Question

In the following exercises, solve using the Square Root Property.

$4b^2-28b+49=25$

Short answer

The required answer is$b=6\mathrm{and}b=1$.

Step-by-step solution

Step 1. Given information.

The given quadratic equation is:

$4b^2-28b+49=25$

Step 2. Solve using the quadratic equation by using Square Root Property. 

$4b^2-28b+49=25$

The left side of the equation is a perfect square trinomial. We will factor it first.

Factor the perfect square trinomial,

${\left(2b-7\right)}^2=25$

Use Square Root Property and Simplify the radical,

$$\begin{gathered} 2b-7=\pm \sqrt{25} \\ 2b-7=\pm 5 \end{gathered}$$

Rewrite to show two equations and solve for the value of b,

$$\begin{gathered} 2b-7=5,2b-7=-5 \\ 2b=12,2b=2 \\ b=6\mathrm{and}b=1 \end{gathered}$$