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:
- \( (-\infty, -3) \)
- \( (-3, 9) \)
- \( (9, \infty) \)
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, ∞).