# How do you solve 4x^4+4<=17x^2 using a sign chart?

Oct 15, 2016

$4 {x}^{4} + 4 \le 17 {x}^{2}$
$4 {x}^{4} - 17 {x}^{2} + 4 \le 0$
$\left(4 {x}^{2} - 1\right) \left({x}^{2} - 4\right) \le 0$
(2x-1)(2x+1)(x+2)(x-2<=0
 x=1/2 ; x=-1/2; x=-2;x=2
Here is the sign chart
x -oo -2 -1/2 1/2 2 +oo
2x-1 - 0 + + +
2x+1 - - - 0 + +
X+2 - 0 + + + +
x-2 - - - - 0 +
(4x^4-17x^2+4) + - + - + +
$- 2 \le x \le - \frac{1}{2}$
and
$\frac{1}{2} \le x \le 2$