Certainly! Let's solve the equation step by step.
We start with:
[tex]\[
\begin{aligned}
\sqrt{(b-5)^2} &= \sqrt{25}
\end{aligned}
\][/tex]
Firstly, we know that [tex]\(\sqrt{x^2} = |x|\)[/tex], where [tex]\(|x|\)[/tex] denotes the absolute value of [tex]\(x\)[/tex]. The absolute value function accounts for both the positive and negative values. Therefore:
[tex]\[
\sqrt{(b-5)^2} = |b-5|
\][/tex]
So, our equation now becomes:
[tex]\[
|b-5| = \sqrt{25}
\][/tex]
Next, we simplify [tex]\(\sqrt{25}\)[/tex]:
[tex]\[
\sqrt{25} = 5
\][/tex]
Thus, our equation is updated to:
[tex]\[
|b-5| = 5
\][/tex]
The absolute value equation [tex]\(|x| = a\)[/tex] implies two separate equations: [tex]\(x = a\)[/tex] or [tex]\(x = -a\)[/tex]. Therefore, we solve for [tex]\(b\)[/tex] in the following two cases:
1. [tex]\(b - 5 = 5\)[/tex]
2. [tex]\(b - 5 = -5\)[/tex]
Case 1: [tex]\(b - 5 = 5\)[/tex]
Add 5 to both sides:
[tex]\[
b - 5 + 5 = 5 + 5
\][/tex]
[tex]\[
b = 10
\][/tex]
Case 2: [tex]\(b - 5 = -5\)[/tex]
Add 5 to both sides:
[tex]\[
b - 5 + 5 = -5 + 5
\][/tex]
[tex]\[
b = 0
\][/tex]
Thus, the solutions to the equation are:
[tex]\[
b = 10 \quad \text{or} \quad b = 0
\][/tex]
So, the final answers are:
[tex]\[
\boxed{b = 10 \text{ or } b = 0}
\][/tex]
