Select all the correct answers.

Which values of [tex][tex]$x$[/tex][/tex] are solutions to this equation?

[tex] -\frac{1}{2} x^2 + 5 x = 8 [/tex]

A. [tex] -2 [/tex]
B. 2
C. [tex] -1.5 [/tex]
D. [tex] -8 [/tex]
E. 8
F. 11.5



Answer :

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]