MHB Simplifying radicals - Help with basic number manipulation

[Comscistudent]

Hi I'm trying to give myself a refresher in Leaving Cert maths and I'm running through some problems. Here's one which has me stumped (sorry I can't figure out how to show the actual symbols on the post, it's just showing as raw LaTEX when I try )

Combine terms and simplify the expression of -

(sqrt6/sqrt7) * sqrt21

The actual answer is that you move the /sqrt7 to the sqrt21 to end up with sqrt6 * sqrt3 = 2sqrt3

I missed this and instead multiplied the right hand term by sqrt7/sqrt7 but my answer is different. Can someone explain why my logic is incorrect?

(sqrt6/sqrt7) * ( (sqrt21*sqrt7)/sqrt7 )
sqrt6/sqrt7 * sqrt147/sqrt7
sqrt882/sqrt7
3sqrt14

 

[Greg]

Hi TubeAlloy and welcome to MHB! :D

$$\dfrac{\sqrt6}{\sqrt7}\cdot\sqrt{21}=\sqrt{18}=3\sqrt2$$

(sqrt6/sqrt7) * ( (sqrt21*sqrt7)/sqrt7 )
sqrt6/sqrt7 * sqrt147/sqrt7
sqrt882/sqrt7
3sqrt14

Your logic is fine but you've made an error in your calculation. Can you spot it?

Quote this post to see how I coded the $\LaTeX$.

 

[Comscistudent]

Oh wow I feel so silly, thanks a million for the help once I knew I wasn't doing something wrong I was able to spot the error.

$$\dfrac{\sqrt6}{\sqrt7}\cdot\dfrac{\sqrt147}{\sqrt7}$$

This is $$\dfrac{\sqrt882}{\sqrt49}$$ not $$\dfrac{\sqrt882}{\sqrt7}$$ as I had thought

So then it's $$\sqrt18$$ == $$3\sqrt2$$

 

[Greg]

Good work!

To get all of the numbers in a radical under the square root sign use \sqrt{123}. Note the curly braces. :)

 

[soroban]

Combine terms and simplify the expression: \frac{\sqrt{6}}{\sqrt{7}}\cdot\sqrt{21}

The actual answer is that you move the /sqrt7 to the sqrt21 to end up with sqrt6 * sqrt3 = 2sqrt3

I missed this and instead multiplied the right hand term by sqrt7/sqrt7 . Why?
but my answer is different. Can someone explain why my logic is incorrect?

(sqrt6/sqrt7) * ( (sqrt21*sqrt7)/sqrt7 )
sqrt6/sqrt7 * sqrt147/sqrt7
sqrt882/sqrt7
3sqrt14

Did some teacher tell you, "To simplify radicals,
introduce more radicals into the expression" ?

Here is the recommended way to simplify it:

. . \begin{array}{ccc}<br /> \dfrac{\sqrt{6}}{\sqrt{7}}\cdot\sqrt{21} &amp;=&amp; \sqrt{6}\cdot\dfrac{\sqrt{21}}{\sqrt{7}} \\ <br /> &amp;= &amp; \sqrt{6}\cdot\sqrt{\dfrac{21}{7}} \\ <br /> &amp; = &amp; \sqrt6\cdot\sqrt{3} \\<br /> &amp; = &amp; \sqrt{18} \\<br /> &amp;=&amp; \sqrt{9\cdot2} \\<br /> &amp;=&amp; \sqrt{9}\cdot\sqrt{2} \\<br /> &amp;=&amp; 3\sqrt{2} \end{array}

 

[Greg]

soroban has made an excellent point: the simpler the better! It's good to see experimentation with other methods though, and I think it's good that the error was spotted. Sometimes the simpler approach is not always realized so it's a good thing to be able to adapt one's skill set to the problem at hand. But at the end of the day, I think striving for simplicity is the best approach.