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.

Guest Dec 9, 2022

#1**+1 **

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.

proyaop Dec 9, 2022