Question

Given the equation 2square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

Answer 1

To solve the equation \( 2\sqrt{x} - 5 = 2 \), we will first isolate the square root term.

1. Add 5 to both sides of the equation:

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

\[
2\sqrt{x} = 7
\]

2. Now divide both sides by 2:

\[
\sqrt{x} = \frac{7}{2}
\]

3. Next, square both sides to eliminate the square root:

\[
x = \left(\frac{7}{2}\right)^2
\]

\[
x = \frac{49}{4}
\]

Now we have found a potential solution: \( x = \frac{49}{4} \).

**Checking if this is an extraneous solution:**

We need to substitute this value back into the original equation to verify that it satisfies the equation:

Original equation:

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

Substituting \( x = \frac{49}{4} \):

\[
2\sqrt{\frac{49}{4}} - 5 = 2
\]

Calculate \( \sqrt{\frac{49}{4}} \):

\[
\sqrt{\frac{49}{4}} = \frac{7}{2}
\]

Now substitute back into the equation:

\[
2 \cdot \frac{7}{2} - 5 = 2
\]

This simplifies to:

\[
7 - 5 = 2
\]

\[
2 = 2
\]

Since both sides of the equation are equal, the solution \( x = \frac{49}{4} \) is valid and is not an extraneous solution.

Thus, the solution is:

\[
\boxed{\frac{49}{4}}
\]