Question

Factorize the following quadratic equation to find the roots:
\[a){x^2} - 3x - 10 = 0\]
\[b)2{x^2} + x - 6 = 0\]
\[c)\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0\]
\[d)2{x^2} - x + \dfrac{1}{8} = 0\]
\[e)100{x^2} - 20x + 1 = 0\]

Answer 1

Hint: We must break the middle term in order to solve this using the factorization method. Divide the middle term such that it can be the product of the first and last term. For example, in the first equation, -3x is split as a sum of -5x and +2x which are factors of the first and last term. i.e., 10 and 1.

Complete step-by-step solution:
\[a){x^2} - 3x - 10 = 0\]
We will find the factors of 10 and 1, then splits middle term in terms of those factors
\[ \Rightarrow {x^2} - 5x + 2x - 10 = 0\]
We take x and 2 common
\[ \Rightarrow x(x - 5) + 2(x - 5) = 0\]
\[ \Rightarrow (x - 5)(x + 2) = 0\]
We have product of them is zero
So, one or both terms should be zero
\[ \Rightarrow (x - 5) = 0\] and \[(x + 2) = 0\]
\[ \Rightarrow x = 5\] and \[ \Rightarrow x = - 2\]
The roots of \[{x^2} - 3x - 10 = 0\] are 5 and -2.
\[b)2{x^2} + x - 6 = 0\]
We will find the factors of -6 and 2, then splits middle term in terms of those factors
$ \Rightarrow 2{x^2} + 4x - 3x - 6 = 0$
We take x and 2 common
$ \Rightarrow 2x(x + 2) - 3(x + 2) = 0$
We have product of them is zero
So, one or both terms should be zero
$ \Rightarrow (2x - 3)(x + 2) = 0$
$ \Rightarrow (2x - 3) = 0$ and $(x + 2) = 0$
$ \Rightarrow x = \dfrac{3}{2}$ and $x = - 2$
 \[c)\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0\]
We will find the factors of 10 (which is product of first and last term), then splits middle term in terms of those factor
\[ \Rightarrow \sqrt 2 {x^2} + 2x + 5x + 5\sqrt 2 = 0\]
\[ \Rightarrow \sqrt 2 x(x + \sqrt 2 ) + 5(x + \sqrt 2 ) = 0\]
\[ \Rightarrow (\sqrt 2 x + 5)(x + \sqrt 2 ) = 0\]
We have product of them is zero
So, one or both terms should be zero
\[ \Rightarrow x = \dfrac{{ - 5}}{{\sqrt 2 }}\] and \[x = - \sqrt 2 \]
\[d)2{x^2} - x + \dfrac{1}{8} = 0\]
We first multiply the whole equation by 8,
\[ \Rightarrow 16{x^2} - 8x + 1 = 0\]
We will find the factors of 16 and 1, then splits middle term in terms of those factor
\[ \Rightarrow 16{x^2} - 4x - 4x + 1 = 0\]
\[ \Rightarrow 4x(4x - 1) - 1(4x - 1) = 0\]
\[ \Rightarrow (4x - 1)(4x - 1) = 0\]
We have product of them is zero
So, one or both terms should be zero
The value of x is same in both terms
\[ \Rightarrow x = \dfrac{1}{4}\]
\[e)100{x^2} - 20x + 1 = 0\]
We will find the factors of 100 and 1, then splits middle term in terms of those factor
\[ \Rightarrow 100{x^2} - 10x - 10x + 1 = 0\]
\[ \Rightarrow 10x(10x - 1) - 1(10x - 1) = 0\]
\[ \Rightarrow (10x - 1)(10x - 1) = 0\]
We have product of them is zero
So, one or both terms should be zero
The value of x is same in both terms
\[ \Rightarrow x = \dfrac{1}{{10}}\]

Note: We can take care about the factors, sometimes we get middle terms split only by finding the factors of first and last term but sometimes we have to produce the first and last terms then find their factors as we did in the third example. We should also try to simplify our equation as we did in the fourth example.