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

Responses

(−∞,−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

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

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

(−∞,−3]∪[5,∞)

1 answer

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

  1. Start by factoring the quadratic \(2x^2 + 7x - 30\).

    We are looking for two numbers that multiply to \(2 \times -30 = -60\) and add to \(7\). The two numbers that satisfy these conditions are \(12\) and \(-5\).

  2. Rewrite the middle term (the term with \(x\)):

    \[ 2x^2 + 12x - 5x - 30 \]

  3. Group the terms:

    \[ (2x^2 + 12x) + (-5x - 30) \]

  4. Factor by grouping:

    \[ 2x(x + 6) - 5(x + 6) \]

    This gives us:

    \[ (2x - 5)(x + 6) \]

  5. Set the factored expression greater than or equal to zero:

    \[ (2x - 5)(x + 6) \geq 0 \]

  6. Find the roots by setting each factor to zero:

    \[ 2x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{2} \]

    \[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \]

  7. The critical points are \(x = -6\) and \(x = \frac{5}{2}\). These points divide the number line into intervals:

    • \( (-\infty, -6) \)
    • \( (-6, \frac{5}{2}) \)
    • \( (\frac{5}{2}, \infty) \)

  8. We will test each interval to determine where the inequality holds.

    • Test an x-value from all three intervals:
      • For \(x = -7\) (in \((- \infty, -6)\)): \[ (2(-7) - 5)((-7) + 6) = (-14 - 5)(-1) = -19 \cdot (-1) > 0 \quad \Rightarrow \quad \text{True} \]
      • For \(x = 0\) (in \((-6, \frac{5}{2}) \)): \[ (2(0) - 5)(0 + 6) = (-5)(6) < 0 \quad \Rightarrow \quad \text{False} \]
      • For \(x = 3\) (in \((\frac{5}{2}, \infty)\)): \[ (2(3) - 5)(3 + 6) = (6 - 5)(9) = 1 \cdot 9 > 0 \quad \Rightarrow \quad \text{True} \]

  9. The intervals where the inequality holds true are:

    • \( (-\infty, -6] \) (because we include \(-6\) where the expression equals zero),
    • \([ \frac{5}{2}, \infty) \) (because we include \(\frac{5}{2}\) where the expression equals zero).

So the solution to the inequality is:

\[ (-\infty, -6] \cup \left[\frac{5}{2}, \infty\right) \]

Thus, the correct answer is:

\[ (-\infty, -6] \cup \left[\frac{5}{2}, \infty\right) \]

Final Answer: \((- \infty, -6] \cup \left[\frac{5}{2}, \infty\right)\)