How do you solve $x^{4} + 35 x^{2} - 36 \geq 0$ $x^4+35x^2-36>=0$ ?

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Answer 1 · 2016-07-21T17:04:51

$x \in (- \infty, - 1] \cup [1, \infty)$ $x in (-oo,-1]uu[1,oo)$ , or, $x in RR-(-1,1)$

$x^4+35x^2-36>=0$

$rArr(x^2-1)(x^2+36)>=0$

Since, $x^2+36>0 rArrx^2-1>=0rArr (x-1)(x+1)>=0$

$x^2-1=0 iff x=+-1.....(1)$

Hence, we need to solve only

$(x^2-1)>0, i.e., (x-1)(x+1)>0$

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

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

Case (1) :-

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

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

$rArr x in (1,oo) nn (-1,oo)$

$rArr x in (1,oo)$

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 $(1)$ ,

we get, $x in (-oo,-1), or, x in (1,oo), or, x=+-1$

i.e., $x in (-oo,-1]uu[1,oo)$ , or, $x in RR-(-1,1)$

Hope, this is Helpful! Enjoy Maths!

How do you solve x^4+35x^2-36>=0x4+35x2−36≥0x^4+35x^2-36>=0?