Question

Solve the radical equation 2x+5−−−−−√−2=2x−3−−−−−√
.(1

Answer 1

To solve the equation \( \sqrt{2x+5} - 2 = \sqrt{2x - 3} \), let's isolate one of the square roots. We can start by adding 2 to both sides:

\[
\sqrt{2x + 5} = \sqrt{2x - 3} + 2
\]

Next, we square both sides of the equation to eliminate the square roots:

\[
(\sqrt{2x + 5})^2 = (\sqrt{2x - 3} + 2)^2
\]

This simplifies to:

\[
2x + 5 = (2x - 3) + 4\sqrt{2x - 3} + 4
\]

Combine like terms:

\[
2x + 5 = 2x + 1 + 4\sqrt{2x - 3}
\]

Subtract \(2x + 1\) from both sides:

\[
5 - 1 = 4\sqrt{2x - 3}
\]

This simplifies to:

\[
4 = 4\sqrt{2x - 3}
\]

Dividing both sides by 4 gives:

\[
1 = \sqrt{2x - 3}
\]

Now, we square both sides again to solve for \(x\):

\[
1^2 = (2x - 3)
\]

This results in:

\[
1 = 2x - 3
\]

Adding 3 to both sides gives:

\[
4 = 2x
\]

Dividing by 2 results in:

\[
x = 2
\]

Lastly, we should check this solution in the original equation to ensure it's valid.

Substituting \(x = 2\):

\[
\sqrt{2(2) + 5} - 2 = \sqrt{2(2) - 3}
\]
\[
\sqrt{4 + 5} - 2 = \sqrt{4 - 3}
\]
\[
\sqrt{9} - 2 = \sqrt{1}
\]
\[
3 - 2 = 1
\]

This is true, so the solution is valid.

Thus, the solution to the equation is:

\[
\boxed{2}
\]