Question

Solve the equation by completing square method.

$a^2-10a=-5$

Short answer

The solutions are$a=5+2\sqrt{5},a=5-2\sqrt{5}$.

Step-by-step solution

Given Information   

The Given equation is$a^2-10a=-5$.

Isolate the variable terms on one side and the constant terms on the other. 

The variable terms are on the left side.

$a^2-10a=-5$

Here,$b=-10$

Find 12b2

$$\begin{gathered} ={\left(\frac{1}{2}(-10)\right)}^2 \\ ={\left(-5\right)}^2 \\ =25 \end{gathered}$$

Add $25$to both sides of equation.

$$\begin{gathered} a^2-10+25=-5+25 \\ {a}^2-10a+25=20 \end{gathered}$$

Factor the perfect square trinomial as a binomial square.   

$(a-5{)}^2=20$

Use the square root property.

$$\begin{gathered} a-5=\pm \sqrt{20} \\ a-5=\pm \sqrt{2\times 2\times 5} \\ a-5=\pm 2\sqrt{5} \end{gathered}$$

Solve for  a.

$a=5\pm 2\sqrt{5}$

Rewrite to show two solutions.

$$\begin{gathered} a=5+2\sqrt{5}, \\ a=5-2\sqrt{5} \end{gathered}$$