To solve the given inequality [tex]\(2x^2 + 2 \geq 5x\)[/tex], we'll proceed step-by-step.
### Step 1: Rewrite the Inequality
First, we rewrite the inequality in standard form by bringing all terms to one side:
[tex]\[2x^2 + 2 \geq 5x\][/tex]
Subtract [tex]\(5x\)[/tex] from both sides:
[tex]\[2x^2 + 2 - 5x \geq 0\][/tex]
### Step 2: Express as a Quadratic Equation
Rearrange the terms:
[tex]\[2x^2 - 5x + 2 \geq 0\][/tex]
### Step 3: Solve the Quadratic Equation
To solve [tex]\(2x^2 - 5x + 2 = 0\)[/tex], we use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]. Here, [tex]\(a = 2\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = 2\)[/tex].
Calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = (-5)^2 - 4(2)(2) = 25 - 16 = 9
\][/tex]
So,
[tex]\[
x = \frac{-(-5) \pm \sqrt{9}}{2 \cdot 2} = \frac{5 \pm 3}{4}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{5+3}{4} = \frac{8}{4} = 2
\][/tex]
[tex]\[
x = \frac{5-3}{4} = \frac{2}{4} = \frac{1}{2}
\][/tex]
### Step 4: Test the Intervals
The solutions [tex]\(x = 2\)[/tex] and [tex]\(x = \frac{1}{2}\)[/tex] divide the real number line into three intervals: [tex]\((-\infty, \frac{1}{2})\)[/tex], [tex]\(\left(\frac{1}{2}, 2\right)\)[/tex], and [tex]\((2, \infty)\)[/tex].
We test each interval to determine where the inequality [tex]\(2x^2 - 5x + 2 \geq 0\)[/tex] holds:
1. Interval [tex]\( (-\infty, \frac{1}{2}) \)[/tex]
Choose a test point, e.g., [tex]\(x = 0\)[/tex]:
[tex]\[
2(0)^2 - 5(0) + 2 = 2 \geq 0 \quad \text{(True)}
\][/tex]
2. Interval [tex]\( \left(\frac{1}{2}, 2\right) \)[/tex]
Choose a test point, e.g., [tex]\(x = 1\)[/tex]:
[tex]\[
2(1)^2 - 5(1) + 2 = 2 - 5 + 2 = -1 \geq 0 \quad \text{(False)}
\][/tex]
3. Interval [tex]\( (2, \infty) \)[/tex]
Choose a test point, e.g., [tex]\(x = 3\)[/tex]:
[tex]\[
2(3)^2 - 5(3) + 2 = 18 - 15 + 2 = 5 \geq 0 \quad \text{(True)}
\][/tex]
### Step 5: Include Boundary Points
Since the inequality is non-strict ([tex]\(\geq\)[/tex]), we include the points where the quadratic expression is zero, which are [tex]\(x = \frac{1}{2}\)[/tex] and [tex]\(x = 2\)[/tex].
### Conclusion
The solution to the inequality [tex]\(2x^2 + 2 \geq 5x\)[/tex] is the union of the intervals where the expression is non-negative, along with the boundary points:
[tex]\[
(-\infty, \frac{1}{2}] \cup [2, \infty)
\][/tex]
Thus, the correct answer is:
d. [tex]\(\left(-\infty, \frac{1}{2}\right] \cup [2, \infty)\)[/tex]
