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 the square root

To solve the equation \(\sqrt{x + 12}=x\), we start by squaring both sides. Squaring the left side \(\sqrt{x + 12}\) gives \(x + 12\), and squaring the right side \(x\) gives \(x^{2}\). So we get the equation:

$$x + 12=x^{2}$$

Step2: Rearrange the equation to 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 \(3\) because \((-4)\times3=-12\) and \(-4 + 3=-1\). So the factored form is:

$$(x - 4)(x+ 3)=0$$

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

Using the zero - product property, if \((x - 4)(x + 3)=0\), then either \(x-4 = 0\) or \(x + 3=0\).

• If \(x-4=0\), then \(x = 4\).
• If \(x + 3=0\), then \(x=-3\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation \(\sqrt{x + 12}=x\) because squaring both sides can introduce extraneous solutions.

• For \(x = 4\): Substitute \(x = 4\) into the original equation. The left side is \(\sqrt{4+12}=\sqrt{16} = 4\), and the right side is \(4\). So \(x = 4\) is a valid solution.
• For \(x=-3\): Substitute \(x=-3\) into the original equation. The left side is \(\sqrt{-3 + 12}=\sqrt{9}=3\), and the right side is \(-3\). Since \(3

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

Answer:

\(x = 4\)