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:
-
Interval: \((-\infty, -3)\)
- Choose \(x = -4\): \((-4)^2 - 6(-4) - 27 = 16 + 24 - 27 = 13\) (which is \(> 0\))
-
Interval: \((-3, 9)\)
- Choose \(x = 0\): \(0^2 - 6(0) - 27 = -27\) (which is \(< 0\))
-
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, ∞).