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

Question

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 \( x = 1 \).