Let \(f\) be the piecewise function such that \(f(x) = \begin{cases} x^2 - 5x - 64, & x \le 0 \\ x^2 + 3x - 38, & x > 0 \end{cases} \) At how many points \(x\) does \(f(x)=2\)?

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Let \(f\) be the piecewise function such that \(f(x) = \begin{cases} x^2 - 5x - 64, & x \le 0 \\ x^2 + 3x - 38, & x > 0 \end{cases} \) At how many points \(x\) does \(f(x)=2\)?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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set each piece equal to 2 and solve for x x^2 - 5x - 64 = 2 leads to x = ?? or x = ?? x^2 + 3x - 38 = 2 leads to x = ?? or x = ??
but the answer isn't 4 (I got for the first equation - -6, 11 and for the second equation 5, -8)
when you solve f(x)=2 you have to make sure the x's fit in the inequality thingy x^2-5x-64=2 when x<=0 so you your x=-6 will only work for that one because 11 is greater than 0 so one solution so far... x^2+3x-38=2 when x>0 so your x=5 will only work for that one because -8 is less than 0 so 1+1=2 you have 2 solutions

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