Question

how does this:

-3 +5√(2k)
-------------------
-2k +4√(5k^2)


simplify to this:

-3 -6√(5) +5√(2k) +10√(10k)
-------------------------------------
38k

Answer 1

even a clue based on some web site to read would be helpful .. anything

Answer 2

Rationalizing the denominator means multiplying by \[\frac{4√(5k^2) }{ 4√(5k^2) }\] I'm assuming this will give you the simplified version.

Answer 3

yes

Answer 4

yeah I just simplified the first problem on paper, it doesn't simplify to the second one any way I can see

Answer 5

if i multiplied

-3 +5√(2k) -2k +4√(5k^2)
------------------- [multiply by] -------------------
-2k +4√(5k^2) -2k +4√(5k^2)

then cancel out "-2k +4√(5k^2) ".
I end up with the the same equation i started with.

-3 +5√(2k)
-------------------
-2k +4√(5k^2)

Answer 6

the equation to simplify was generated by a windows program that creates work sheets in mathematics. it may be an erroneous answer. i would still like to know how to simplify the equation.

Answer 7

no no you don't multiply by the -2k when rationalizing the denominator. That's already rational. You multiply by the radical number which is 4sqrt(5x^2) over itself only.

Answer 8

thanks for attempting hoa :)

Answer 9

\[\frac{ -3 +5√(2k) }{ -2k +4√(5k^2)} * \frac{ 4√(5k^2)}{ 4√(5k^2) }\] This will rationalize the denominator, which is the only simplification i would make.

Answer 10

then the resulting multiplication becomes:



-3 +5√(2k) 4√(5k^2)
------------------- [multiply by] -------------------
-2k +4√(5k^2) 4√(5k^2)

-3 +5√(2k)
--------------------------------------
-2k [multiplied by] 4√(5k^2)

Answer 11

that's wrong. You have to distribute in the numerator, and when you multiply two square roots like you're doing in the denominator it erases the square root symbol.

Answer 12

Which is the entire point of "rationalizing" usually you don't want a radical number in the denominator.

Answer 13

i clearly do not know how to do this at all.

Answer 14

let me work it out on paper and post my own result to rationalizing and you can try and follow the logic

Answer 15

thanks :)

Answer 16

this stuff is new to me, i have only been studying for less then a month. i have a test in a couple of weeks.

Answer 17

\[\frac{ -3+5\sqrt2k }{ -2k+4\sqrt(5k^2) } * \frac{4\sqrt(5k^2) }{ 4\sqrt(5k^2) } =\]

\[\frac{ -12\sqrt(5k^2) + 20(\sqrt(10k^3)}{ -2k+16*5k^2} =\]

\[\frac{ -12\sqrt(5k^2) + 20(\sqrt(10k^3)}{ -2k+80k^2} =\]

Something like this, I'm sure I made a computation error somewhere but this is basically it. You distribute in the numerator and the square root cancels out in the bottom because obviously \[\sqrt(2) * \sqrt(2) = 2\]

Answer 18

I should have distributed in the bottom as well, I'm retarded. Hopefully you get the principal though.

Answer 19

yeah, it makes sense :) i have never multiplied square-roots before. i had not known it was possible until i reviewed your example.

would the bottom distribute out to be :

-8k√(5k^2) + 80k^2