Question

solve the radical equation. check for extraneous solutions.
$x = \sqrt{x + 6}$

select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. $x = \square$
(type your answer(s) as integers or simplified fractions. use a comma to separate answers as

b. there are no solutions.

Explanation:

Step1: Square both sides to eliminate the square root.

To solve the equation \( x = \sqrt{x + 6} \), we start by squaring both sides. Squaring the left side gives \( x^2 \), and squaring the right side gives \( (\sqrt{x + 6})^2 = x + 6 \). So we get the quadratic equation:
\( x^2 = x + 6 \)

Step2: Rearrange into standard quadratic form.

Subtract \( x \) and \( 6 \) from both sides to set the equation to zero:
\( x^2 - x - 6 = 0 \)

Step3: Factor the quadratic equation.

We need to find two numbers that multiply to \( -6 \) and add to \( -1 \). The numbers \( -3 \) and \( 2 \) work. So we can factor the quadratic as:
\( (x - 3)(x + 2) = 0 \)

Step4: Solve for \( x \) using the zero - product property.

If \( (x - 3)(x + 2)=0 \), then either \( x - 3 = 0 \) or \( x + 2 = 0 \). Solving \( x - 3 = 0 \) gives \( x = 3 \), and solving \( x + 2 = 0 \) gives \( x=-2 \).

Step5: Check for extraneous solutions.

• For \( x = 3 \): Substitute into the original equation \( x=\sqrt{x + 6} \). The left side is \( 3 \), and the right side is \( \sqrt{3+6}=\sqrt{9} = 3 \). So \( x = 3 \) is a valid solution.
• For \( x=-2 \): Substitute into the original equation. The left side is \( -2 \), and the right side is \( \sqrt{-2 + 6}=\sqrt{4}=2 \). Since \( -2

eq2 \), \( x=-2 \) is an extraneous solution.

Answer:

\( x = 3 \)