Question

What is the solution set to the quadratic inequality −x2+6x+25≤−2 ?(1 point)
Responses

(−∞,−3)∪(9, ∞)
the union of open paren infinity close paren and open paren 9 close paren

[−3, 9]
left bracket negative 3 comma 9 right bracket

(−3, 9)
open paren negative 3 comma 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:

\[
-x^2 + 6x + 25 + 2 \leq 0
\]

This simplifies to:

\[
-x^2 + 6x + 27 \leq 0
\]

Next, we multiply through by \(-1\) (noting that this reverses the inequality):

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

Now we need to find the roots of the associated equation \(x^2 - 6x - 27 = 0\). We use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = -6\), and \(c = -27\).

Calculating the discriminant:

\[
b^2 - 4ac = (-6)^2 - 4(1)(-27) = 36 + 108 = 144
\]

Now we find the roots:

\[
x = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2}
\]

This gives us two roots:

\[
x_1 = \frac{18}{2} = 9 \quad \text{and} \quad x_2 = \frac{-6}{2} = -3
\]

Thus, the roots are \(x = 9\) and \(x = -3\). The quadratic \(x^2 - 6x - 27\) opens upwards (since the coefficient of \(x^2\) is positive).

Now we determine where the quadratic is greater than or equal to zero by testing intervals defined by the roots:

1. **Interval**: \((-\infty, -3)\)
- Choose \(x = -4\): \((-4)^2 - 6(-4) - 27 = 16 + 24 - 27 = 13\) (which is \(> 0\))

2. **Interval**: \((-3, 9)\)
- Choose \(x = 0\): \(0^2 - 6(0) - 27 = -27\) (which is \(< 0\))

3. **Interval**: \((9, \infty)\)
- Choose \(x = 10\): \(10^2 - 6(10) - 27 = 100 - 60 - 27 = 13\) (which is \(> 0\))

Thus, the quadratic is non-negative in the intervals \((-\infty, -3]\) and \([9, \infty)\).

Therefore, the solution set to the inequality \(-x^2 + 6x + 25 \leq -2\) is:

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

So the correct response is:

**(−∞, −3]∪[9, ∞)**.