To solve the quadratic inequality [tex]\( 8x^2 \geq 3 + 10x \)[/tex], we should follow these steps:
1. Rewrite the Inequality in Standard Form:
Start by moving all terms to one side of the inequality:
[tex]\[
8x^2 - 10x - 3 \geq 0
\][/tex]
2. Find the Roots of the Corresponding Equation:
To identify the critical points, we solve the equation:
[tex]\[
8x^2 - 10x - 3 = 0
\][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 8 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = -3 \)[/tex], we find the roots:
[tex]\[
x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} = \frac{10 \pm \sqrt{100 + 96}}{16} = \frac{10 \pm \sqrt{196}}{16} = \frac{10 \pm 14}{16}
\][/tex]
So the roots are:
[tex]\[
x = \frac{24}{16} = \frac{3}{2} \quad \text{and} \quad x = \frac{-4}{16} = -\frac{1}{4}
\][/tex]
3. Determine the Intervals:
The roots divide the number line into three intervals:
[tex]\[
(-\infty, -\frac{1}{4}), \quad \left(-\frac{1}{4}, \frac{3}{2}\right), \quad \text{and} \quad \left(\frac{3}{2}, \infty\right)
\][/tex]
4. Test Values in Each Interval:
Check the sign of [tex]\( 8x^2 - 10x - 3 \)[/tex] in each interval:
- For [tex]\( x \in (-\infty, -\frac{1}{4}) \)[/tex], choose [tex]\( x = -1 \)[/tex]:
[tex]\[
8(-1)^2 - 10(-1) - 3 = 8 + 10 - 3 = 15 > 0
\][/tex]
- For [tex]\( x \in \left(-\frac{1}{4}, \frac{3}{2}\right) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
8(0)^2 - 10(0) - 3 = -3 < 0
\][/tex]
- For [tex]\( x \in \left(\frac{3}{2}, \infty\right) \)[/tex], choose [tex]\( x = 2 \)[/tex]:
[tex]\[
8(2)^2 - 10(2) - 3 = 32 - 20 - 3 = 9 > 0
\][/tex]
5. Include the Endpoints:
Since the inequality is [tex]\(\geq\)[/tex], the critical points where [tex]\(8x^2 - 10x - 3 = 0\)[/tex] (i.e., [tex]\( x = -\frac{1}{4} \)[/tex] and [tex]\( x = \frac{3}{2} \)[/tex]) are included in the solution set.
6. Combine the Intervals:
The solution to the inequality [tex]\( 8x^2 \geq 3 + 10x \)[/tex] is therefore:
[tex]\[
(-\infty, -\frac{1}{4}] \cup [\frac{3}{2}, \infty)
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{(-\infty, -\frac{1}{4}] \cup [\frac{3}{2}, \infty)}
\][/tex]
