How do you solve $16 p^{2} - 81 p^{6} \leq 0$ $16p^2-81p^6<=0$ using a sign chart?

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Answer 1 · 2017-01-27T11:06:21

The answer is $p \in] - \infty, - \frac{2}{3}] \cup [\frac{2}{3}, + \infty [$ $p in ]-oo, -2/3] uu [2/3, +oo [$

We need

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

Let's factorise the inequality

$16 p^{2} - 81 p^{6} = p^{2} (16 - 81 p^{4})$ $16p^2-81p^6=p^2(16-81p^4)$

$= p^{2} (4 - 9 p^{2}) (4 + 9 p^{2})$ $=p^2(4-9p^2)(4+9p^2)$

$= p^{2} (2 + 3 p) (2 - 3 p) (4 + 9 p^{2})$ $=p^2(2+3p)(2-3p)(4+9p^2)$

Let $f (p) = p^{2} (2 + 3 p) (2 - 3 p) (4 + 9 p^{2})$ $f(p)=p^2(2+3p)(2-3p)(4+9p^2)$

$p^2>=0 , AA p in RR$

$4+9p^2>0 , AA p in RR$

Now we can build the sign chart

$color(white)(aaaa)$ $p$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaaa)$ $-2/3$ $color(white)(aaaaaa)$ $2/3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $2+3p$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaa)$ $+$

$color(white)(aaaa)$ $2-3p$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaa)$ $-$

$color(white)(aaaa)$ $f(p)$ $color(white)(aaaaaaaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaa)$ $-$

Therefore,

$f(p)<=0$ when $p in ]-oo, -2/3] uu [2/3, +oo [$

How do you solve 16p^2-81p^6<=016p2−81p6≤016p^2-81p^6<=0 using a sign chart?