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Question 1209885: Let f be the piecewise function such that
f(x) =
x^2 - 5x - 64 & \text{if} \ x \le 0, \\
x^2 + 3x - 38 & \text{if} \ x > 0.
Find all x such that f(x) = 50.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! We are given the piecewise function:
$$ f(x) = \begin{cases} x^2 - 5x - 64 & \text{if } x \le 0 \\ x^2 + 3x - 38 & \text{if } x > 0 \end{cases} $$
We want to find all $x$ such that $f(x) = 50$.
**Case 1: $x \le 0$**
We have $x^2 - 5x - 64 = 50$, which simplifies to $x^2 - 5x - 114 = 0$.
We can use the quadratic formula to solve for $x$:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
In this case, $a = 1$, $b = -5$, and $c = -114$.
$$ x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-114)}}{2(1)} = \frac{5 \pm \sqrt{25 + 456}}{2} = \frac{5 \pm \sqrt{481}}{2} $$
We have two possible solutions:
$$ x_1 = \frac{5 + \sqrt{481}}{2} \approx \frac{5 + 21.93}{2} \approx 13.465 $$
$$ x_2 = \frac{5 - \sqrt{481}}{2} \approx \frac{5 - 21.93}{2} \approx -8.465 $$
Since we are considering $x \le 0$, only $x_2$ is a valid solution.
Thus, $x = \frac{5 - \sqrt{481}}{2}$ is a solution.
**Case 2: $x > 0$**
We have $x^2 + 3x - 38 = 50$, which simplifies to $x^2 + 3x - 88 = 0$.
We can factor the quadratic as $(x - 8)(x + 11) = 0$.
The solutions are $x = 8$ and $x = -11$.
Since we are considering $x > 0$, only $x = 8$ is a valid solution.
Thus, $x = 8$ is a solution.
Therefore, the solutions are $x = \frac{5 - \sqrt{481}}{2}$ and $x = 8$.
Final Answer: The final answer is $\boxed{8, \frac{5 - \sqrt{481}}{2}}$
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