Question

$\sqrt{2x + 2} - \sqrt{x + 4} = 0$

Explanation:

Step1: Isolate one square root

We start with the equation \(\sqrt{2x + 2}-\sqrt{x + 4}=0\). Add \(\sqrt{x + 4}\) to both sides of the equation to isolate \(\sqrt{2x+2}\).
\(\sqrt{2x + 2}=\sqrt{x + 4}\)

Step2: Square both sides

To eliminate the square roots, we square both sides of the equation. Recall that \((\sqrt{a})^2=a\) for \(a\geq0\). So, squaring both sides gives:
\((\sqrt{2x + 2})^2=(\sqrt{x + 4})^2\)
\(2x+2=x + 4\)

Step3: Solve for \(x\)

Subtract \(x\) from both sides:
\(2x - x+2=x - x + 4\)
\(x+2 = 4\)
Then subtract 2 from both sides:
\(x+2-2=4 - 2\)
\(x=2\)

Step4: Check the solution

We need to check if \(x = 2\) satisfies the original equation because squaring both sides can introduce extraneous solutions.
For the left - hand side of the original equation \(\sqrt{2x + 2}-\sqrt{x + 4}\), substitute \(x = 2\):
\(\sqrt{2\times2+2}-\sqrt{2 + 4}=\sqrt{4 + 2}-\sqrt{6}=\sqrt{6}-\sqrt{6}=0\)
Which is equal to the right - hand side (0). So \(x = 2\) is a valid solution.

Answer:

\(x = 2\)