To find the value of [tex]\(x\)[/tex] in the given problem, we need to use the fact that the sum of the interior angles of an octagon is always 1080 degrees. The interior angles provided are [tex]\(2x^\circ\)[/tex], [tex]\(\frac{1}{2}x(x + 40)^\circ\)[/tex], [tex]\(110^\circ\)[/tex], [tex]\(135^\circ\)[/tex], [tex]\(160^\circ\)[/tex], [tex]\((2x + 10)^\circ\)[/tex], and [tex]\(185^\circ\)[/tex].
Here is the step-by-step solution:
1. Write down the sum of the angles:
We know the sum of these angles must equal 1080 degrees.
2. Form the equation:
[tex]\[
2x + \frac{1}{2}x(x + 40) + 110 + 135 + 160 + (2x + 10) + 185 = 1080
\][/tex]
3. Combine like terms:
Combine all the constants together:
[tex]\[
110 + 135 + 160 + 10 + 185 = 600
\][/tex]
So, the equation now becomes:
[tex]\[
2x + \frac{1}{2}x(x + 40) + 2x + 600 = 1080
\][/tex]
4. Simplify further:
Combine the [tex]\(x\)[/tex] terms:
[tex]\[
4x + \frac{1}{2}x(x + 40) + 600 = 1080
\][/tex]
5. Clear the parenthesis:
Distribute [tex]\(\frac{1}{2}x\)[/tex] inside the parentheses:
[tex]\[
\frac{1}{2}x^2 + 20x
\][/tex]
Therefore, the equation becomes:
[tex]\[
4x + \frac{1}{2}x^2 + 20x + 600 = 1080
\][/tex]
6. Combine like terms:
Combine the [tex]\(x\)[/tex] terms again:
[tex]\[
\frac{1}{2}x^2 + 24x + 600 = 1080
\][/tex]
7. Move all terms to one side to set the equation to 0:
[tex]\[
\frac{1}{2}x^2 + 24x + 600 - 1080 = 0
\][/tex]
This simplifies to:
[tex]\[
\frac{1}{2}x^2 + 24x - 480 = 0
\][/tex]
8. Multiply through by 2 to clear the fraction:
[tex]\[
x^2 + 48x - 960 = 0
\][/tex]
9. Solve the quadratic equation:
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 48\)[/tex], and [tex]\(c = -960\)[/tex]:
[tex]\[
x = \frac{-(48) \pm \sqrt{48^2 - 4 \cdot 1 \cdot (-960)}}{2 \cdot 1}
\][/tex]
Simplify under the square root:
[tex]\[
x = \frac{-48 \pm \sqrt{2304 + 3840}}{2}
\][/tex]
[tex]\[
x = \frac{-48 \pm \sqrt{6144}}{2}
\][/tex]
[tex]\[
x = \frac{-48 \pm 78.431}{2}
\][/tex]
10. Solve for the two potential values of [tex]\(x\)[/tex]:
[tex]\[
x_1 = \frac{-48 + 78.431}{2}
\][/tex]
[tex]\[
x_1 \approx 15.1918358845308
\][/tex]
[tex]\[
x_2 = \frac{-48 - 78.431}{2}
\][/tex]
[tex]\[
x_2 \approx -63.1918358845308
\][/tex]
However, since [tex]\(x\)[/tex] represents an angle measure, it cannot be negative. Thus, the valid solution for [tex]\(x\)[/tex] is:
[tex]\[
x \approx 15.1918358845308
\][/tex]
