How do you solve #x^4-5x^2<=-4#?

2 Answers
Jan 31, 2017

Answer:

The answer is #x in [-2, -1 ] uu [1, 2 ]#

Explanation:

We need

#a^2-b^2=(a+b)(a-b)#

Let rewrite the inequality

#x^4-5x^2+4<=0#

Let's factorise

#(x^2-1)(x^2-4)<=0#

#(x+1)(x-1)(x+2)(x-2)<=0#

Let #f(x)=(x+1)(x-1)(x+2)(x-2)#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##-1##color(white)(aaaa)##+1##color(white)(aaaa)##+2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in [-2, -1 ] uu [1, 2 ]#

Jan 31, 2017

Answer:

By graphical method, #x in [-1, -2] and x in [1. 2]#

Explanation:

graph{y-x^4+5x^2-4>=0 [-3.03, 3.048, -1.528, 1.51]}

Let y-(x^4-5x^2+4)>=0.

The section of the graph by the x-axis ( y = 0 ) depicts the solution

-#(x^2-5x+4)>=0#, giving

#x^2-5x+4<=0#