Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
radical equations and extraneous roots
what is the solution of \\(\sqrt{2x + 4} = 16\\)?
\\(x = 6\\)
\\(x = 72\\)
\\(x = 126\\)
no solution

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{2x + 4}=16\), we first square both sides of the equation. Squaring the left side will eliminate the square root, and squaring the right side will give us a new equation to solve for \(x\).
\[
(\sqrt{2x + 4})^2 = 16^2
\]
Simplifying both sides, we get:
\[
2x + 4 = 256
\]

Step2: Subtract 4 from both sides

Next, we want to isolate the term with \(x\). So we subtract 4 from both sides of the equation.
\[
2x + 4 - 4 = 256 - 4
\]
Simplifying both sides, we have:
\[
2x = 252
\]

Step3: Divide both sides by 2

Finally, to solve for \(x\), we divide both sides of the equation by 2.
\[
\frac{2x}{2}=\frac{252}{2}
\]
Simplifying both sides, we find:
\[
x = 126
\]

Answer:

\(x = 126\)