Surds Equation Help

[shaoen01]

Hi all,

How do you do the question below?

Express \(\displaystyle (\frac{\sqrt{5} - 1}{\sqrt{5} + 2})^2\) in the form of \(\displaystyle A + B \sqrt(5)\), where A and B are integers.

\(\displaystyle (\frac{\sqrt{5} - 1}{\sqrt{5} + 2})^2
= [\frac{(\sqrt{5} - 1)(\sqrt{5} - 2)}{(\sqrt{5} + 2)(\sqrt{5} - 2)}]^2
= [\frac{(5 - 2\sqrt{5} - \sqrt{5} + 2)}{(\sqrt{5} - 2)^2}]^2
= [\frac{5 - 3\sqrt{5} + 2}{5-4}]^2
= (\frac{7-3\sqrt{5}}{1})^2
=49 - 9\sqrt{5}\)

I can't seem to be able to get \(\displaystyle 94 - 42\sqrt{5}\)
Thanks

 

[stapel]

Note: (sqrt[5] - 2)(sqrt[5] + 2) = 5 - 4, not 5 - 2.

Eliz.

 

[soroban]

Hello, shaoen01!

Express \(\displaystyle \L\,\left(\frac{\sqrt{5}\,-\,1}{\sqrt{5}\,+\,2}\right)^2\) in the form of \(\displaystyle A\,+\,B\sqrt{5}\), where \(\displaystyle A\) and \(\displaystyle B\) are integers.

Answer: \(\displaystyle \,94\,-\,42\sqrt{5}\)

Rationalize the "inside" first . . .

\(\displaystyle \L\;\frac{\sqrt{5}\,-\,1}{\sqrt{5}\,+\,2}\,\cdot\,\frac{\sqrt{5}\,-\,2}{\sqrt{5}\,-\,2} \;=\;\frac{5\,-\,2\sqrt{5}\,-\,\sqrt{5}\,+\,2}{5\,-\,2\sqrt{5}\,+\,2\sqrt{5}\,-\,4} \;=\;\frac{7\,-\,3\sqrt{5}}{1}\)\(\displaystyle \;=\;7\,-\,3\sqrt{5}\)


Then square:

\(\displaystyle \;\left(7\,-\,3\sqrt{5}\right)^2 \;=\;\left(7\,-3\sqrt{5}\right)\left(7\,-\,3\sqrt{5}\right) \;=\;7\cdot7\,-\,21\sqrt{5}\,-\,21\sqrt{5} \,+\,\left(3\sqrt{5}\right)\left(3\sqrt{5}\right)\)

\(\displaystyle \;\;\;\;=\;49 \,-\,42\sqrt{5}\,+\,45\;=\;\L94\,-\,42\sqrt{5}\)

 

[shaoen01]

Hello, shaoen01!

Rationalize the "inside" first . . .

\(\displaystyle \L\;\frac{\sqrt{5}\,-\,1}{\sqrt{5}\,+\,2}\,\cdot\,\frac{\sqrt{5}\,-\,2}{\sqrt{5}\,-\,2} \;=\;\frac{5\,-\,2\sqrt{5}\,-\,\sqrt{5}\,+\,2}{5\,-\,2\sqrt{5}\,+\,2\sqrt{5}\,-\,4} \;=\;\frac{7\,-\,3\sqrt{5}}{1}\)\(\displaystyle \;=\;7\,-\,3\sqrt{5}\)


Then square:

\(\displaystyle \;\left(7\,-\,3\sqrt{5}\right)^2 \;=\;\left(7\,-3\sqrt{5}\right)\left(7\,-\,3\sqrt{5}\right) \;=\;7\cdot7\,-\,21\sqrt{5}\,-\,21\sqrt{5} \,+\,\left(3\sqrt{5}\right)\left(3\sqrt{5}\right)\)

\(\displaystyle \;\;\;\;=\;49 \,-\,42\sqrt{5}\,+\,45\;=\;\L94\,-\,42\sqrt{5}\)

Hi Soroban,

You mentioned that i should rationalize inside first. But i am just curious about something, \(\displaystyle (7 - 3\sqrt{5})^2 = (7 - 3\sqrt{5})(7 - 3\sqrt{5})\) shouldn't it be \(\displaystyle (7 - 3\sqrt{5})^2 = (7 - 3\sqrt{5})(7 + 3\sqrt{5})\)? But after working that out, it does not seem right.

I mean since
\(\displaystyle [\frac{(\sqrt{5} - 1)(\sqrt{5} - 2)}{(\sqrt{5} + 2)(\sqrt{5} - 2)}]^2
= [\frac{(5 - 2\sqrt{5} - \sqrt{5} + 2)}{(\sqrt{5} - 2)^2}]^2\)

What i am referring to is the \(\displaystyle (\sqrt{5} - 2)^2\)

I've done my working this way and not sure if i got the answer by sheer luck so let me if i am wrong.

\(\displaystyle (7 - 3\sqrt{5})^2
= (7)^2 - 2(7)(3\sqrt{5}) + (3\sqrt{5})^2)
= 49 - 42\sqrt{5} + 45
= 94 - 42\sqrt{5}\)

 

[Gene]

Nope, you have no reason to change the sign.
(a-b)^2 = (a-b)(a-b) not (a-b)(a+b).
Then (a-b)(a+b) = a^2-b^2 not (a-b)^2

 

[Denis]

Hey Gene, you old bean, where you been :?:

 

[shaoen01]

Hi Gene,

I think i must have mistook a^2-b^2 as the same as (a-b)^2. Thanks for the clarification.

 

[shaoen01]

Hi,

I could also take (a-b)^2 = a^2 - 2(a)(b) + b^2 right?

 

[Denis]

I could also take (a-b)^2 = a^2 - 2(a)(b) + b^2 right?

YES; (a-b)^2 = a^2 - 2ab + b^2, and NOTHING else!

AND (a+b)^2 = a^2 + 2ab + b^2