Question

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

(−∞, −3]∪[9, ∞)
the union of left parenthesis infinity right bracket and left bracket 9 right parenthesis

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

(−∞,−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 can first rearrange it into standard form:

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

This simplifies to:

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

Next, we can multiply through by -1 (which reverses the inequality):

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

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

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

Where \(a = 1\), \(b = -6\), and \(c = -27\):

\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-27)}}{2(1)}
\]

\[
x = \frac{6 \pm \sqrt{36 + 108}}{2}
\]

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

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

This gives us two roots:

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

Now we have the roots \(x = -3\) and \(x = 9\).

Next, we analyze the sign of the quadratic \(x^2 - 6x - 27\) in the intervals defined by these roots:

1. \( (-\infty, -3) \)
2. \( (-3, 9) \)
3. \( (9, \infty) \)

We can test points from each interval:

- For \(x < -3\) (e.g., \(x = -4\)):
\[
(-4)^2 - 6(-4) - 27 = 16 + 24 - 27 = 13 \geq 0
\]
- For \(-3 < x < 9\) (e.g., \(x = 0\)):
\[
0^2 - 6(0) - 27 = -27 < 0
\]
- For \(x > 9\) (e.g., \(x = 10\)):
\[
(10)^2 - 6(10) - 27 = 100 - 60 - 27 = 13 \geq 0
\]

Thus, the quadratic \(x^2 - 6x - 27\) is positive in the intervals \((-∞, -3]\) and \([9, ∞)\).

Since we are looking for where the expression is greater than or equal to zero, we include the boundary points:

The solution set to the inequality is:

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

So the correct response is:

**(−∞, −3]∪[9, ∞)**.