How do you simplify $\sqrt[5]{- t^{10}}$ $root5(-t^10 )$ ?

Question

1247 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-06T23:19:01

$\sqrt[5]{- t^{10}} = - t^{2}$ $root(5)(-t^10) = -t^2$

If $a > 0$ $a > 0$ and $b, c \geq 0$ $b, c >= 0$ , then $a^{b c} = {(a^{b})}^{c}$ $a^(bc) = (a^b)^c$

Also $\sqrt[n]{- a} = - \sqrt[n]{a}$ $root(n)(-a) = -root(n)(a)$ if $n in ZZ$ is odd.

So $root(5)(-t^10) = -root(5)(t^10) = -(t^10)^(1/5) = -t^(10*1/5) = -t^2$

We can also see that $(-t^2)^5 = (-1)^5*(t^2)^5 = -t^10$ , so $-t^2$ is a fifth root of $-t^10$ .

How do you simplify root5(-t^10 )5√−t10root5(-t^10 )?