Question

Solve the equation by completing square method.

$r^2-4r=12$

Short answer

The solutions are$r=-2,6$.

Step-by-step solution

Given Information   

The Given equation is $r^2-4r=12$

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

The variable terms are on the left side.

$r^2-4r=12$

Here, $b=-4$

So,

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

Add $4$both sides of equation.

$$\begin{gathered} r^2-4r+4=12+4 \\ {r}^2-4r+4=16 \end{gathered}$$

Factor the perfect square trinomial as a binomial square.    

$(r-2{)}^2=16$

Use the square root property.

$$\begin{gathered} r-2=\pm \sqrt{16} \\ r-2=\pm 4 \\ r=2\pm 4 \end{gathered}$$

Rewrite to show two solutions.

$$\begin{gathered} r=2+4=6 \\ r=2-4=-2 \end{gathered}$$

Thus,$r=-2,6$.

Check the solutions.

Substitute $r=-2$

$$\begin{gathered} r^2-4r=12 \\ (-2{)}^2-4(-2)\stackrel{?}{=}12 \\ 4+8\stackrel{?}{=}12 \\ 12=12True \end{gathered}$$

Substitute $r=6$

$$\begin{gathered} r^2-4r=12 \\ (6{)}^2-4(6)\stackrel{?}{=}12 \\ 36-24\stackrel{?}{=}12 \\ 12=12True \end{gathered}$$

Thus, the solutions are$r=-2,r=6$.