Question

Solve the quadratic inequality 2x2+7x−30≥0
by factoring.(1 point)
Responses

[−5,3]
left bracket negative 5 comma 3 right bracket

(−∞,−3]∪[5,∞)
left parenthesis negative infinity comma negative 3 right bracket union left bracket 5 comma infinity right parenthesis

(−∞,−6]∪[52,∞)
left parenthesis negative infinity comma negative 6 right bracket union left bracket Start Fraction 5 over 2 End Fraction comma infinity right parenthesis

[−52,6]
left bracket negative Start Fraction 5 over 2 End Fraction comma 6 right bracket

Answer 1

To solve the quadratic inequality \(2x^2 + 7x - 30 \geq 0\) by factoring, we first need to factor the quadratic expression.

### Step 1: Factor the quadratic
We can begin by rewriting the equation in standard form:
\[ 2x^2 + 7x - 30 = 0 \]

Next, we can factor this expression. We are looking for two numbers that multiply to \(2 \cdot (-30) = -60\) and add to \(7\). The numbers that satisfy these conditions are \(12\) and \(-5\).

Now we can rewrite the middle term using these values:
\[ 2x^2 + 12x - 5x - 30 = 0 \]

Now, group the terms:
\[ (2x^2 + 12x) + (-5x - 30) = 0 \]
\[ 2x(x + 6) - 5(x + 6) = 0 \]

Now factor out the common factor \((x + 6)\):
\[ (2x - 5)(x + 6) = 0 \]

### Step 2: Find the critical points
To find where the expression is equal to zero, we set:
\[ (2x - 5)(x + 6) = 0 \]

This gives us the critical points:
1. \(2x - 5 = 0\) ⟹ \(x = \frac{5}{2}\)
2. \(x + 6 = 0\) ⟹ \(x = -6\)

### Step 3: Test intervals
Now, we need to test the intervals defined by these critical points:
- \((-∞, -6)\)
- \((-6, \frac{5}{2})\)
- \((\frac{5}{2}, ∞)\)

Choose a test point from each interval:
1. For \(x = -7\) in \((-∞, -6)\):
\[ (2(-7) - 5)(-7 + 6) = (-14 - 5)(-1) = (-19)(-1) = 19 \quad (\text{positive}) \]

2. For \(x = 0\) in \((-6, \frac{5}{2})\):
\[ (2(0) - 5)(0 + 6) = (-5)(6) = -30 \quad (\text{negative}) \]

3. For \(x = 3\) in \((\frac{5}{2}, ∞)\):
\[ (2(3) - 5)(3 + 6) = (6 - 5)(9) = (1)(9) = 9 \quad (\text{positive}) \]

### Step 4: Write the solution
From the test points, we see that the inequality \(2x^2 + 7x - 30 \geq 0\) is positive (or zero) in the intervals:
- \((-∞, -6]\)
- \([ \frac{5}{2}, ∞)\)

Thus, the solution to the original inequality is:
\[
(-\infty, -6] \cup \left[\frac{5}{2}, \infty\right)
\]

### Final Answer
So the correct option is:
\[
(-\infty, -6] \cup \left[\frac{5}{2}, \infty\right)
\]