Question

which of the following is an extraneous solution of \\(\sqrt{-3x - 2} = x + 2\\)?
\\(\bigcirc\\ x = -6\\)
\\(\bigcirc\\ x = -1\\)
\\(\bigcirc\\ x = 1\\)
\\(\bigcirc\\ x = 6\\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{-3x - 2}=x + 2\), we first square both sides. Squaring the left side gives \((\sqrt{-3x - 2})^2=-3x - 2\), and squaring the right side gives \((x + 2)^2=x^2+4x + 4\). So the equation becomes:

$$ -3x - 2=x^2+4x + 4 $$

Step2: Rearrange into standard quadratic form

Move all terms to one side to form a quadratic equation. Add \(3x\) and \(2\) to both sides:

$$ 0=x^2+4x + 4+3x + 2 $$

Simplify the right side:

$$ x^2+7x + 6 = 0 $$

Step3: Factor the quadratic equation

Factor the quadratic \(x^2+7x + 6\). We need two numbers that multiply to \(6\) and add to \(7\). Those numbers are \(1\) and \(6\), so:

$$ (x + 1)(x + 6)=0 $$

Step4: Solve for \(x\)

Set each factor equal to zero:

• For \(x + 1 = 0\), we get \(x=-1\).
• For \(x + 6 = 0\), we get \(x=-6\).

Step5: Check for extraneous solutions

Now we need to check these solutions in the original equation \(\sqrt{-3x - 2}=x + 2\) because squaring both sides can introduce extraneous solutions.

Check \(x=-1\):

Left side: \(\sqrt{-3(-1)-2}=\sqrt{3 - 2}=\sqrt{1}=1\)
Right side: \(-1 + 2 = 1\)
Since left side = right side, \(x = - 1\) is a valid solution.

Check \(x=-6\):

Left side: \(\sqrt{-3(-6)-2}=\sqrt{18 - 2}=\sqrt{16}=4\)
Right side: \(-6 + 2=-4\)
Since \(4
eq - 4\), \(x=-6\) does not satisfy the original equation, so it is an extraneous solution.

Answer:

\(x = - 6\) (corresponding to the option "x = -6")