To solve the quadratic equation [tex]\(-8x^2 - 5x + 3 = 0\)[/tex] using the quadratic formula, follow these steps:
1. Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] from the equation [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[
a = -8, \quad b = -5, \quad c = 3
\][/tex]
2. Compute the discriminant [tex]\(\Delta\)[/tex] using the formula [tex]\(\Delta = b^2 - 4ac\)[/tex]:
[tex]\[
\Delta = (-5)^2 - 4 \cdot (-8) \cdot 3
\][/tex]
[tex]\[
\Delta = 25 + 96
\][/tex]
[tex]\[
\Delta = 121
\][/tex]
3. Determine the nature of the roots by examining the discriminant:
- If [tex]\(\Delta > 0\)[/tex], there are two distinct real roots.
- If [tex]\(\Delta = 0\)[/tex], there is one real root (a repeated root).
- If [tex]\(\Delta < 0\)[/tex], there are no real roots.
Since [tex]\(\Delta = 121 > 0\)[/tex], there are two distinct real roots.
4. Use the quadratic formula to find the roots:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
5. Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex] into the formula:
[tex]\[
x = \frac{-(-5) \pm \sqrt{121}}{2 \cdot (-8)}
\][/tex]
[tex]\[
x = \frac{5 \pm 11}{-16}
\][/tex]
6. Calculate the two solutions:
[tex]\[
x_1 = \frac{5 + 11}{-16} = \frac{16}{-16} = -1.0
\][/tex]
[tex]\[
x_2 = \frac{5 - 11}{-16} = \frac{-6}{-16} = 0.375 \approx 0.38
\][/tex]
7. Round the answers to two decimal places, if necessary.
The solutions to the equation [tex]\(-8x^2 - 5x + 3 = 0\)[/tex] are:
[tex]\[
x_1 = -1.0 \quad \text{and} \quad x_2 = 0.38
\][/tex]
The discriminant value [tex]\(\Delta\)[/tex] is [tex]\(121\)[/tex], confirming that there are two real roots for the given quadratic equation.
