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.

1 answer

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:

\[ \boxed{23}. \]