Question

Solve the quadratic inequality 2x2+7x−30≥0 by factoring.(1 point)%0D%0AResponses%0D%0A%0D%0A(−∞,−3]∪[5,∞)%0D%0Aleft parenthesis negative infinity comma negative 3 right bracket union left bracket 5 comma infinity right parenthesis%0D%0A%0D%0A[−52,6]%0D%0Aleft bracket negative Start Fraction 5 over 2 End Fraction comma 6 right bracket%0D%0A%0D%0A(−∞,−6]∪[52,∞)%0D%0Aleft parenthesis negative infinity comma negative 6 right bracket union left bracket Start Fraction 5 over 2 End Fraction comma infinity right parenthesis%0D%0A%0D%0A[−5,3]

Answer 1

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

1. We start by finding two numbers that multiply to \( 2 \times -30 = -60 \) and add up to \( 7 \). The numbers \( 12 \) and \( -5 \) fit this requirement since \( 12 \times -5 = -60 \) and \( 12 + (-5) = 7 \).

2. We can rewrite the middle term using these two numbers:
\[
2x^2 + 12x - 5x - 30
\]

3. Next, we group the terms:
\[
(2x^2 + 12x) + (-5x - 30)
\]
\[
= 2x(x + 6) - 5(x + 6)
\]

4. Now, we can factor out the common factor \( (x + 6) \):
\[
= (2x - 5)(x + 6)
\]

Now we rewrite the inequality:
\[
(2x - 5)(x + 6) \geq 0
\]

5. We find the zeros of the expression by setting each factor to zero:
- \( 2x - 5 = 0 \) gives \( x = \frac{5}{2} \)
- \( x + 6 = 0 \) gives \( x = -6 \)

6. Next, we analyze the intervals defined by these critical points: \( (-\infty, -6) \), \( (-6, \frac{5}{2}) \), and \( (\frac{5}{2}, \infty) \).

7. We test a point from each interval:
- For \( x = -7 \) (in \( (-\infty, -6) \)):
\((2(-7) - 5)(-7 + 6) = (-14 - 5)(-1) = 19 > 0\) (satisfied)
- For \( x = 0 \) (in \( (-6, \frac{5}{2}) \)):
\((2(0) - 5)(0 + 6) = (-5)(6) = -30 < 0\) (not satisfied)
- For \( x = 3 \) (in \( (\frac{5}{2}, \infty) \)):
\((2(3) - 5)(3 + 6) = (6 - 5)(9) = 1 \cdot 9 > 0\) (satisfied)

8. Since we are interested in where the product is greater than or equal to zero, we include the points where the product equals zero (the boundary points \( -6 \) and \( \frac{5}{2} \)).

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

Therefore, the answer is:
\[
\boxed{(-\infty, -6] \cup \left[\frac{5}{2}, \infty\right)}
\]