Solve inequality?
|x-12|>x^2
Could one of you please solve this inequality with detailed solution method I would like to see how you do it.
I've solved the absolute value by making a case distinction, and in the end, I end up with a quadratic inequality once the absolute value is gone. And I'd like to know if I still have to use the greater-than sign afterward, or can I use the equal-to sign afterward?
Because if in the second case the amount is less than 0 I get to the point
(x+4)(x-3) I have to make =0 or 0>(x+4)(x-3) here because if I now determine the zero places, I cannot simply jump from the > sign to x=-4 and x=3
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-(x-12) > x2
-x + 12 > x2
0 > x2 + x – 12
xse are +3 -4
.
x 12 > x 2
0 > x2 -x + 12
xse are not there
.
Ergo ( -4 ; 3 )
Did you jump from 0 > x2 + x – 12 to 0 = x2 + x – 12 to say +3 -4? That keeps me confused in the first block you constantly use the bigger sign In the second case, you jump from bigger to equal.
And should the solution not be an open interval??
twice right for you . Corrected