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# How do you solve x^2-x-12=0?

May 19, 2015

We can Split the Middle Term of this equation and factorise it to find the solution.

In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of two numbers such that:

${N}_{1} \times {N}_{2} = a \times c = 1 \times \left(- 12\right) = - 12$

and

${N}_{1} + {N}_{2} = b = - 1$

After trying out a few numbers we get ${N}_{1} = - 4$ and ${N}_{2} = 3$
$\left(- 4\right) \times 3 = - 12$, and $\left(- 4\right) + 3 = - 1$

${x}^{2} - x - 12 = {x}^{2} - 4 x + 3 x - 12$
$= x \left(x - 4\right) + 3 \left(x - 4\right)$
$= \left(x + 3\right) \left(x - 4\right)$

the solution is $\textcolor{b l u e}{x} = - 3 , \textcolor{b l u e}{x} = 4$

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