Question

Solve the following quadratic equation by completing the squares :

$x^2-x-6=0$.

Answer 1

Step 1: Convert the given equation into perfect square

We have, $x^2-x-6=0$.

Comparing the given equation with standard form $a{x}^2+bx+c=0$.

Here,

$$\begin{gathered} a=1, \\ b=-1, \\ c=-6. \end{gathered}$$

Now, add ${\left(\frac{b}{2a}\right)}^2$ to both sides of the given equation, to make it a perfect square,

Here,

${\left(\frac{b}{2a}\right)}^2={\left(\frac{{\displaystyle -1}}{2\times 1}\right)}^2=\frac{1}{4}$

We get,

$$\begin{gathered} x^2-x=6 \\ \Rightarrow x^2-x+\frac{1}{4}=6+\frac{1}{4} \\ \Rightarrow {\left(x-\frac{1}{2}\right)}^2=\frac{24+1}{4} \\ \Rightarrow {\left(x-\frac{1}{2}\right)}^2=\frac{25}{4} \end{gathered}$$

Step 2: Solve equation for variable $x$ to find the roots

Solve for variable $x$,

$$\begin{gathered} \Rightarrow \sqrt{{\left(x-\frac{1}{2}\right)}^2}=\sqrt{\frac{25}{4}} \\ \Rightarrow x-\frac{1}{2}=\pm \frac{5}{2} \\ \Rightarrow x=\frac{1}{2}+\frac{5}{2},\frac{1}{2}-\frac{5}{2} \\ \Rightarrow x=\frac{6}{2},-\frac{4}{2} \end{gathered}$$

After simplifying,

$x=3,-2$

Therefore, the roots of the given equation are $3$ and $-2$.