Sure, let's solve the quadratic inequality [tex]\(2x^2 + 5x - 3 > 0\)[/tex] step-by-step.
### Step 1: Formulate the corresponding quadratic equation
To solve the inequality, we first need to find the roots of the corresponding quadratic equation [tex]\(2x^2 + 5x - 3 = 0\)[/tex]. The roots will help us understand the intervals where the inequality holds true.
### Step 2: Use the quadratic formula
Recall that the quadratic formula to find the roots [tex]\(x\)[/tex] of the equation [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -3\)[/tex].
### Step 3: Calculate the discriminant
The discriminant (the part under the square root in the quadratic formula) is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = 5^2 - 4(2)(-3) \][/tex]
[tex]\[ \Delta = 25 + 24 \][/tex]
[tex]\[ \Delta = 49 \][/tex]
### Step 4: Find the roots of the quadratic equation
Now, we can find the roots using the quadratic formula:
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
Substituting the values:
[tex]\[ x_1 = \frac{-5 + \sqrt{49}}{2 \cdot 2} \][/tex]
[tex]\[ x_1 = \frac{-5 + 7}{4} \][/tex]
[tex]\[ x_1 = \frac{2}{4} \][/tex]
[tex]\[ x_1 = 0.5 \][/tex]
And for [tex]\(x_2\)[/tex]:
[tex]\[ x_2 = \frac{-5 - \sqrt{49}}{2 \cdot 2} \][/tex]
[tex]\[ x_2 = \frac{-5 - 7}{4} \][/tex]
[tex]\[ x_2 = \frac{-12}{4} \][/tex]
[tex]\[ x_2 = -3 \][/tex]
### Step 5: Determine the intervals
The quadratic equation [tex]\(2x^2 + 5x - 3 = 0\)[/tex] has roots at [tex]\(x = 0.5\)[/tex] and [tex]\(x = -3\)[/tex]. To determine where the inequality [tex]\(2x^2 + 5x - 3 > 0\)[/tex] holds, we need to consider the sign of the quadratic expression in the intervals defined by these roots:
1. [tex]\( x < -3 \)[/tex]
2. [tex]\( -3 < x < 0.5 \)[/tex]
3. [tex]\( x > 0.5 \)[/tex]
### Step 6: Test the intervals
We need to test each interval to see where the inequality [tex]\(2x^2 + 5x - 3 > 0\)[/tex] holds true.
- For [tex]\(x < -3\)[/tex]:
Let's take [tex]\(x = -4\)[/tex]:
[tex]\[
2(-4)^2 + 5(-4) - 3 = 2(16) - 20 - 3 = 32 - 20 - 3 = 9 > 0
\][/tex]
Thus, the inequality holds for [tex]\(x < -3\)[/tex].
- For [tex]\(-3 < x < 0.5\)[/tex]:
Let's take [tex]\(x = 0\)[/tex]:
[tex]\[
2(0)^2 + 5(0) - 3 = -3 < 0
\][/tex]
Thus, the inequality does not hold for [tex]\(-3 < x < 0.5\)[/tex].
- For [tex]\(x > 0.5\)[/tex]:
Let's take [tex]\(x = 1\)[/tex]:
[tex]\[
2(1)^2 + 5(1) - 3 = 2 + 5 - 3 = 4 > 0
\][/tex]
Thus, the inequality holds for [tex]\(x > 0.5\)[/tex].
### Conclusion
The quadratic inequality [tex]\(2x^2 + 5x - 3 > 0\)[/tex] is satisfied when [tex]\(x < -3\)[/tex] or [tex]\(x > 0.5\)[/tex]. Therefore, the solution to the inequality is:
[tex]\[ x \in (-\infty, -3) \cup (0.5, \infty) \][/tex]
