Question

5 numeric 1 point solve for the extraneous solution for \\(\sqrt{x + 2} = x\\)

Explanation:

Step1: Square both sides to eliminate the square root

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

Step2: Rearrange the quadratic equation

Rearrange the equation \(x + 2=x^2\) into standard quadratic form \(ax^2+bx + c = 0\). Subtract \(x\) and \(2\) from both sides:
\(x^2-x - 2=0\)

Step3: Factor the quadratic equation

Factor the quadratic \(x^2-x - 2\). We need two numbers that multiply to \(-2\) and add to \(-1\). Those numbers are \(-2\) and \(1\). So the factored form is:
\((x - 2)(x + 1)=0\)

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

Using the zero - product property, if \((x - 2)(x + 1)=0\), then either \(x - 2=0\) or \(x + 1=0\). Solving \(x - 2=0\) gives \(x = 2\), and solving \(x + 1=0\) gives \(x=-1\).

Step5: Check for extraneous solutions

Now we check these solutions in the original equation \(\sqrt{x + 2}=x\).

• For \(x = 2\): Substitute \(x = 2\) into the left side, \(\sqrt{2 + 2}=\sqrt{4}=2\), and the right side is \(2\). So \(x = 2\) is a valid solution.
• For \(x=-1\): Substitute \(x=-1\) into the left side, \(\sqrt{-1 + 2}=\sqrt{1}=1\), and the right side is \(-1\). Since \(1

eq - 1\), \(x=-1\) does not satisfy the original equation. So \(x=-1\) is an extraneous solution.

Answer:

\(-1\)