How do you solve $3 w^{4} - 27 w^{2} > 0$ $3w^4-27w^2>0$ using a sign chart?

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Answer 1 · 2016-12-20T15:36:00

You can factorize into $3 w^{2} (w^{2} - 9) > 0$ $3w^2(w^2-9)>0$

Since $3 w^{2} \geq 0$ $3w^2>=0$ you may divide by it without changing the $>$ $>$ sign.
If $w = 0$ $w=0$ it won't fit the bill, so $w \neq 0$ $w!=0$

Which leaves us with $w^{2} - 9 > 0 \to w^{2} > 9$ $w^2-9>0->w^2>9$

So either $w < - 3 or w > + 3$ $w<-3orw>+3$
graph{3x^4-27x^2 [-16.04, 16, -8.03, 8]}

Answer 2 · 2016-12-20T15:50:53

The answer is $w \in] - \infty, - 3 [\cup] 3, + \infty [$ $w in ] -oo,-3 [ uu ] 3,+ oo[$

Let $f (w) = 3 w^{4} - 27 w^{2}$ $f(w)=3w^4-27w^2$

$= 3 w^{2} (w + 3) (w - 3)$ $=3w^2(w+3)(w-3)$

$w^2 >0, AA win RR$

Let's do a sign chart

$color(white)(aaaa)$ $w$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-3$ $color(white)(aaaa)$ $3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $w+3$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $w-3$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(w)$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(w)>0$ , when $w in ] -oo,-3 [ uu ] 3,+ oo[$

How do you solve 3w^4-27w^2>03w4−27w2>03w^4-27w^2>0 using a sign chart?