How do you solve x^4+35x^2-36>=0?

Jul 21, 2016

$x \in \left(- \infty , - 1\right] \cup \left[1 , \infty\right)$, or, $x \in \mathbb{R} - \left(- 1 , 1\right)$

Explanation:

${x}^{4} + 35 {x}^{2} - 36 \ge 0$

$\Rightarrow \left({x}^{2} - 1\right) \left({x}^{2} + 36\right) \ge 0$

Since, ${x}^{2} + 36 > 0 \Rightarrow {x}^{2} - 1 \ge 0 \Rightarrow \left(x - 1\right) \left(x + 1\right) \ge 0$

${x}^{2} - 1 = 0 \iff x = \pm 1. \ldots . \left(1\right)$

Hence, we need to solve only

$\left({x}^{2} - 1\right) > 0 , i . e . , \left(x - 1\right) \left(x + 1\right) > 0$

:.[(x-1)>0, &, (x+1)>0]..............Case (1), or,

[(x-1)<0, &, (x+1)<0].................Case(2).

Case (1) :-

(x-1)>0, &, (x+1)>0 rArr x>1, &, x> -1

rArrx in (1,oo) & x in (-1,oo)

$\Rightarrow x \in \left(1 , \infty\right) \cap \left(- 1 , \infty\right)$

$\Rightarrow x \in \left(1 , \infty\right)$

Case (2) :-

Working on the same lines as above [in Case (1)], in this case, we get

#x<-1, i.e., x in (-oo,-1)

Combining these Cases with $\left(1\right)$,

we get, $x \in \left(- \infty , - 1\right) , \mathmr{and} , x \in \left(1 , \infty\right) , \mathmr{and} , x = \pm 1$

i.e., $x \in \left(- \infty , - 1\right] \cup \left[1 , \infty\right)$, or, $x \in \mathbb{R} - \left(- 1 , 1\right)$

Hope, this is Helpful! Enjoy Maths!