Algebra Question Bank #4

[cliffordlfoster]

If (sqrt x + sqrt y)/(x-y) = (2sqrt x + 2sqrt y)/(x + 2sqrt xy +y), what is the ratio of x to y?

A) 1/2
B) 2
C) 4
D) 7
E) 9

In the answer explanation, the denominator of (x - y) in (sqrt x + sqrt y)/(x-y) is converted to (sqrt x - sqrt y)(sqrt x + sqrt y). Can someone explain this step?

[RonPurewal]

If (sqrt x + sqrt y)/(x-y) = (2sqrt x + 2sqrt y)/(x + 2sqrt xy +y), what is the ratio of x to y?

A) 1/2
B) 2
C) 4
D) 7
E) 9

In the answer explanation, the denominator of (x - y) in (sqrt x + sqrt y)/(x-y) is converted to (sqrt x - sqrt y)(sqrt x + sqrt y). Can someone explain this step?

that's the "difference of squares" pattern. it's the same factoring pattern as, e.g., x^2 - 25 = (x + 5)(x - 5).

* if you don't know this factoring pattern, look in our foundations of math book and/or just type "difference of squares" into google.

* if you know this factoring pattern, just notice how it works -- the things in the (this + that)(this - that) pattern are the square roots of the original things on the left.
for instance, when you look at
x^2 - 25 = (x + 5)(x - 5)
note that the x's are the square root of x^2, and that the 5's are the square root of 25.
in this problem, the same thing is happening: the square root of the original x is √x, and the square root of y is √y.

it might seem a bit weird to factor a nice-looking thing like (x - y) into a mess of ugly square roots. indeed, if there were no larger context, and no problem to solve, that would be a positively insane thing to do.
on the other hand, there are already a bunch of √x's and √y's flying around this problem, so that's a pretty solid clue indicating that you should do this.

[griffin.811]

Hi again Ron.

So for this question, I actually noticed the "like quadratics" but my inclination was to square each side of the equation.

This worked out fine for the left side of the equation (assuming I performed my functions properly), and I arrived at 1/(x-y). I did however run into trouble when trying this with the right side of the equation.

I think I decided to square each because when I looked at the denominator of the right side of the equation, neither the x, nor y was squared, and I knew that if they were, I'd have a "sq. of a sum" and would try my luck from there.

Are there any signs in the question that I should've picked up on that would have led me to your method of putting everything in radical version, v. trying to square everything as I did?

Thanks

[RonPurewal]

Hi again Ron.

So for this question, I actually noticed the "like quadratics" but my inclination was to square each side of the equation.

This worked out fine for the left side of the equation (assuming I performed my functions properly), and I arrived at 1/(x-y).

If you squared the left-hand side and got 1/(x - y), then you squared the left-hand side incorrectly.
You can determine this without even finding your mistake, by the way: If you could square the left-hand side and get 1/(x - y), then the original must have been 1/√(x - y). It's not, so, whoops, something went wrong.

The left-hand side simplifies to 1/(√x - √y). So, when you square it, you should get something equivalent to
1/(√x - √y)^2, or, equivalently, 1/(x - 2√xy + y). Pretty nasty, that.
(Since you didn't simplify the left-hand side to 1/(√x - √y), you should have gotten something infinitely nastier, which would simplify to the still-horrible mess up there.)

[RonPurewal]

Are there any signs in the question that I should've picked up on that would have led me to your method of putting everything in radical version, v. trying to square everything as I did?

This will probably disappoint you if you're looking for something brilliant, but:
Simplifying stuff is better than un-simplifying stuff.

[griffin.811]

I definitely did square the left side denominator incorrectly. I took x-y and turned it into x^2 - y^2, instead of (x-y)^2 and recognizing the "square of a diff." I think that would have solved everything.

I feel let down about the lack of brilliance on the last post you made, I've come to expect nothing but the best insight...

Joking, but thanks again, much appreciated as always!

[tim]

Brilliant.

[RonPurewal]

I feel let down about the lack of brilliance on the last post you made, I've come to expect nothing but the best insight...

In my experience, the most profound insights are very, very rarely profound-looking statements.