Let's solve the equation [tex]\(\sqrt{x+12} = x\)[/tex] step-by-step.
1. Isolate the square root:
[tex]\[\sqrt{x+12} = x\][/tex]
2. Square both sides to eliminate the square root:
[tex]\[(\sqrt{x+12})^2 = x^2\][/tex]
[tex]\[x + 12 = x^2\][/tex]
3. Rearrange the equation to set it to zero:
[tex]\[x^2 - x - 12 = 0\][/tex]
4. Factor the quadratic equation:
We need to find two numbers that multiply to [tex]\(-12\)[/tex] and add to [tex]\(-1\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(3\)[/tex].
Thus, we factorize:
[tex]\[(x - 4)(x + 3) = 0\][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[x - 4 = 0 \quad \text{or} \quad x + 3 = 0\][/tex]
[tex]\[x = 4 \quad \text{or} \quad x = -3\][/tex]
6. Verify the solutions in the original equation [tex]\(\sqrt{x+12} = x\)[/tex]:
- For [tex]\(x = 4\)[/tex]:
[tex]\[\sqrt{4+12} = \sqrt{16} = 4\][/tex]
The left side equals the right side, so [tex]\(x = 4\)[/tex] is a valid solution.
- For [tex]\(x = -3\)[/tex]:
[tex]\[\sqrt{-3+12} = \sqrt{9} = 3\][/tex]
The left side does not equal the right side ([tex]\(3 \ne -3\)[/tex]), so [tex]\(x = -3\)[/tex] is not a valid solution.
7. Conclusion:
The only valid solution to [tex]\(\sqrt{x+12} = x\)[/tex] is [tex]\(x = 4\)[/tex].
Therefore, the correct answer is:
[tex]\[x = 4\][/tex]
