Question

Directions - Solve the radical equation. Check for extraneous solutions.




3
x
+
12
=
x
+
8
3x+12
​
=
x+8
​

Answer 1

To solve the equation \( 3\sqrt{x} + 12 = x + 8 \), we first isolate the square root term.

1. Subtract 12 from both sides:

\[
3\sqrt{x} = x + 8 - 12
\]

This simplifies to:

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

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

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

3. Now we square both sides to remove the square root:

\[
x = \left(\frac{x - 4}{3}\right)^2
\]

4. Expanding the right side:

\[
x = \frac{(x - 4)(x - 4)}{9} = \frac{x^2 - 8x + 16}{9}
\]

5. To eliminate the fraction, multiply both sides by 9:

\[
9x = x^2 - 8x + 16
\]

6. Rearranging the equation gives:

\[
x^2 - 8x - 9x + 16 = 0
\]

This simplifies to:

\[
x^2 - 17x + 16 = 0
\]

7. Next, we can factor this quadratic equation:

\[
(x - 1)(x - 16) = 0
\]

8. Setting each factor equal to zero gives us potential solutions:

\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

\[
x - 16 = 0 \quad \Rightarrow \quad x = 16
\]

Now we have the potential solutions \( x = 1 \) and \( x = 16 \). We need to check these for extraneous solutions by substituting them back into the original equation.

### Checking \( x = 1 \):

Substituting into the original equation:

\[
3\sqrt{1} + 12 = 1 + 8
\]

This simplifies to:

\[
3(1) + 12 = 1 + 8 \quad \Rightarrow \quad 3 + 12 = 9
\]

This is not true, so **\( x = 1 \) is an extraneous solution**.

### Checking \( x = 16 \):

Substituting into the original equation:

\[
3\sqrt{16} + 12 = 16 + 8
\]

This simplifies to:

\[
3(4) + 12 = 16 + 8 \quad \Rightarrow \quad 12 + 12 = 24
\]

This simplifies to:

\[
24 = 24
\]

This is true, so **\( x = 16 \) is a valid solution**.

### Conclusion:

The only solution to the equation \( 3\sqrt{x} + 12 = x + 8 \) is

\[
\boxed{16}
\]

No extraneous solutions are present other than \( x = 1 \).