# 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.$100{x^2} - 20x + 1 = 0$

Last updated date: 21st Mar 2023

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Answer

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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.

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