Certainly! Let's solve the quadratic equation step-by-step:
1. Given Equation:
[tex]\[
x^2 + 13 = 8x + 37
\][/tex]
2. Rearrange into Standard Quadratic Form:
Move all terms to one side to get:
[tex]\[
x^2 - 8x + 13 - 37 = 0
\][/tex]
Simplify the constants:
[tex]\[
x^2 - 8x - 24 = 0
\][/tex]
3. Identify Coefficients:
The quadratic equation is now in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[
a = 1, \quad b = -8, \quad c = -24
\][/tex]
4. Calculate the Discriminant:
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = (-8)^2 - 4(1)(-24)
\][/tex]
Calculate:
[tex]\[
\Delta = 64 + 96 = 160
\][/tex]
5. Check the Discriminant:
Since the discriminant is positive ([tex]\( \Delta > 0 \)[/tex]), there are two distinct real solutions.
6. Quadratic Formula:
The solutions [tex]\(x\)[/tex] are given by:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substitute [tex]\(a = 1\)[/tex], [tex]\(b = -8\)[/tex], and [tex]\(\Delta = 160\)[/tex]:
[tex]\[
x = \frac{-(-8) \pm \sqrt{160}}{2 \cdot 1}
\][/tex]
Simplify:
[tex]\[
x = \frac{8 \pm \sqrt{160}}{2}
\][/tex]
7. Simplify Further:
Notice that [tex]\(\sqrt{160}\)[/tex] can be simplified:
[tex]\[
\sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10}
\][/tex]
So the expression for [tex]\(x\)[/tex] becomes:
[tex]\[
x = \frac{8 \pm 4\sqrt{10}}{2}
\][/tex]
Divide through by 2:
[tex]\[
x = 4 \pm 2\sqrt{10}
\][/tex]
Thus, the solutions to the quadratic equation are:
[tex]\[
x = 4 \pm 2\sqrt{10}
\][/tex]
Therefore, the correct answer is:
D. [tex]\(x = 4 \pm 2\sqrt{10}\)[/tex]
