 QUESTION

# Find the roots of the following quadratic equation by factorisation:${{x}^{2}}+x-6=0$

Hint: To solve this problem, we will rearrange the above expression ${{x}^{2}}+x-6$ which would help us to factorise this expression. In particular, we will modify the coefficient of x, which will help us to factorise the equation. It would then help us to get the required values of x (which are the roots).

Complete step-by-step solution -
To solve this problem, we try to understand a few basics of solving this problem. We should know that to factorise this problem, we need 4 terms. Thus, we try to re-group these 4 terms in groups of 2 each. Thus, to do this we rewrite the expression as ${{x}^{2}}+(b+a)x-6$ (where, b + a should be 1 so that the expression should not be altered). Thus, re-grouping, we have,
$={{x}^{2}}+bx+ax-6$
$=x(x+b)+a\left( x-\dfrac{6}{a} \right)$ -- (A)
Now, we want (x+b) and $\left( x-\dfrac{6}{a} \right)$ to be the same. This is so that if these terms are the same, we can re-group and thus factorise. Thus, equating, we have,
$b=-\dfrac{6}{a}$
ab = -6 -- (1)
Also, b + a = 1 -- (2)
Since, (1) is a nonlinear equation, we start by listing the divisors of 6 (1, 2, 3, 6 or -1, -2, -3, -6). We need to basically ensure that sum of b and a to be 1. Thus, for that to happen, we only have 2 possibilities (that is b = -2 and a = 3). Coincidentally, this pair also satisfies ab = -6. Thus, substituting in (A), we get,
\begin{align} & =x(x-2)+a\left( x-\dfrac{6}{3} \right) \\ & =x(x-2)+3\left( x-2 \right) \\ \end{align}
= (x+3) (x-2)
Now, we have to solve for ${{x}^{2}}+x-6=0$, thus, we have,
(x+3) (x-2) = 0
Thus, the values of x are –
x + 3 = 0
x = -3 -- (3)
Also, x – 2 = 0, thus, we have,
x = 2 -- (4)
Hence, the roots are 2 and -3.

Note: Another way to find the roots of the equation $a{{x}^{2}}+bx+c=0$ is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. From this formula, we would get two values of x. One can use this and verify that the value of x we get would be the same as that in the solution.