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

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

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Answer 1 · 2017-01-31T13:53:15

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 $x in [-2, -1 ] uu [1, 2 ]$

Answer 2 · 2017-01-31T16:26:05

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

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

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

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