Find the exact value of the radical expression in simplest form. √p^6+√9p^6+3√p^6-√p^2
\[\large \sqrt{p^6}+\sqrt{9p^6}+3\sqrt{p^6}-\sqrt{p^2}\]
Alright, we know that \[\large \sqrt{p^6}=\sqrt{p*p*p*p*p*p}=\sqrt{p^4}\sqrt{p^2}=p^2\sqrt{p^2}\]
\[\large p^2\sqrt{p^2}=p^2*p=p^3\]
Let's apply this for all \(\large \sqrt{p^6}\)
\[\large \begin{align}&=\sqrt{p^6}+\sqrt{9p^6}+3\sqrt{p^6}-\sqrt{p^2}\\&=p^3+\sqrt{9}p^3+3p^3-\sqrt{p^2} \end{align}\]
\[\sqrt{9}=3;~\sqrt{p^2}=p\]
\[\large \begin{align}&=p^3+\sqrt{9}p^3+3p^3-\sqrt{p^2}\\&= p^3+3p^3+3p^3-p\end{align}\] Group similar terms \[\large =10p^3-p\] And then you can simply factor one p out \[\large = p(10p^2-1)\]
Do you understand the process above?
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 \(\large p^3\)
yes sir. thank you
Good! :)
Join our real-time social learning platform and learn together with your friends!