To solve the equation [tex]\(\sqrt{x-5} = x - 11\)[/tex], follow these steps:
1. Isolate the square root: The square root [tex]\(\sqrt{x-5}\)[/tex] is already isolated on one side of the equation.
2. Square both sides: To eliminate the square root, square both sides of the equation:
[tex]\[
(\sqrt{x-5})^2 = (x-11)^2
\][/tex]
This simplifies to:
[tex]\[
x - 5 = (x - 11)^2
\][/tex]
3. Expand the right side: Expand the squared term on the right side:
[tex]\[
x - 5 = (x-11)(x-11)
\][/tex]
[tex]\[
x - 5 = x^2 - 22x + 121
\][/tex]
4. Move all terms to one side: Subtract [tex]\(x\)[/tex] and add 5 to both sides to set the equation to zero:
[tex]\[
0 = x^2 - 22x + 121 - x + 5
\][/tex]
Combine like terms:
[tex]\[
0 = x^2 - 23x + 126
\][/tex]
5. Solve the quadratic equation: The quadratic equation [tex]\(x^2 - 23x + 126 = 0\)[/tex] can be solved using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -23\)[/tex], and [tex]\(c = 126\)[/tex].
Calculate the discriminant:
[tex]\[
b^2 - 4ac = (-23)^2 - 4(1)(126) = 529 - 504 = 25
\][/tex]
So, the solutions are:
[tex]\[
x = \frac{23 \pm \sqrt{25}}{2(1)} = \frac{23 \pm 5}{2}
\][/tex]
This gives two solutions:
[tex]\[
x = \frac{28}{2} = 14 \quad \text{and} \quad x = \frac{18}{2} = 9
\][/tex]
6. Verify the solutions: Substitute both solutions back into the original equation to check if they are valid:
For [tex]\(x = 14\)[/tex]:
[tex]\[
\sqrt{14 - 5} = \sqrt{9} = 3 \quad \text{and} \quad 14 - 11 = 3
\][/tex]
This holds true.
For [tex]\(x = 9\)[/tex]:
[tex]\[
\sqrt{9 - 5} = \sqrt{4} = 2 \quad \text{and} \quad 9 - 11 = -2
\][/tex]
This does not hold true as it results in [tex]\(\sqrt{4} = -2\)[/tex], which is not correct for the square root.
Therefore, the valid solution to the equation [tex]\(\sqrt{x-5} = x - 11\)[/tex] is:
[tex]\[x = 14\][/tex]
So the correct answer to the given equation is:
D. [tex]\(x = 14\)[/tex]
