Algebra question!?

Question

radical x - radical y
------------------------- =
radical x + radical y

I don't get why when you multiply by the conjugate (radical x - radical y) you get x- 2 radical x, radical y plus y, can someone explain how this works?

Answer 1

The reason why you have to multiply by the conjugate is b/c a fraction can only have an integer or something with out all that junk as the denominator.

So in order to get the integer you need to multiply the fraction by a form of one that'll do that which happens to be (sqrt(x) - sqrt(y))/(sqrt(x) - sqrt(y)).

Multiplying you get:

(x + 2sqrt(xy) + y)/(x-y) >>> this is a correct fraction (remember when multiplying radicals you multiply what's underneath 'em!)

Answer 2

There may be something missing. Your numerator is a result of (a-b)x(a-b)= a2-2ab+b2, just substitute radical x and radical y for a and b respectively. There is still the matter of the denominator. If that is also multiplied by the conjucate, you have the case of (a+b)(a-b)=a2-b2. Substituting again, the denomintor would become x-y

Answer 3

it is just the rule for expanding a square as follows:
(a+b)^2 = a^2 + 2ab + c^2
remember that (√a)^2 = a

so, √x - √y √x - √y √x - √y x -2(√x √y) +y
---------- = ---------- x --------- = -------------------
√x + √y √x + √y √x - √y x -y

Answer 4

rationalise the denominator with the conjugate
rtx-rty(rtx-rty)/(rtx+rty)(rtx-rty)
now for the Nr use the identity (a-b)^2 ,expand and for the denominator (a+b)(a-b) =a2-b^2
=>x+y-2rtxy/x-y

Answer 5

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

Answer 6

1
rad x -rady
--------------
radx+ rady

=
radx
-------------------
radx+rady-rady

=
radx
-------
radx

=
1