Question

Find the exact value of the radical expression in simplest form.
√p^6+√9p^6+3√p^6-√p^2

Answer 1

\[\large \sqrt{p^6}+\sqrt{9p^6}+3\sqrt{p^6}-\sqrt{p^2}\]

Answer 2

Alright, we know that \[\large \sqrt{p^6}=\sqrt{p*p*p*p*p*p}=\sqrt{p^4}\sqrt{p^2}=p^2\sqrt{p^2}\]

Answer 3

\[\large p^2\sqrt{p^2}=p^2*p=p^3\]

Answer 4

Let's apply this for all \(\large \sqrt{p^6}\)

Answer 5

\[\large \begin{align}&=\sqrt{p^6}+\sqrt{9p^6}+3\sqrt{p^6}-\sqrt{p^2}\\&=p^3+\sqrt{9}p^3+3p^3-\sqrt{p^2} \end{align}\]

Answer 6

\[\sqrt{9}=3;~\sqrt{p^2}=p\]

Answer 7

\[\large \begin{align}&=p^3+\sqrt{9}p^3+3p^3-\sqrt{p^2}\\&= p^3+3p^3+3p^3-p\end{align}\]

Group similar terms

\[\large =10p^3-p\]
And then you can simply factor one p out
\[\large = p(10p^2-1)\]

Answer 8

Do you understand the process above?

Answer 9

And this could simply be

\[\large \sqrt{p^6}=\sqrt{p*p*p*p*p*p}=\sqrt{p*p*p}*\sqrt{p*p*p}=(\sqrt{p^3})^2\]
A square root and a square cancels out, leaves \(\large p^3\)

Answer 10

yes sir.
thank you

Answer 11

Good! :)