Solve for quadratic inequality using interval method:  2x^2 > 5x + 3

To solve the quadratic inequality \( 2x^2 > 5x + 3 \) using the interval method, follow these steps:

### Step 1: Rearrange the inequality
First, express the inequality in standard quadratic form by moving all terms to one side:

\[
2x^2 - 5x - 3 > 0
\]

### Step 2: Solve the corresponding quadratic equation
Solve the equation \( 2x^2 - 5x - 3 = 0 \) by using the quadratic formula. The quadratic formula is:

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

For \( 2x^2 - 5x - 3 = 0 \), the coefficients are \( a = 2 \), \( b = -5 \), and \( c = -3 \).

Substitute these 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, the two solutions are:

\[
x = \frac{5 + 7}{4} = 3 \quad \text{and} \quad x = \frac{5 - 7}{4} = -\frac{1}{2}
\]

### Step 3: Determine the intervals
The two solutions, \( x = 3 \) and \( x = -\frac{1}{2} \), divide the real number line into three intervals:

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

### Step 4: Test each interval
Now test a point from each interval in the original inequality \( 2x^2 - 5x - 3 > 0 \).

- For \( (-\infty, -\frac{1}{2}) \), pick \( x = -1 \):

\[
2(-1)^2 - 5(-1) - 3 = 2(1) + 5 - 3 = 4 > 0
\]

So, the inequality holds in this interval.

- For \( (-\frac{1}{2}, 3) \), pick \( x = 0 \):

\[
2(0)^2 - 5(0) - 3 = -3 < 0
\]

So, the inequality does **not** hold in this interval.

- For \( (3, \infty) \), pick \( x = 4 \):

\[
2(4)^2 - 5(4) - 3 = 2(16) - 20 - 3 = 32 - 23 = 9 > 0
\]

So, the inequality holds in this interval.

### Step 5: Conclusion
The inequality \( 2x^2 - 5x - 3 > 0 \) holds in the intervals \( (-\infty, -\frac{1}{2}) \) and \( (3, \infty) \).

Thus, the solution to the inequality is:

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

Would you like more details on any of these steps?

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Here are some related questions:

1. How would the solution change if the inequality was \( \geq \) instead of \( > \)?
2. What is the graphical interpretation of the quadratic inequality solution?
3. Can you solve a cubic inequality using the interval method?
4. How can you apply this method to inequalities involving absolute values?
5. What are the differences between solving quadratic equations and quadratic inequalities?

**Tip:** Always check your intervals using test points to ensure the sign of the inequality holds in each region!

Solve for quadratic inequality using interval method: 2x^2 > 5x + 3

Learn how to solve the quadratic inequality 2x^2 > 5x + 3 using the interval method. This detailed solution includes steps on rearranging the inequality, solving the quadratic equation, and testing intervals. Suitable for students in Grades 9-12.

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