Question: solve this quadratic inequality with these steps: 1: set one side of the inequality to zero (standard form) 2: factor the quadratic expression 3: Equate each (linear) factor to zero to find the critical points 4: create intervals for each interval 5: use sign test, and 6: write the solutions (it's $2x^2$ is greater than or equal to $5x -3$)
1. **State the problem:** Solve the quadratic inequality $$2x^2 \geq 5x - 3$$.
2. **Set one side to zero (standard form):**
$$2x^2 - 5x + 3 \geq 0$$
3. **Factor the quadratic expression:**
We look for factors of $2x^2 - 5x + 3$.
Try to factor:
$$2x^2 - 5x + 3 = (2x - 3)(x - 1)$$
4. **Find critical points by setting each factor to zero:**
$$2x - 3 = 0 \implies x = \frac{3}{2}$$
$$x - 1 = 0 \implies x = 1$$
5. **Create intervals based on critical points:**
The critical points split the real line into three intervals:
- $(-\infty, 1)$
- $(1, \frac{3}{2})$
- $(\frac{3}{2}, \infty)$
6. **Use sign test on each interval:**
- For $x < 1$, pick $x=0$:
$$(2(0) - 3)(0 - 1) = (-3)(-1) = 3 > 0$$
- For $1 < x < \frac{3}{2}$, pick $x=1.2$:
$$(2(1.2) - 3)(1.2 - 1) = (2.4 - 3)(0.2) = (-0.6)(0.2) = -0.12 < 0$$
- For $x > \frac{3}{2}$, pick $x=2$:
$$(2(2) - 3)(2 - 1) = (4 - 3)(1) = 1 > 0$$
7. **Write the solution:**
The inequality $2x^2 - 5x + 3 \geq 0$ holds where the product is non-negative, i.e., on intervals where the sign test is positive or zero.
Include critical points since inequality is \geq:
$$x \in (-\infty, 1] \cup \left[\frac{3}{2}, \infty\right)$$