Question

what is the solution of \\(\sqrt{x + 12} = x\\)?
\\(x = -3\\)
\\(x = 4\\)
\\(x = -3\\) or \\(x = 4\\)
no solution

Explanation:

Step1: Square both sides to eliminate radical

To solve the equation \(\sqrt{x + 12}=x\), we first square both sides of the equation. Squaring the left side \(\sqrt{x + 12}\) gives \(x + 12\), and squaring the right side \(x\) gives \(x^{2}\). So we have the equation:
\[x + 12=x^{2}\]

Step2: Rearrange into standard quadratic form

Rearrange the equation \(x + 12=x^{2}\) to the standard quadratic form \(ax^{2}+bx + c = 0\). Subtract \(x\) and \(12\) from both sides:
\[x^{2}-x - 12=0\]

Step3: Factor the quadratic equation

Factor the quadratic equation \(x^{2}-x - 12 = 0\). We need two numbers that multiply to \(- 12\) and add up to \(-1\). The numbers are \(-4\) and \(3\). So the factored form is:
\[(x - 4)(x+ 3)=0\]

Step4: Solve for x

Set each factor equal to zero and solve for \(x\):

• For \(x - 4=0\), we get \(x = 4\).
• For \(x + 3=0\), we get \(x=-3\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• For \(x = 4\): Substitute \(x = 4\) into \(\sqrt{x + 12}\), we have \(\sqrt{4+12}=\sqrt{16} = 4\), which is equal to \(x = 4\). So \(x = 4\) is a valid solution.
• For \(x=-3\): Substitute \(x=-3\) into \(\sqrt{x + 12}\), we have \(\sqrt{-3 + 12}=\sqrt{9}=3\), but \(3

eq-3\). So \(x=-3\) is an extraneous solution.

Answer:

\(x = 4\) (corresponding to the option "x = 4")