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

1 Answer
Jul 21, 2016

Answer:

#x in (-oo,-1]uu[1,oo)#, or, #x in RR-(-1,1)#

Explanation:

#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!