Question

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

Answer 1

To solve the equation

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

we start by isolating one of the square root terms. We can do this by adding 2 to both sides:

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

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

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

This simplifies to:

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

Combining like terms on the right side gives us:

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

Now we can simplify the equation:

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

Subtract 1 from both sides:

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

Dividing both sides by 4 results in:

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

Next, we square both sides again to eliminate the square root:

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

Adding 3 to both sides gives us:

\[
4 = 2x.
\]

Dividing by 2 results in:

\[
x = 2.
\]

Finally, it’s important to check if this solution satisfies the original equation. We substitute \(x = 2\) back into the original equation:

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

Calculating both sides:

\[
\sqrt{4 + 5} - 2 = \sqrt{4 - 3} \implies \sqrt{9} - 2 = \sqrt{1} \implies 3 - 2 = 1.
\]

Both sides equal 1, meaning our solution is verified.

Thus, the solution to the equation is

\[
\boxed{2}.
\]