Answer
$[-3/2, -1/2]$ U $[1/2, 3/2]$
Work Step by Step
$16x^4-40x^2+9\leq 0$
$(4x^2-1)(4x^2-9)\leq 0$
$4x^2-1\leq 0$
$4x^2-1+1 \leq 0+1$
$4x^2 \leq 1$
$4x^2/4 \leq 1/4$
$x^2 \leq 1/4$
$\sqrt {x^2} \leq \sqrt {1/4}$
$x \leq ±1/2$
$x \leq 1/2$, $x\leq -1/2$
$4x^2-9\leq 0$
$4x^2-9+9 \leq 0+9$
$4x^2 \leq 9$
$4x^2/4 \leq 9/4$
$x^2 \leq 9/4$
$\sqrt {x^2} \leq \sqrt {9/4}$
$x \leq ±3/2$
$x \leq 3/2$, $x\leq -3/2$
We have five sections: $(−∞,-3/2)$, $(-3/2, -1/2)$, $(-1/2, 1/2)$, $(1/2, 3/2)$, and $(3/2,∞)$. We need to test one value for x in each section to determine if the section would be a solution set. Since we have the $\leq$ sign, we include the end points and use brackets instead of parentheses.
Let $x=-3$, $x=-1$, $x=0$, $x=1$, and $x=3$
$x=-3$
$16x^4-40x^2+9\leq 0$
$16(-3)^4-40(-3)^2+9\leq 0$
$16*81-40*9+9\leq 0$
$1296-360+9 \leq 0$
$945 \leq 0$ (false)
$x=-1$
$16x^4-40x^2+9\leq 0$
$16(-1)^4-40(-1)^2+9\leq 0$
$16*1-40+9 \leq 0$
$16-31 \leq 0$
$-15 \leq 0$ (true)
$x=0$
$16x^4-40x^2+9\leq 0$
$16*0^4-40*0^2+9\leq 0$
$16*0-40*0+9 \leq 0$
$0-0+9 \leq 0$
$9 \leq 0$ (false)
$x=1$
$16x^4-40x^2+9\leq 0$
$16(1)^4-40(1)^2+9\leq 0$
$16*1-40+9 \leq 0$
$16-31 \leq 0$
$-15 \leq 0$ (true)
$x=3$
$16x^4-40x^2+9\leq 0$
$16(3)^4-40(3)^2+9\leq 0$
$16*81-40*9+9\leq 0$
$1296-360+9 \leq 0$
$945 \leq 0$ (false)
