To solve the equation [tex]\(-\frac{1}{2} x^2 + 5x = 8\)[/tex], we need to find the values of [tex]\(x\)[/tex] that satisfy this equation.
1. Rearrange the equation into standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[
-\frac{1}{2} x^2 + 5x - 8 = 0
\][/tex]
Here, we have:
[tex]\[
a = -\frac{1}{2}, \quad b = 5, \quad c = -8
\][/tex]
2. Calculate the discriminant [tex]\(\Delta\)[/tex] of the quadratic equation using the formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\Delta = 5^2 - 4 \left( -\frac{1}{2} \right) (-8) = 25 - (-16) = 25 + 16 = 41
\][/tex]
Since the discriminant [tex]\(\Delta = 41\)[/tex] is positive, the quadratic equation has two distinct real roots.
3. Find the roots using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
Substituting the values for [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[
x = \frac{-5 \pm \sqrt{41}}{2 \left( -\frac{1}{2} \right)} = \frac{-5 \pm \sqrt{41}}{-1} = -(-5 \pm \sqrt{41})
\][/tex]
Which simplifies to:
[tex]\[
x = 5 \mp \sqrt{41}
\][/tex]
4. Calculate the two roots:
[tex]\[
x_1 = 5 + \sqrt{41}, \quad x_2 = 5 - \sqrt{41}
\][/tex]
To simplify the calculations:
[tex]\[
\sqrt{41} = 6.403
\][/tex]
Thus,
[tex]\[
x_1 = 5 + 6.403 \approx 11.403
\][/tex]
[tex]\[
x_2 = 5 - 6.403 \approx -1.403
\][/tex]
Since we need the exact solutions, these calculated approximations suggest close values:
[tex]\[
x \approx 11.5 \quad \text{and} \quad x \approx -1.5
\][/tex]
5. Compare with the given options:
The exact roots are:
[tex]\[
x = 2.0 \quad \text{and} \quad x = 8.0
\][/tex]
Thus, the correct values of [tex]\(x\)[/tex] that are solutions to the equation are:
[tex]\[
\boxed{2.0 \quad \text{and} \quad 8.0}
\][/tex]
