Question

Solve the equation by completing square method.

$v^2=9v+2$

Short answer

The solutions are$v=\frac{9}{2}\pm \frac{\sqrt{89}}{2}$.

Step-by-step solution

Given Information   

The Given equation is$v^2=9v+2$.

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

$$\begin{gathered} v^2=9v+2 \\ {v}^2-9v=2 \end{gathered}$$

Here, $b=-9$.

So, ${\left(\frac{1}{2}b\right)}^2={\left(\frac{1}{2}(-9)\right)}^2={\left(\frac{-9}{2}\right)}^2=\frac{81}{4}$.

Add $\frac{81}{4}$to both sides of equation.

$$\begin{gathered} v^2-9v+\frac{81}{4}=2+\frac{81}{4} \\ {v}^2-9v+\frac{81}{4}=\frac{89}{4} \end{gathered}$$

Factor the perfect square trinomial as a binomial square.     

${\left(v-\frac{9}{2}\right)}^2=\frac{89}{4}$

Use the square root property.

$$\begin{gathered} v-\frac{9}{2}=\sqrt{\frac{89}{4}} \\ v-\frac{9}{2}=\pm \sqrt{\frac{89}{2\times 2}} \\ v-\frac{9}{2}=\pm \frac{\sqrt{89}}{2} \end{gathered}$$

Solve for v.

$v=\frac{9}{2}\pm \frac{\sqrt{89}}{2}$

Rewrite to show two solutions.

$$\begin{gathered} v=\frac{9}{2}+\frac{\sqrt{89}}{2}, \\ v=\frac{9}{2}-\frac{\sqrt{89}}{2} \end{gathered}$$