Question

Solve for x and show your steps. Is the solution extraneous? Check your work to show how you determined if the solution is extraneous or not.

The square root of the quantity 3 x plus 12 end quantity equals 9.

Answer 1

To solve the equation

\[
\sqrt{3x + 12} = 9,
\]

we start by squaring both sides to eliminate the square root:

\[
(\sqrt{3x + 12})^2 = 9^2.
\]

This simplifies to:

\[
3x + 12 = 81.
\]

Next, we solve for \(x\) by isolating it. First, subtract 12 from both sides:

\[
3x = 81 - 12,
\]

which simplifies to:

\[
3x = 69.
\]

Now, divide both sides by 3:

\[
x = \frac{69}{3} = 23.
\]

Now, let's check if this solution is extraneous by substituting \(x = 23\) back into the original equation:

Calculate \(3x + 12\) when \(x = 23\):

\[
3(23) + 12 = 69 + 12 = 81.
\]

Now, take the square root:

\[
\sqrt{81} = 9.
\]

Since the left side equals the right side of the original equation, we confirm:

\[
\sqrt{3x + 12} = 9.
\]

Thus, the solution \(x = 23\) is valid and not extraneous.

So, the final solution is:

\[
\boxed{23}.
\]