# [SOLVED] roots

• March 28th 2009, 02:45 AM
princess_21
[SOLVED] roots
here's the problem.
i don't know how to solve this (Shake), can you help me? thanks.
$\frac{(^4\sqrt{4b^4})+(^4\sqrt{9b^4}-(\sqrt{12a^2}-(^4\sqrt{64a^4}+(\sqrt{20a^2}-(\sqrt{5b^2})}{b-2a}^{-1}$
• March 28th 2009, 03:21 AM
earboth
Quote:

Originally Posted by princess_21
here's the problem.
i don't know how to solve this (Shake), can you help me? thanks.

The first exemple must be an olympic exercise:

$\left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} =$

$\left(\dfrac{b\sqrt{2} + b\sqrt{3}-2a\sqrt{3}-2a\sqrt{2} + 2a\sqrt{5} - b\sqrt{5}}{b-2a} \right)^{-1} =$

$\left(\dfrac{b \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)-2a \left(\sqrt{3}+\sqrt{2} - \sqrt{5}\right) }{b-2a} \right)^{-1} =$

$\left(\dfrac{ \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)(b-2a) }{b-2a} \right)^{-1} = \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}}$

Now eliminate the square roots in the denominator:

$\dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}} \cdot \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{\sqrt{2} + \sqrt{3}+\sqrt{5}} = \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2+2\sqrt{6}+3-5} =$

$\dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30} \right)$

phooo!

The next exemple is much easier. Show your work and I'M going to help you if you don't succeed. (You should get $2\sqrt{6}$)
• March 28th 2009, 03:39 AM
princess_21
Quote:

Originally Posted by earboth
The first exemple must be an olympic exercise:

$\left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} =$

$\left(\dfrac{b\sqrt{2} + b\sqrt{3}-2a\sqrt{3}-2a\sqrt{2} + 2a\sqrt{5} - b\sqrt{5}}{b-2a} \right)^{-1} =$ how did it become like this?

$\left(\dfrac{b \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)-2a \left(\sqrt{3}+\sqrt{2} - \sqrt{5}\right) }{b-2a} \right)^{-1} =$

$\left(\dfrac{ \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)(b-2a) }{b-2a} \right)^{-1} = \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}}$

Now eliminate the square roots in the denominator:

$\dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}} \cdot \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{\sqrt{2} + \sqrt{3}+\sqrt{5}} = \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2+2\sqrt{6}+3-5} =$

$\dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30} \right)$

phooo!

The next exemple is much easier. Show your work and I'M going to help you if you don't succeed. (You should get $2\sqrt{6}$)

i'm still confused... can you help me again?
• March 28th 2009, 05:19 AM
earboth
Quote:

Originally Posted by princess_21
i'm still confused... can you help me again?

I'll show you how to do the second question, but if I shall help you I must know what exactly is confusing you.

Ooooh, I finally found your question in the quoted calculations.

If you have $\left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} =$ then you can simplify

$\sqrt[4]{4b^4} = (2^2 \cdot b^4)^{\frac14} = 2^{\frac24} \cdot b^{\frac44} = b\cdot 2^{\frac12} = b\cdot \sqrt{2}$

Apply similar transformations with all summands in the numerator and you'll get the next line of my calculations.

And now the 2nd question:

$\left(\dfrac{a}{\sqrt{2}} \div (2a\sqrt{3})\right)^{-1} = \left(\dfrac{a}{\sqrt{2}} \cdot \dfrac1{2a\sqrt{3}}\right)^{-1} =$ $\left(\dfrac1{2\sqrt{6}} \right)^{-1} = 2\sqrt{6}$
• March 28th 2009, 05:31 AM
princess_21
yes understand the second equation. I don't understand the first equation. My teacher told me that I need to change all the index so they can be combined. I change them all as fourth root by raising all square root to 2. is that correct?
• March 28th 2009, 05:39 AM
earboth
Quote:

Originally Posted by princess_21
yes understand the second equation. I don't understand the first equation. My teacher told me that I need to change all the index so they can be combined. I change them all as fourth root by raising all square root to 2. is that correct?

That is possible, of course. But you'll deal with quite large numbers. For instance:

$\sqrt{12a^2} = \sqrt[4]{144a^4}$

while I prefer the transform:

$\sqrt{12a^2} = 2a\sqrt{3}$

To me this form is much easier to handle.
• March 28th 2009, 05:50 AM
princess_21
http://www.mathhelpforum.com/math-he...4ef637fc-1.gif

Ok I got it. but I don't understand why $\frac{\sqrt{2} + \sqrt{3} +\sqrt{5}}{2\sqrt{6}}= \frac{1}{12}$
• March 28th 2009, 05:57 AM
earboth
Quote:

Originally Posted by princess_21
http://www.mathhelpforum.com/math-he...4ef637fc-1.gif

Ok I got it. but I don't understand why $\frac{\sqrt{2} + \sqrt{3} +\sqrt{5}}{2\sqrt{6}}= \frac{1}{12}$

Unfortunately you quoted me wrong:

$
\dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac{{\color{red}\bold{\sqrt{6}}} (\sqrt{2} + \sqrt{3}+\sqrt{5})}{2\sqrt{6} \cdot {\color{red}\bold{\sqrt{6}}}} = \dfrac{\sqrt{12}+\sqrt{18} + \sqrt{30}}{12} =$
$\dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30} \right)
$
• March 28th 2009, 06:26 AM
princess_21
yeah I got it thanks
I'll try to continue from here

$\frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})$

$\frac{4\sqrt{18}+6\sqrt{12}+2\sqrt{180}}{12}$

$\frac{12\sqrt{2}+12\sqrt{3}+12\sqrt{5}}{12}$

is this correct?
• March 28th 2009, 06:34 AM
earboth
Quote:

Originally Posted by princess_21
yeah I got it thanks
I'll try to continue from here

$\frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})$

$\frac{4\sqrt{18}+6\sqrt{12}+2{180}}{12}$

$\frac{12\sqrt{2}+12\sqrt{3}+6\sqrt{5}}{12}$

is this correct?

Honestly, I don't know what you did and why?

All my calculations were necessary to get a rational denominator. Thus I multiplied numerator and denominator by $\sqrt{6}$ which I marked in red. That means I didn't change the value of the quotient.

$2\cdot \sqrt{6}\cdot \sqrt{6} = 12$

The product of the numerator and $\sqrt{6}$ is

$\sqrt{12}+\sqrt{18} + \sqrt{30} = \sqrt{4 \cdot 3}+\sqrt{9 \cdot 2} + \sqrt{30}$ which I simplified to

$2\sqrt{3}+3\sqrt{2} + \sqrt{30}$

That's all.
• March 28th 2009, 06:55 AM
princess_21

the two attachments are just one question the first one multiplied by the second one.
this is the original equation. I multiplied the answers.
• March 28th 2009, 07:01 AM
earboth
Quote:

Originally Posted by princess_21
yeah I got it thanks
I'll try to continue from here

$\frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})$

$\frac{4\sqrt{18}+6\sqrt{12}+2{180}}{12}$

$\frac{12\sqrt{2}+12\sqrt{3}+6\sqrt{5}}{12}$

is this correct?

Got it - better late then never :D

Simplify this expression a little bit:

$\frac{12\sqrt{2}+12\sqrt{3}+6\sqrt{5}}{12} = \sqrt{2} + \sqrt{3} + \dfrac{1}{2}\sqrt{5}$