Let's solve the inequality $2x^2 > 5x + 3$ using the interval method. First, we'll rearrange the inequality into standard form:

$2x^2 - 5x - 3 > 0$

Step 1: Solve the corresponding equality

To use the interval method, we first solve the equation:

$2x^2 - 5x - 3 = 0$

We will apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the quadratic $2x^2 - 5x - 3 = 0$, we have:

• $a = 2$
• $b = -5$
• $c = -3$

Now, applying the values into the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$ $x = \frac{5 \pm \sqrt{25 + 24}}{4}$ $x = \frac{5 \pm \sqrt{49}}{4}$ $x = \frac{5 \pm 7}{4}$

Thus, we have two solutions: $x_1 = \frac{5 + 7}{4} = 3$ $x_2 = \frac{5 - 7}{4} = -\frac{1}{2}$

Step 2: Determine the intervals

Now that we know the roots $x = 3$ and $x = -\frac{1}{2}$, we can divide the real number line into three intervals:

1. $(-\infty, -\frac{1}{2})$
2. $\left(-\frac{1}{2}, 3\right)$
3. $(3, \infty)$

Step 3: Test values in each interval

We need to test the sign of the quadratic expression $2x^2 - 5x - 3$ in each interval.

1. For $x = -1$ (in the interval $(-\infty, -\frac{1}{2})$): $2(-1)^2 - 5(-1) - 3 = 2 + 5 - 3 = 4 \quad (\text{positive})$
2. For $x = 0$ (in the interval $\left(-\frac{1}{2}, 3\right)$): $2(0)^2 - 5(0) - 3 = -3 \quad (\text{negative})$
3. For $x = 4$ (in the interval $(3, \infty)$): $2(4)^2 - 5(4) - 3 = 32 - 20 - 3 = 9 \quad (\text{positive})$

Step 4: Solution

The quadratic expression is positive on the intervals $(-\infty, -\frac{1}{2})$ and $(3, \infty)$. Therefore, the solution to the inequality $2x^2 - 5x - 3 > 0$ is:

$x \in (-\infty, -\frac{1}{2}) \cup (3, \infty)$

Would you like a detailed explanation on any step or have any questions?

Here are 5 related questions:

1. How do you solve quadratic inequalities using the test point method?
2. What is the difference between solving quadratic equations and inequalities?
3. How do you factor quadratic expressions?
4. Can the interval method be used for any degree of polynomial?
5. How does the quadratic formula work for non-real solutions?

Tip: Always check the direction of the inequality before solving; this determines whether to test for positive or negative values in each interval.