Question

What is the solution set to the quadratic inequality −x2+6x+25≤−2 ?(1 point)Responses[−3, 9]left bracket negative 3 comma 9 right bracket(−3, 9)open paren negative 3 comma 9 close paren(−∞,−3)∪(9, ∞)the union of open paren infinity close paren and open paren 9 close paren(−∞, −3]∪[9, ∞)

Answer 1

To solve the quadratic inequality \(-x^2 + 6x + 25 \leq -2\), we first rearrange it into standard form:

\[ -x^2 + 6x + 25 + 2 \leq 0 \] \[ -x^2 + 6x + 27 \leq 0 \] Multiplying the entire inequality by -1 (which reverses the inequality sign) gives us:

\[ x^2 - 6x - 27 \geq 0 \]

Next, we will find the roots of the corresponding quadratic equation \(x^2 - 6x - 27 = 0\) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \(a = 1\), \(b = -6\), and \(c = -27\):

\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-27)}}{2(1)} \] \[ x = \frac{6 \pm \sqrt{36 + 108}}{2} \] \[ x = \frac{6 \pm \sqrt{144}}{2} \] \[ x = \frac{6 \pm 12}{2} \]

Calculating the two possible values:

\[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{-6}{2} = -3 \]

So the roots of the equation are \(x = -3\) and \(x = 9\). Now, we test intervals to determine the sign of the quadratic \(x^2 - 6x - 27\):

1. Test \(x < -3\) (e.g., \(x = -4\)): \[ (-4)^2 - 6(-4) - 27 = 16 + 24 - 27 = 13 \quad (\text{positive}) \]

2. Test \(-3 < x < 9\) (e.g., \(x = 0\)): \[ 0^2 - 6(0) - 27 = -27 \quad (\text{negative}) \]

3. Test \(x > 9\) (e.g., \(x = 10\)): \[ 10^2 - 6(10) - 27 = 100 - 60 - 27 = 13 \quad (\text{positive}) \]

The quadratic \(x^2 - 6x - 27\) is positive when \(x < -3\) and \(x > 9\), and negative (or zero) between the roots \([-3, 9]\).

Therefore, the solution set for the inequality \(x^2 - 6x - 27 \geq 0\) is:

\[ (-\infty, -3] \cup [9, \infty) \]

So the correct answer is:

\[ (-\infty, -3] \cup [9, \infty) \]