# How do you solve the quadratic equation by completing the square: v^2 - 2v = 3?

##### 1 Answer
Jul 28, 2015

${v}_{1} = 3$, ${v}_{2} = - 1$.

#### Explanation:

Your starting quadratic equation looks like this

${v}^{2} \textcolor{b l u e}{- 2} v = 3$

Now, to solve quadratic equations by completing the square you need to add a term to both sides of the equation such that the left side of the equation becomes the aquare of a binomial.

To do that, divide the coefficient of the $x$-term by 2 while keeping the sign, square it, and add the result to both sides of the equation.

In your case, you have

$\frac{\textcolor{b l u e}{- 2}}{2} = - 1$, then

${\left(- 1\right)}^{2} = + 1$

The quadratic becomes

${v}^{2} - 2 v + 1 = 3 + 1$

${v}^{2} - 2 v + 1 = 4$

The left side of the equation is equivalent to

${v}^{2} - 2 v + 1 = {\left(v - 1\right)}^{2}$

You thus have

${\left(v - 1\right)}^{2} = 4$

To solve this equation, take the square root from both sides of the equation

$\sqrt{{\left(v - 1\right)}^{2}} = \sqrt{4}$

$v - 1 = \pm 2 \implies {v}_{1 , 2} = \pm 2 + 1 = \left\{\begin{matrix}{v}_{1} = + 2 + 1 = 3 \\ {v}_{2} = - 2 + 1 = - 1\end{matrix}\right.$

The two solutions to your equation are

${v}_{1} = \textcolor{g r e e n}{3}$ and ${v}_{2} = \textcolor{g r e e n}{- 1}$