Question

Solve the equation by completing square method.

$z^2+12z=-11$

Short answer

The solutions are $z=-1,z=-11$.

Step-by-step solution

Given Information.   

The Given equation is$z^2+12z=-11$

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

The variable terms are on the left side.

$z^2+12z=-11$

The coefficient of $zis12$.

So,$b=12$

Find 12b2and add to the both sides of equation.

$$\begin{gathered} ={\left(\frac{1}{2}\left(12\right)\right)}^2 \\ ={\left(6\right)}^2=36 \end{gathered}$$

Add $36$both sides.

$$\begin{gathered} z^2+12z+36=-11+36 \\ {z}^2+12z+36=25 \end{gathered}$$

Factor the perfect square trinomial as a binomial square. 

$(z+6{)}^2=25$

Use square root property.

$$\begin{gathered} z+6=\pm \sqrt{25} \\ z+6=\pm 5 \end{gathered}$$

$z=-6\pm 5$

Rewrite to show two solutions.

$$\begin{gathered} z=-6+5=-1 \\ z=-6-5=-11 \end{gathered}$$

Thus, the solutions are$z=-1,-11$.

Check the solutions. 

Substitute $z=-1$

$$\begin{gathered} z^2+12z=-11 \\ (-1{)}^2+12(-1)\stackrel{?}{=}-11 \\ 1-12\stackrel{?}{=}-11 \\ -11=-11True \end{gathered}$$

Substitute $z=-11$

$$\begin{gathered} z^2+12z=-11 \\ (-11{)}^2+12(-11)\stackrel{?}{=}-11 \\ 121-132\stackrel{?}{=}-11 \\ -11=-11True \end{gathered}$$