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:

\(x = 126\)