Question

Solve the equation by completing square method.$a^2=3a+8$

Short answer

The solutions are$a=\frac{-\sqrt{41}+3}{2},\frac{\sqrt{41}+3}{2}$

Step-by-step solution

Given Information.   

The Given equation is $a^2=3a+8$

Isolate the variable terms on one side and the constant terms on the other.The variable terms are on the left side.

$a^2-3a=8$

The coefficient of $a$is $-3$

So ,$b=-3$

Find 12b2and add to the both sides of equation.

$$\begin{gathered} ={\left(\frac{1}{2}\left(-3\right)\right)}^2 \\ ={\left(\frac{-3}{2}\right)}^2 \\ =\frac{9}{4} \end{gathered}$$

Add$\frac{9}{4}$on both side of equation

$$\begin{gathered} a^2-3a+\frac{9}{4}=8+\frac{9}{4} \\ {a}^2-3a+\frac{9}{4}=\frac{41}{4} \end{gathered}$$

Factor the perfect square trinomial as a binomial square  

${\left(a-\frac{3}{2}\right)}^2=\frac{41}{4}$

Use square root property

$$\begin{gathered} \left(a-\frac{3}{2}\right)=\pm \sqrt{\frac{41}{4}} \\ a=\pm \frac{\sqrt{41}}{2}+\frac{3}{2} \\ a=\frac{-\sqrt{41}+3}{2},\frac{\sqrt{41}+3}{2} \end{gathered}$$