Question

Solve the quadratic equation using square root property.

$4z^2+4z+1=49$

Short answer

The solutions are$z=-4,3$.

Step-by-step solution

Given Information.   

The Given equation is $4z^2+4z+1=49$.

Factor the trinomial.    

$$\begin{gathered} (2z{)}^2+2(2z)+(1{)}^2=49 \\ (2z+1{)}^2=49 \end{gathered}$$

Use the square root property.

$$\begin{gathered} 2z+1=\pm \sqrt{49} \\ 2z+1=\pm 7 \end{gathered}$$

Isolate the term.  

$$\begin{gathered} 2z=-1\pm 7 \\ z=\frac{-1\pm 7}{2} \end{gathered}$$

Rewrite to show two solutions.

$$\begin{gathered} z=\frac{-1+7}{2}=\frac{6}{2}=3 \\ z=\frac{-1-7}{2}=\frac{-8}{2}=-4 \end{gathered}$$

The solutions are$z=-4,3$.