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# Algebra

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Find all real values of x such that (x^2 + 5x + 5 + 11x - 17)(3x^2 - x - 4 - 8x + 12) >= 0.

Note: Your solution should be thorough, and explain why all your vaues are the only values that satisfy the inequality.

Dec 9, 2022

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This is a longer grueling problem: By the zero product property, we have:

x^2 + 16x - 12 = 0    => -16 +- sqrt(256 + 48) over 2, -16 +- 4sqrt(19) over 2, -8 +/- 2sqrt(19)

3x^2 - 9x + 8 = 0      => 9 +- sqrt(81 - 4*8*3), but we see its imaginary, so only consider the above case to be real.

If multiply to greater than 0, by parity, we have:

x^2 + 16x - 12x > 0    =>   x = (-infinity, -8 - 2sqrt(19)) U (-8 + 2sqrt(19), infinity)

and (3x^2 - 9x + 8) > 0 at the same time. => x = (-infinity, infinity)

By negative parities, we have:

x^2 + 16x - 12 < 0         => x = (-8 - 2sqrt(19), -8 + 2sqrt(19))

and 3x^2 - 9x + 8 < 0    => no solutions for x

Now we have to satisfy the original inequality, so we look for the most constricting ranges, obviously (-infinity, infinity) isn't helpful, and we can see that most of our bolded revolve around -8 +/- 2sqrt(19), meaning that those two values are our values of x.

Dec 9, 2022
edited by proyaop  Dec 9, 2022