Question

Solve the equation by completing square method.

$c^2+4c=5$

Short answer

The solutions are$c=1,c=-5$.

Step-by-step solution

Given Information     

The Given equation is$c^2+4c=5$.

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

The variable terms are on the left side.

$c^2+4c=5$

Here, $b=4$.

So,

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

Add $4$to both sides of equation.

$$\begin{gathered} c^2+4c+4=5+4 \\ {c}^2+4c+4=9 \end{gathered}$$

Factor the perfect square trinomial as a binomial square.   

$(c+2{)}^2=9$

Use the square root property.

$$\begin{gathered} c+2=\pm \sqrt{9} \\ c+2=\pm 3 \\ c=-2\pm 3 \end{gathered}$$

Rewrite to show two solutions.

$$\begin{gathered} c=-2+3=1 \\ c=-2-3=-5 \end{gathered}$$

Thus, the solutions are$c=1,c=-5$.