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Mathematics 22 Online
OpenStudy (nosnip):

2(3+ sqrt(5)) / (3- sqrt(5))? Express in the form b+c sqrt(5) I got the answer 17 +3 sqrt5 but the correct answer is 7+3 sqrt 5?

zepdrix (zepdrix):

\[\large\rm \frac{2(3+\sqrt5)}{3-\sqrt5}\color{royalblue}{\left(\frac{3+\sqrt5}{3+\sqrt5}\right)}=\frac{2(3+\sqrt5)^2}{3^2-\sqrt5^2}\]

zepdrix (zepdrix):

\[\large\rm =\frac{2(9+6\sqrt5+5)}{9-5}\]

zepdrix (zepdrix):

\[\large\rm =\frac{2(14+6\sqrt5)}{4}\]

zepdrix (zepdrix):

Maybe you can look at those steps and see where you made a boo boo :O

zepdrix (zepdrix):

Hmm I'm not sure quite sure where you came up with 17 :o Would have to see your steps.

OpenStudy (nosnip):

I started by expanding the bracket in the numerator of the first fraction, so I had 6+2 sqrt(5) and multiplied this by 3+ sqrt(5) --> Is this why I got the incorrect answer?

zepdrix (zepdrix):

No, that should lead to the same result :)

zepdrix (zepdrix):

\[\large\rm \frac{(6+2\sqrt5)(3+\sqrt5)}{4}=\frac{18+9\sqrt5+2\cdot5}{4}\]Is this what you got for the expansion on top?

zepdrix (zepdrix):

Or was your bottom different maybe?

OpenStudy (nosnip):

For some reason I have 18+ 12 sqrt(5) +50 on top

OpenStudy (nosnip):

I did 6 x 3, then 6 x sqrt(5), then 3 x 2sqrt(5), then 5 x 2 sqrt 5

zepdrix (zepdrix):

ya ya 12root5 in the middle, my bad! :) woops!

zepdrix (zepdrix):

\[\large\rm \frac{(6+2\sqrt5)(3+\sqrt5)}{4}=\frac{18+12\sqrt5+2\cdot5}{4}\]

OpenStudy (nosnip):

What about the 2.5 at the end?

OpenStudy (nosnip):

How did you get that?

OpenStudy (nosnip):

shoudln't it be 10? as it is 2 x 5 (sqrt5)^5)

OpenStudy (nosnip):

Ohhh that was my mistake!

zepdrix (zepdrix):

Yes, good :) I wrote 2*5, not 2.5. Hopefully that wasn't confusing.

zepdrix (zepdrix):

Ya you did sqrt5*sqrt5 = 25. Silly little mistake there :D

OpenStudy (nosnip):

Thank you very much I understand :))

zepdrix (zepdrix):

yay team \c:/

OpenStudy (nosnip):

:D

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