
Find the roots of the following quadratic equations by factorisation:
i.${x^2} - 3x - 10 = 0$
ii.$2{x^2} + x - 6 = 0$
iii.$\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0$
iv.$2{x^2} - x + \dfrac{1}{8} = 0$
v.$100{x^2} - 20x + 1 = 0$
Answer
534.3k+ views
Hint: If we have a quadratic equation $a{x^2} + bx + c = 0$ and we need to solve by factorisation method then we’ll try to break b into two parts namely $\alpha {\text{ and }}\beta $ such that, $\alpha + \beta = b$ and $\alpha \beta = ac$. Use this information to solve.
1. ${x^2} - 3x - 10 = 0$. We’ll break 3 into -5 and 2 so that $ - 5 + 2 = 3$ and $ - 5 \times 2 = 10$.
On solving we get,
$
{x^2} - 3x - 10 = 0 \\
\Rightarrow {x^2} - 5x + 2x - 10 = 0 \\
\Rightarrow x(x - 5) + 2(x - 5) = 0 \\
\Rightarrow (x - 5)(x + 2) = 0 \\
\Rightarrow x = 5, - 2 \\
$
2. $2{x^2} + x - 6 = 0$. We’ll break 1 into 4 and -3 so that $4 - 3 = 1$and $4 \times ( - 3) = - 12$.
On solving we get,
$
2{x^2} + x - 6 = 0 \\
\Rightarrow 2{x^2} + 4x - 3x - 6 = 0 \\
\Rightarrow 2x(x + 2) - 3(x + 2) = 0 \\
\Rightarrow (x + 2)(2x - 3) = 0 \\
\Rightarrow x = \dfrac{3}{2}, - 2 \\
$
3. $\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0$. We’ll break 7 into 2 and 5 so that $2 + 5 = 7$ and $2 \times 5 = 10 = \sqrt 2 \times 5\sqrt 2 $. On solving we get,
$
\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0 \\
\Rightarrow \sqrt 2 {x^2} + 2x + 5x + 5\sqrt 2 = 0 \\
\Rightarrow \sqrt 2 x(x + \sqrt 2 ) + 5(x + \sqrt 2 ) = 0 \\
\Rightarrow (x + \sqrt 2 )(\sqrt 2 x + 5) = 0 \\
\Rightarrow x = - \sqrt 2 ,\dfrac{{ - 5}}{{\sqrt 2 }} \\
$
4. $2{x^2} - x + \dfrac{1}{8} = 0$. It can also be written as $16{x^2} - 8x + 1 = 0$. We’ll break -8 into -4 and -4 so that $( - 4) + ( - 4) = - 8$ and $( - 4) \times ( - 4) = 16$. On solving we get,
$
16{x^2} - 8x + 1 = 0 \\
\Rightarrow 16{x^2} - 4x - 4x + 1 = 0 \\
\Rightarrow 16x\left( {x - \dfrac{1}{4}} \right) - 4\left( {x - \dfrac{1}{4}} \right) = 0 \\
\Rightarrow \left( {x - \dfrac{1}{4}} \right)(16x - 4) = 0 \\
\Rightarrow x = \dfrac{1}{4},\dfrac{1}{4} \\
$
5. $100{x^2} - 20x + 1 = 0$. We’ll break -20 into -10 and -10 to that \[( - 10) + ( - 10) = - 20\]also $( - 10) \times ( - 10) = 100$. On solving we get,
$
100{x^2} - 20x + 1 = 0 \\
\Rightarrow 100{x^2} - 10x - 10x + 1 = 0 \\
\Rightarrow 10x(10x - 1) - 1(10x - 1) = 0 \\
\Rightarrow (10x - 1)(10x - 1) = 0 \\
\Rightarrow x = \dfrac{1}{{10}},\dfrac{1}{{10}} \\
$
Note: Factorisation is nothing but the process by which we can solve not only quadratic but cubic equations as well.
1. ${x^2} - 3x - 10 = 0$. We’ll break 3 into -5 and 2 so that $ - 5 + 2 = 3$ and $ - 5 \times 2 = 10$.
On solving we get,
$
{x^2} - 3x - 10 = 0 \\
\Rightarrow {x^2} - 5x + 2x - 10 = 0 \\
\Rightarrow x(x - 5) + 2(x - 5) = 0 \\
\Rightarrow (x - 5)(x + 2) = 0 \\
\Rightarrow x = 5, - 2 \\
$
2. $2{x^2} + x - 6 = 0$. We’ll break 1 into 4 and -3 so that $4 - 3 = 1$and $4 \times ( - 3) = - 12$.
On solving we get,
$
2{x^2} + x - 6 = 0 \\
\Rightarrow 2{x^2} + 4x - 3x - 6 = 0 \\
\Rightarrow 2x(x + 2) - 3(x + 2) = 0 \\
\Rightarrow (x + 2)(2x - 3) = 0 \\
\Rightarrow x = \dfrac{3}{2}, - 2 \\
$
3. $\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0$. We’ll break 7 into 2 and 5 so that $2 + 5 = 7$ and $2 \times 5 = 10 = \sqrt 2 \times 5\sqrt 2 $. On solving we get,
$
\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0 \\
\Rightarrow \sqrt 2 {x^2} + 2x + 5x + 5\sqrt 2 = 0 \\
\Rightarrow \sqrt 2 x(x + \sqrt 2 ) + 5(x + \sqrt 2 ) = 0 \\
\Rightarrow (x + \sqrt 2 )(\sqrt 2 x + 5) = 0 \\
\Rightarrow x = - \sqrt 2 ,\dfrac{{ - 5}}{{\sqrt 2 }} \\
$
4. $2{x^2} - x + \dfrac{1}{8} = 0$. It can also be written as $16{x^2} - 8x + 1 = 0$. We’ll break -8 into -4 and -4 so that $( - 4) + ( - 4) = - 8$ and $( - 4) \times ( - 4) = 16$. On solving we get,
$
16{x^2} - 8x + 1 = 0 \\
\Rightarrow 16{x^2} - 4x - 4x + 1 = 0 \\
\Rightarrow 16x\left( {x - \dfrac{1}{4}} \right) - 4\left( {x - \dfrac{1}{4}} \right) = 0 \\
\Rightarrow \left( {x - \dfrac{1}{4}} \right)(16x - 4) = 0 \\
\Rightarrow x = \dfrac{1}{4},\dfrac{1}{4} \\
$
5. $100{x^2} - 20x + 1 = 0$. We’ll break -20 into -10 and -10 to that \[( - 10) + ( - 10) = - 20\]also $( - 10) \times ( - 10) = 100$. On solving we get,
$
100{x^2} - 20x + 1 = 0 \\
\Rightarrow 100{x^2} - 10x - 10x + 1 = 0 \\
\Rightarrow 10x(10x - 1) - 1(10x - 1) = 0 \\
\Rightarrow (10x - 1)(10x - 1) = 0 \\
\Rightarrow x = \dfrac{1}{{10}},\dfrac{1}{{10}} \\
$
Note: Factorisation is nothing but the process by which we can solve not only quadratic but cubic equations as well.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 10 biology CBSE

Dr BR Ambedkars fathers name was Ramaji Sakpal and class 10 social science CBSE

Discuss the main reasons for poverty in India

Write an application to the principal requesting five class 10 english CBSE

What is the past participle of wear Is it worn or class 10 english CBSE

What is the next number in the sequence 77493618 class 10 maths CBSE
