25x^2+30x+5$

$17x^2-6x-11=0$

$(17x+11)(x-1)=0$

Valid solution: $x=1$

$x=1$

Q10 SA-II 2016

00:00

Solve the quadratic equation: $x^2+\left(\frac{a+b}{a}+\frac{a}{a+b}\right)x+1=0$

Let $\alpha=\frac{a+b}{a}$ and $\beta=\frac{a}{a+b}$

Then equation becomes: $x^2+(\alpha+\beta)x+1=0$

Since $\alpha\beta=1$, roots are $-\alpha$ and $-\beta$

$x=-\frac{a+b}{a}$ or $x=-\frac{a}{a+b}$

Q11 2024

00:00

Case Study: A rectangular floor can be tiled using 200 square tiles. If the side of each tile is increased by 1 unit, only 128 tiles are required.

(i) Form the quadratic equation.
(ii) Write it in standard form.
(iii) Find the value of $x$ by factorisation.

Let side of tile = $x$ units

Area of floor = $200x^2$

New area = $128(x+1)^2$

$200x^2=128(x+1)^2$

$25x^2=16(x^2+2x+1)$

$9x^2-32x-16=0$

$(9x+4)(x-4)=0$

$x=4$ (reject $x=-\frac49$)

Side of tile = 4 units

Q12 SA-I 2023

00:00

Solve the quadratic equation: $2x^2-2\sqrt2x+1=0$

$a=2, b=-2\sqrt2, c=1$

$D=b^2-4ac=8-8=0$

Roots are real and equal

$x=\frac{-b}{2a}=\frac{\sqrt2}{2}$

$x=\frac{\sqrt2}{2}$

Q13 SA-II 2016

00:00

Solve for $x$: $\sqrt{2x+9}+x=13$

$\sqrt{2x+9}=13-x$

Squaring both sides

$2x+9=(13-x)^2$

$x^2-28x+160=0$

$(x-8)(x-20)=0$

Valid solution: $x=8$

$x=8$

Q14 SA-II 2016

00:00

Solve for $x$: $\sqrt{6x+7}-(2x-7)=0$

$\sqrt{6x+7}=2x-7$

Squaring both sides

$6x+7=4x^2-28x+49$

$2x^2-17x+21=0$

$(2x-7)(x-3)=0$

Valid solution: $x=\frac72$

$x=\frac72$

00:00

The roots of the equation $x^2 + 3x - 10 = 0$ are:

(a)$2, -5$

(b)$-2, 5$

(c)$2, 5$

(d)$-2, -5$

Factorising:

$x^2+5x-2x-10=0$

$(x+5)(x-2)=0$

$x=-5, 2$

Option (a) $2, -5$

Q2 SA-I 2021–22

00:00

Solve for $x$: $x^2 - 2ax - (4b^2 - a^2) = 0$

Rewrite equation:

$x^2-2ax+a^2-4b^2=0$

$(x-a)^2-(2b)^2=0$

$(x-a-2b)(x-a+2b)=0$

$x=a\pm 2b$

Q3 SA-I 2021–22

00:00

Solve for $x$: $2x^2-2\sqrt2x+1=0$

$a=2, b=-2\sqrt2, c=1$

$D=b^2-4ac=8-8=0$

Roots are real and equal

$x=\frac{-b}{2a}=\frac{2\sqrt2}{4}=\frac{\sqrt2}{2}$

$x=\frac{\sqrt2}{2}$

Q4 SA-II 2016

00:00

Solve for $x$: $\sqrt{2x+9}+x=13$

$\sqrt{2x+9}=13-x$

Squaring both sides:

$2x+9=(13-x)^2$

$x^2-28x+160=0$

$(x-8)(x-20)=0$

Check: $x=8$ valid, $x=20$ rejected

$x=8$

Q5 SA-II 2016

00:00

Solve for $x$: $\sqrt{6x+7}-(2x-7)=0$

$\sqrt{6x+7}=2x-7$

Squaring both sides:

$6x+7=4x^2-28x+49$

$4x^2-34x+42=0$

$(2x-7)(2x-6)=0$

Valid solution: $x=\frac{7}{2}$

$x=\frac{7}{2}$

Q6 SA-II 2016

00:00

Solve: $\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}=\dfrac{3}{2}$ ($x\neq1,2,3$)

Taking LCM and simplifying

$2(2x-4)=3(x-1)(x-3)$

$3x^2-16x+17=0$

$x= \frac{17}{3}, 1$

$x=\frac{17}{3}$

Q7 SA-II 2016

00:00

Solve for $x$: $\dfrac{1}{x}+\dfrac{2}{2x-3}=\dfrac{1}{x-2}$

Taking LCM and simplifying

$2x^2-7x+6=0$

$(2x-3)(x-2)=0$

Valid solution: $x=\frac{3}{2}$

$x=\frac{3}{2}$

Q8 SA-II 2019

00:00

Solve: $\dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{7}{x+5}$ ($x\neq-1,-2,-5$)

Taking LCM and simplifying

$x^2+4x+1=0$

$x=-2\pm\sqrt3$

$x=-2\pm\sqrt3$

Q9 SA-II 2017

00:00

Solve for $x$: $\dfrac{1}{x+1}+\dfrac{3}{5x+1}=\dfrac{5}{x+4}$ ($x\neq-1,-\frac15,-4$)

Taking LCM: $(x+1)(5x+1)(x+4)$

$(5x+1)(x+4)+3(x+1)(x+4)=5(x+1)(5x+1)$

$5x^2+21x+4+3x^2+15x+12=25x^2+30x+5$

$17x^2-6x-11=0$

$(17x+11)(x-1)=0$

Valid solution: $x=1$

$x=1$

Q10 SA-II 2016

00:00

Solve the quadratic equation: $x^2+\left(\frac{a+b}{a}+\frac{a}{a+b}\right)x+1=0$

Let $\alpha=\frac{a+b}{a}$ and $\beta=\frac{a}{a+b}$

Then equation becomes: $x^2+(\alpha+\beta)x+1=0$

Since $\alpha\beta=1$, roots are $-\alpha$ and $-\beta$

$x=-\frac{a+b}{a}$ or $x=-\frac{a}{a+b}$

Q11 2024

00:00

Let side of tile = $x$ units

Area of floor = $200x^2$

New area = $128(x+1)^2$

$200x^2=128(x+1)^2$

$25x^2=16(x^2+2x+1)$

$9x^2-32x-16=0$

$(9x+4)(x-4)=0$

$x=4$ (reject $x=-\frac49$)

Side of tile = 4 units

Q12 SA-I 2023

00:00

Solve the quadratic equation: $2x^2-2\sqrt2x+1=0$

$a=2, b=-2\sqrt2, c=1$

$D=b^2-4ac=8-8=0$

Roots are real and equal

$x=\frac{-b}{2a}=\frac{\sqrt2}{2}$

$x=\frac{\sqrt2}{2}$

Q13 SA-II 2016

00:00

Solve for $x$: $\sqrt{2x+9}+x=13$

$\sqrt{2x+9}=13-x$

Squaring both sides

$2x+9=(13-x)^2$

$x^2-28x+160=0$

$(x-8)(x-20)=0$

Valid solution: $x=8$

$x=8$

Q14 SA-II 2016

00:00

Solve for $x$: $\sqrt{6x+7}-(2x-7)=0$

$\sqrt{6x+7}=2x-7$

Squaring both sides

$6x+7=4x^2-28x+49$

$2x^2-17x+21=0$

$(2x-7)(x-3)=0$

Valid solution: $x=\frac72$

$x=\frac72$