Question

can someone please help explain..
how to simplify and write in exact radical form
√15x^3 (times) √3x^2

Answer 1

@jdoe0001 do you get how to do this?

Answer 2

one sec

Answer 3

okay

Answer 4

\(\bf \sqrt{15x^3}\cdot \sqrt{3x^2}\implies \sqrt{15\cdot 3\cdot x^{\color{brown}{ 3}}\cdot x^{\color{brown}{ 2}}}\implies \sqrt{45\cdot x^{{\color{brown}{ 3+2}}}}
\\ \quad \\
{\color{blue}{ 45\to 3\cdot 3\cdot 5\to 3^2\cdot 5}} \qquad thus
\\ \quad \\
\sqrt{45\cdot x^{{\color{brown}{ 3+2}}}}\implies \sqrt{3^2\cdot 5\cdot x^5}\implies 3\sqrt{5x^5}
\\ \quad \\
{\color{blue}{ x^5\to x^{2+2+1}\to x^2\cdot x^2\cdot x^1}}\qquad thus
\\ \quad \\
3\sqrt{5x^5}\implies 3\sqrt{5x^2x^2x}\implies 3\cdot x\cdot x\sqrt{5x}\implies 3x^2\sqrt{5x}\)

Answer 5

alright I see how you got the answer but I am unsure about your 3rd line of work where you break up the exponent. Could you explain it ?

Answer 6

@jdoe0001

Answer 7

well... tis a simple exponent rule \(\bf a^n\cdot a^m\implies a^{n+m}\)
and from there you can use the "commutative property"

so... in say... for example if we had.... \(\large a^3\cdot a^{11}\implies a^{3+11}\implies a^{14}
\begin{cases}
a^{9+5}\to a^9\cdot a^5\\
a^{10+4}\to a^{10}\cdot a^4\\
a^{7+7}\to a^7\cdot a^7\\
a^{20-6}\to a^{10}\cdot a^{-6}\\
\textit{and so on ....}
\end{cases}\)

Answer 8

hmmm well \(\large a^3\cdot a^{11}\implies a^{3+11}\implies a^{14}
\begin{cases}
a^{9+5}\to a^9\cdot a^5\\
a^{10+4}\to a^{10}\cdot a^4\\
a^{7+7}\to a^7\cdot a^7\\
a^{20-6}\to a^{20}\cdot a^{-6}\\
\textit{and so on ....}
\end{cases}\)

Answer 9

in the case above, you had \(x^5\)

and the root is square, or root 2
so we took out all "2" possibles out of 5
so 2 + 2 + 1

and those come out of the radical

we pretty much did the same with 45 if you notice
broke down to 3 * 3 * 5
so we took out all possible squares, thus \(3^2\)

Answer 10

thank you so much!

Answer 11

yw