Please help me! can anyone teach me to answer radicals step by step?
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)
$\displaystyle \sqrt{\frac{3}{16}} = \frac{\sqrt{3}}{\sqrt{16}} = \frac{\sqrt{3}}{\sqrt{4^2}}$$\displaystyle = \frac{\sqrt{3}}{4}$
2 - 5 are already in simplest form and
6. $\displaystyle \sqrt[4]{m^{2a}n^4} = \sqrt[4]{m^2a} \sqrt[4]{n^4} = n \sqrt[4]{m^{2a}}$
(Technically, since $\displaystyle \sqrt[4]{x^2} = (x^2)^{1/4} = x^{2/4} = |x|^{1/2} = \sqrt{|x|}$ this last can also be written $\displaystyle n \sqrt{|m|^a}$. Similarly 4 can be written as $\displaystyle \sqrt{2|xy|}$ and 5 as $\displaystyle \sqrt{|a - 3b|}$.)
-Dan
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:
$\displaystyle \sqrt[4]{81m^4}=\sqrt[4]{3^4 \cdot m^4}=3m$
to #3:
$\displaystyle \sqrt[4]{81a^2}=\sqrt{\sqrt{3^4 \cdot a^2}}=3 \cdot \sqrt{a}$
I used the property that $\displaystyle \sqrt[4]{x}=\sqrt{\sqrt{x}}$
to #4:
I assume that your problem reads:
$\displaystyle 5(a+b) \cdot \sqrt{a+b}$
You know that $\displaystyle x = \sqrt{x^2},\ x>0$
$\displaystyle 5(a+b) \cdot \sqrt{a+b}=5 \sqrt{(a+b)^2} \cdot \sqrt{a+b}= 5 \cdot \sqrt{(a+b)^3}$
EB