A little help?

[WTF?]

I have a little problem to simplify that I'm a bit confused about,

I try it, and I get this:

I don't know if I'm right, so please help me.

I also have this one that's confusing as well:

 

[Matt]

I have a little problem to simplify that I'm a bit confused about,

I try it, and I get this:

Absolutely correct.

I also have this one that's confusing as well:

Both of those are the same, yes.

 

[WTF?]

Thank you Sir.

 

[zerodegrees]

Remember, roots are still exponents.

They are instead FRACTIONAL POWERS.

So...a square root would be RAISED TO THE ONE-HALF POWER.

A cube root would be RAISED TO THE ONE-THIRD POWER.

Hope this helps.

 

[WTF?]

Hello, I'm in need of another push again. :lol:

I have several more like it, so I hope you can help me, and thanks btw.

 

[pka]

\(\displaystyle \L
\sqrt[5]{{96p^5 q^6 }} = 2pq\sqrt[5]{{3q}}\)

\(\displaystyle \L
\sqrt[3]{{128m^9 }} = \sqrt[3]{{2^7 m^9 }} = 2^2 \cdot m^3 \cdot \sqrt[3]{2}\)

 

[WTF?]

Hmm...the other two are right?

 

[pka]

Hmm...the other two are right?

I did not mean to imply that!
I expected you to study and see if they are correct.

 

[WTF?]

Hmm...the other two are right?

I did not mean to imply that!
I expected you to study and see if they are correct.

Ah, ok.

But I still don't know how you got the first two. :x Why is there two q's?

But when I try a value, it seems to work. :x

 

[Eva Barragan]

Mr. Danielzyck?

 

[pka]

Look at this example \(\displaystyle \L
\sqrt[3]{{2^{10} a^7 b^6 c^4 }} = 2^3 a^2 b^2 c\sqrt[3]{{2ac}}\).

See divide by the root and take the remainder. Do you see how it works?

 

[WTF?]

Do you see how it works?

:?

The book doesn't say that, it just leaves the power as is if it doesn't divide by the root evenly. :?:

 

[pka]

It is a sorry state of mathematical notation!
The ‘root’ notation is the most confused!
Many who write about this have no clear idea themselves what it means
All I can say is “Go with your book or your instructor.”
Know that either may be incorrect!

 

[WTF?]

It is a sorry state of mathematical notation!
The ‘root’ notation is the most confused!
Many who write about this have no clear idea themselves what it means
All I can say is “Go with your book or your instructor.”
Know that either may be incorrect!

I tried substituting a value into yours and I got two distinct answers though. :?:

 

[WTF?]

Last but not least, I have another question.

"Find: a. all the real 6th root of 15,625 b. \(\displaystyle \sqrt[6]{{15,625}}\)

c. 15,625^1/6

:?: :?:

 

[pka]

I tried substituting a value into yours and I got two distinct answers though. :?:

You must have made mistakes in the way you calculated!

 

[pka]

I began working on this problem in 1964.
I have given many talks and workshops to/for in-service teachers.
I can tell you that I have had the most trouble with radicals.

It is so simple I think: the nth root of k is a number such that the nth power is k.
That is if \(\displaystyle \L
\sqrt[n]{b} = c\quad \Rightarrow \quad \left( c \right)^n = b\) .
Therefore, \(\displaystyle \L
\sqrt[n]{{x^m }}\quad \Rightarrow \quad \left( {x^m } \right)^{1/n} \quad \Rightarrow \quad x^{m/n}\) .

\(\displaystyle \L
\frac{{10}}{3} = 3 + \frac{1}{3}\) so that \(\displaystyle \L
\sqrt[3]{{x^{10} }} = 3\sqrt[3]{x}\) .

But we must be very careful a domain issuers.
Look at this \(\displaystyle \L
\sqrt[4]{{x^8 y^6 }} = x^2 |y|\sqrt {|y|}\) .

You must see that the absolute value insures the proper domain

 

[WTF?]

okay thanks.

\(\displaystyle \sqrt[6]{{x}}\)

X has two solutions right? if the root is even?