Question

ack - radicals in the denominator: simplify (-L * sqrt (5/2)) / (1- sqrt (5/2))

Answer 1

\[\frac{ -L \sqrt{5/2} }{1-\sqrt{5/2} }\]

Answer 2

wolfram says the answer is \[\frac{ 1 }{ 3 }L(5-\sqrt{10}) \]

and no idea how it gets there from here

Answer 3

\[\Large \frac{-L\sqrt{\frac{5}{2}}}{1-\sqrt{\frac{5}{2}}}\]
Mmmm this one is a tad tricky due to all the fractions.
I would recommend first multiplying through by 2/2.
Err actually let's multiply through by \(\large \sqrt2/\sqrt2\).
Ya ya ya, I think that will work out nicely.

Answer 4

\[\Large \frac{-L\sqrt5}{\sqrt2-\sqrt5}\]Understand that step?

Answer 5

ok because sqrt 5/2 is also sqrt 5/sqrt 2

Answer 6

true story :D

Answer 7

then can multiply by sqrt 2 - sqrt 5

Answer 8

Mmm close! :) We want to multiply by the `conjugate` of our denominator.

\[\Large \frac{-L\sqrt5}{\sqrt2-\sqrt5}\color{royalblue}{\left(\frac{\sqrt2+\sqrt5}{\sqrt2+\sqrt5}\right)}\]

Answer 9

that conjugate trick always seems like cheating to me

Answer 10

why's that? :x

Answer 11

sure go ahead and just change signs all willynilly any other time and see how far you get =)

Answer 12

I mean, the only sneaky thing we're doing is multiplying by 1 in a strange way.\[\Large \color{royalblue}{\left(\frac{\sqrt2+\sqrt5}{\sqrt2+\sqrt5}\right)} \quad=\quad 1\]

Answer 13

The reason we multiply by the conjugate is because it produces the `difference of squares` when you multiply out all the terms.\[\Large (a-b)(a+b)=a^2-b^2\]
And that gets rid of the roots for us.

Answer 14

I agree that it works...it is just....not an obvious move and therefore sneaky

so that gets a negative 3 in the denominator and (-L)(sqrt 5) (sqrt 2 + sqrt5) on top

negative L and negative 3 explains Wolfram's 1/3

Answer 15

Once you get into the math groove, it becomes a rather obvious move :\
But maybe that's just my experience.. :p

Answer 16

distribute the sqrt 5 gets (sqrt 5 * sqrt 2) = sqrt 10
and sqrt 5 * sqrt 5 is 5 and look the answer pops right out

Answer 17

Speaking of which, it's NOT a move you have to make.
You could instead just do the subtraction in the denominator instead of fussing with the conjugate multiplication.

It's just a little trickier to do it that way.

Answer 18

my mileage will vary...
well thanks a ton hated leaving the answer half done especially since I knew I'd set it up right

one medal for you

Answer 19

:3