kindly click on the attachment

I really need your help urgently! after understanding your lesson I have 30 more items of that remaining to answer

thank you

2. Originally Posted by Aussieboy

I really need your help urgently! after understanding your lesson I have 30 more items of that remaining to answer...
Hello,

please tell us what are you supposed to do with those terms!

EB

3. Originally Posted by Aussieboy

kindly click on the attachment

I really need your help urgently! after understanding your lesson I have 30 more items of that remaining to answer

thank you
What exactly are you supposed to do with them? I can guess at the first one (if I'm reading it correctly)
$\sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{\sqrt{16}} = \frac{\sqrt{3}}{\sqrt{4^2}}$ $= \frac{\sqrt{3}}{4}$

2 - 5 are already in simplest form and

6. $\sqrt[4]{m^{2a}n^4} = \sqrt[4]{m^2a} \sqrt[4]{n^4} = n \sqrt[4]{m^{2a}}$
(Technically, since $\sqrt[4]{x^2} = (x^2)^{1/4} = x^{2/4} = |x|^{1/2} = \sqrt{|x|}$ this last can also be written $n \sqrt{|m|^a}$. Similarly 4 can be written as $\sqrt{2|xy|}$ and 5 as $\sqrt{|a - 3b|}$.)

-Dan

4. Originally Posted by earboth
Hello,

please tell us what are you supposed to do with those terms!

EB

5. Originally Posted by topsquark
What exactly are you supposed to do with them? I can guess at the first one (if I'm reading it correctly)
$\sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{\sqrt{16}} = \frac{\sqrt{3}}{\sqrt{4^2}}$ $= \frac{\sqrt{3}}{4}$

2 - 5 are already in simplest form and

6. $\sqrt[4]{m^{2a}n^4} = \sqrt[4]{m^2a} \sqrt[4]{n^4} = n \sqrt[4]{m^{2a}}$
(Technically, since $\sqrt[4]{x^2} = (x^2)^{1/4} = x^{2/4} = |x|^{1/2} = \sqrt{|x|}$ this last can also be written $n \sqrt{|m|^a}$. Similarly 4 can be written as $\sqrt{2|xy|}$ and 5 as $\sqrt{|a - 3b|}$.)

-Dan

Kindly elaborate all the steps and how to get the rationalizing factor?
try this other examples

6. Originally Posted by Aussieboy
Kindly elaborate all the steps and how to get the rationalizing factor?
try this other examples
Hello,

I'm refering to the 2nd collection of problems (Radicals 2.doc)
(By the way: Do us and yourself a favour and start a new thread if you have new and other problems to solve. Otherwise you risk that noboby will notice these new problems)

to #1: This is in simplest form already.

to #2:
$\sqrt[4]{81m^4}=\sqrt[4]{3^4 \cdot m^4}=3m$

to #3:
$\sqrt[4]{81a^2}=\sqrt{\sqrt{3^4 \cdot a^2}}=3 \cdot \sqrt{a}$
I used the property that $\sqrt[4]{x}=\sqrt{\sqrt{x}}$

to #4:
$5(a+b) \cdot \sqrt{a+b}$
You know that $x = \sqrt{x^2},\ x>0$
$5(a+b) \cdot \sqrt{a+b}=5 \sqrt{(a+b)^2} \cdot \sqrt{a+b}= 5 \cdot \sqrt{(a+b)^3}$