Question

how do I rationalize the denominator of (x-y)/(x+2(√xy)+y) ?

Answer 1

I know the answer to be x-2(√xy)+y, but I do not know how to get there. I know that I need to multiply top and bottom by the conjugate, but I've never done that on a denominator with three terms.

Answer 2

whoops, sorry, the answer is (x-2(√xy)+y)/(x-y), again, I do not exactly know the steps in between.

Answer 3

hints:

\[\large x = \left(\sqrt{x}\right)^2\]

\[\large y = \left(\sqrt{y}\right)^2\]

---------------------------------------

so

\[\large x + 2\sqrt{xy} + y\]

turns into

\[\large \left(\sqrt{x}\right)^2 + 2\sqrt{xy} + \left(\sqrt{y}\right)^2\]

Answer 4

the last expression I wrote is of the form

a^2 + 2ab + b^2

Answer 5

so are you saying I need to rationalize (√x+√y)(√x+√y)? How would its conjugate look like?

Answer 6

the conjugate of a+b is a-b

Answer 7

so the conjugate of √x+√y is √x-√y

Answer 8

you need to apply this conjugate twice (once per factor of √x+√y)

Answer 9

ok, so I need to multiply the numerator (which x-y) by (√x-√y)(√x-√y)?

Answer 10

yes you do

Answer 11

and to balance that out, you multiply the denominator by (√x-√y)(√x-√y)

Answer 12

Ok, here goes, I'm trying it out now.

Answer 13

doing that will rationalize each factor √x+√y

Answer 14

so effectively each conjugate pair will match up

in general if you multiply a+b with a-b, you get

(a+b)(a-b) = a^2 - b^2 ... difference of squares law


notice we now have squares, which undo square roots, so this is the reason why we're doing all this

Answer 15

I got a really ugly answer, there must be something I'm not understanding here. I multiplied the numerator by the conjugate [\[(x-y)*(x-2\sqrt{xy}+y) \] which gives me \[x ^{2}-2x ^{3/2}\sqrt{y}-2\sqrt{x}y ^{3/2}-y\] in the numerator.

Answer 16

let me check that

Answer 17

In the denominator I got \[x ^{2}-2xy+y ^{2}\]

Answer 18

this is what I'm getting for the numerator


\[\large (x-y)(x - 2\sqrt{xy} + y)\]


\[\large x(x - 2\sqrt{xy} + y)-y(x - 2\sqrt{xy} + y)\]


\[\large x^2 - 2x\sqrt{xy} + xy-xy + 2y\sqrt{xy} - y^2\]


\[\large x^2 - 2x\sqrt{xy} + 2y\sqrt{xy} - y^2\]


\[\large x^2 + (-2x + 2y)\sqrt{xy} - y^2\]

Answer 19

the denominator of \[\large x^{2}-2xy+y^{2}\] is correct

Answer 20

and I've just confirmed that

\[\large \frac{x^2 + (-2x + 2y)\sqrt{xy} - y^2}{x^{2}-2xy+y^{2}}\]

is indeed equivalent to the original expression (ie the answer is confirmed)

Answer 21

So essentially we found the answer, just written in a different form?

Answer 22

yes there are multiple ways to write the answer

what did you get?

Answer 23

In the back of the book the answer is (x-2(√xy)+y)/(x-y)

Answer 24

hmm something cancels to make this a lot easier

Answer 25

one sec

Answer 26

Thank you , sir, you are a life saver. I appreciate your time and effort.

Answer 27

oh wow, I made a typo, but let me fix it (it's near the end)

Answer 28

\[\large \frac{x-y}{x + 2\sqrt{xy} + y}\]



\[\large \frac{x-y}{\left(\sqrt{x}\right)^2 + 2\sqrt{xy} + \left(\sqrt{y}\right)^2}\]



\[\large \frac{x-y}{\left(\sqrt{x}+\sqrt{y}\right)^2}\]



\[\large \frac{x-y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\]



\[\large \frac{(x-y)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\]



\[\large \frac{(x-y)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\]



\[\large \frac{(x-y)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left((\sqrt{x})^2-(\sqrt{y})^2\right)\left((\sqrt{x})^2-(\sqrt{y})^2\right)}\]



\[\large \frac{(x-y)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(x-y\right)\left(x-y\right)}\]



\[\large \frac{\cancel{(x-y)}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\cancel{(x-y)}\left(x-y\right)}\]



\[\large \frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{x-y}\]



\[\large \frac{(\sqrt{x})^2-2\sqrt{x}*\sqrt{y}+(\sqrt{y})^2}{x-y}\]



\[\large \frac{x-2\sqrt{xy}+y}{x-y}\]

Answer 29

ok all fixed