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Mathematics 14 Online
OpenStudy (anonymous):

What is the geometric mean of 12 and 18? a. 1.5 b. √6 c. 15 d. 6√6 Also, if you could explain how to find a geometric mean, that would be awesome! Thanks! x

OpenStudy (jdoe0001):

\(\bf \cfrac{12}{{\color{red}{ x}}}=\cfrac{{\color{red}{ x}}}{18}\qquad {\color{red}{ \textit{geometric mean}}}\)

OpenStudy (jdoe0001):

solve for "x"

OpenStudy (anonymous):

I got x=108.

OpenStudy (jdoe0001):

108? hmm how did you get 108?

OpenStudy (anonymous):

I cross multiplied? It's been a while since I've had to do problems like this, so I am probably doing it wrong….

OpenStudy (jdoe0001):

well, cross-multiplying is ... correct.. so \(\bf \cfrac{12}{{\color{red}{ x}}}=\cfrac{{\color{red}{ x}}}{18}\implies 12\cdot 18={\color{red}{ x\cdot x}}\implies 216=x^2\\ \quad \\ \textit{taking }\sqrt{\qquad }\textit{ to both sides}\\ \quad \\ \sqrt{216}=\sqrt{x^2}\implies \sqrt{216}=x\\ \quad \\ {\color{blue}{ 216\implies 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3\implies 2\cdot 2^2\cdot 3\cdot 3^2}}\qquad thus\\ \quad \\ \sqrt{216}=x\implies \sqrt{2\cdot 2^2\cdot 3\cdot 3^2}=x\implies \large \sqrt[{\color{red}{ 2}}]{2\cdot 2^{\color{red}{ 2}}\cdot 3\cdot 3^{\color{red}{ 2}}}=x\)

OpenStudy (jdoe0001):

take what you can from the radical, and leave the rest :)

OpenStudy (anonymous):

Ah, that's what I did wrong! I said 2x, not x squared. This is amazing, thank you so much! So, when I did the square root of 216, I got 14.696938… Would I just round up to 15?

OpenStudy (jdoe0001):

well, is not one of your choices for one, though is correct decimally

OpenStudy (jdoe0001):

notice the radicand, recall that we take out from the radical what MATCHES the root

OpenStudy (anonymous):

Ok, I guess I don't know how to do that, although I know what you're saying. Thank you so so much, by the way.

OpenStudy (jdoe0001):

one sec

OpenStudy (jdoe0001):

\(\Large \sqrt[{\color{red}{ 2}}]{2\cdot {\color{blue}{ 2}}^{\color{red}{ 2}}\cdot 3\cdot {\color{blue}{ 3}}^{\color{red}{ 2}}}=x\implies {\color{blue}{ 2\cdot 3}}\sqrt{2\cdot 3}\)

OpenStudy (jdoe0001):

you take out anything that matches the "root", thus

OpenStudy (anonymous):

Ohh… Thank you sooo much!! You've been so much help, I really appreciate it! One last question, if you don't mind:) I'm working on the same type of question in my assignment: What is the geometric mean of 2 and 36. So I got to finding the square root of 70. How do I break it down like you did to the square root of 216?

OpenStudy (jdoe0001):

70? where did you get 70 from?

OpenStudy (anonymous):

Oh oops, I meant 72

OpenStudy (jdoe0001):

got a calculator?

OpenStudy (anonymous):

yes

OpenStudy (jdoe0001):

ok... try dividing 72 by 2, if you can, again by 2, and on if you can't try some other small number, like 3 and on

OpenStudy (jdoe0001):

what factors would that give you?

OpenStudy (anonymous):

Ok, so it would be broken down into 2·2·2·3·3

OpenStudy (jdoe0001):

\(\bf \cfrac{2}{x}=\cfrac{x}{36}\implies 72=x^2\implies \sqrt{72}=x\\ \quad \\ {\color{blue}{ 72\implies 2\cdot 2\cdot 2\cdot 3\cdot 3\implies 2\cdot 2^2\cdot 3^2}}\\ \quad \\ \sqrt{72}=x\implies \sqrt{2\cdot 2^2\cdot 3^2}=x\)

OpenStudy (jdoe0001):

\(\bf \sqrt{72}=x\implies \sqrt{2\cdot 2^2\cdot 3^2}=x\implies 2\cdot 3\sqrt{2}=x\implies 6\sqrt{2}=x\)

OpenStudy (anonymous):

Ok, that's what I got. So I would have 6√2?

OpenStudy (anonymous):

Awesome. Thank you so much again!

OpenStudy (jdoe0001):

yw

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