Simplifying radicals: can’t identify the error I’m making

[JeffM]

sorry for the delay in getting back to you.

So the denominator has to be factored so that [MATH]sqrt[/MATH]3 can be divided by [MATH]sqrt[/MATH]3?

Yes.

[MATH]\sqrt{ab} \equiv \sqrt{a} * \sqrt{b}.[/MATH]
Here is an example:

[MATH]\sqrt{4 * 25} = \sqrt{100} = 10 = 2 * 5 = \sqrt{4} * \sqrt{25}.[/MATH]
And this is true also

[MATH]a \ge 0 \text { and } b > 0 \implies \sqrt{\dfrac{a}{b}} \equiv \dfrac{\sqrt{a}}{\sqrt{b}}.[/MATH]
Here is an example:

[MATH]\sqrt{\dfrac{36}{4}} = \sqrt{9} = 3 = \dfrac{6}{2} = \dfrac{\sqrt{36}}{\sqrt{4}}.[/MATH]
It is very easy to make mistakes with these identities if you do not go step by step. I take it step by step myself, and I have been working with radicals long before you were born.

So let's see you finish up this problem.

 

[Audentes]

So let's see you finish up this problem.

4n²√(1/2n)

 

[JeffM]

4n²√(1/2n)

That is correct mathematically. But again it is probably not what they expect. They almost certainly want you to simplify to this.

[MATH]4n^2 * \dfrac{1}{\sqrt{2n}} = 4n^2 * \dfrac{1}{\sqrt{2n}} * \dfrac{\sqrt{2n}}{\sqrt{2n}} =[/MATH]
[MATH]4n^2 * \dfrac{\sqrt{2n}}{2n} = 2n\sqrt{2n}.[/MATH]
This whole process is called "rationalizing the denominator." The point is that we do not want radicals in the denominator if they can be avoided without too much trouble. At one time (back when I rode the orange-painted mammoth to school), there were very good reasons for that. Today, it just ensures that answers are easily comparable.

 

[Audentes]

great, thank you so much for the help!
@JeffM and also @Dr.Peterson

 

[tkhunny]

View attachment 25500

I didn't see any other comment tot he very first question. Are those expressions the same?

[math]\dfrac{\sqrt{a}}{\sqrt{b}}\;and\;\sqrt{\dfrac{a}{b}}[/math]
Try a = -2 and b = -4 and see if they are the same.

 

[Audentes]

View attachment 25500

I didn't see any other comment tot he very first question. Are those expressions the same?

[math]\dfrac{\sqrt{a}}{\sqrt{b}}\;and\;\sqrt{\dfrac{a}{b}}[/math]
Try a = -2 and b = -4 and see if they are the same.

Oh it looks like I missed his post while posting a reply to JeffM. thank you for alerting me @tkhunny

@lex
thank you for pointing that out with the root on the whole equation and when it is on the numerator and the denominator. That adds to the clarity for me.
Sorry for missing it, thank you too for the help!

 

[lex]

You're welcome

 

[lex]

I didn't see any other comment tot he very first question. Are those expressions the same?

[math]\dfrac{\sqrt{a}}{\sqrt{b}}\;and\;\sqrt{\dfrac{a}{b}}[/math]
Try a = -2 and b = -4 and see if they are the same.

 

[Audentes]

View attachment 25521

Nice comparison