(sqrt x + sqrt y) / (sqrt x - sqrt y) = ???

P

Patrick_John

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Hello to everyone

I'm new to this site, nice to meet you :)

Not sure if this is intermediate/advanced algebra.

If x + y = 5, xy = 1, and x > y, and if "sqrt[x]" is taken to mean "the square root of x", then:

. . .( sqrt[x] + sqrt[y] ) / ( sqrt[x] - sqrt[y] ) = ???

And the four answer options are:

. . .a) sqrt[21] / 3

. . .b) 7 / sqrt[21]

. . .c) sqrt[7/3]

. . .d) sqrt[7] / sqrt[3]

What's the process to get those results?

Thanks in advance.
__________________________________
Edited by stapel -- Reason for edit: clarity of formatting.
 
[(x)^(1/2) + (y)^(1/2)] / [(x)^(1/2) - (y)^(1/2)]

= [(x)^(1/2) + (y)^(1/2)] / [(x)^(1/2) - (y)^(1/2)] * [(x)^(1/2) + (y)^(1/2)] / [(x)^(1/2) + (y)^(1/2)]

= [(x)^(1/2) + (y)^(1/2)] (^2)/ [(x) - (y)]

= [x + y + 2(xy)^(1/2)] / (x-y)

Now continue....
 
Thanks for your help.

Sorry I just realized that the spaces I inputed in the answers disappeared, making them look weirdly, I'll try posting a pic instead

If x+y=5, xy = 1 and x>y, then

t_QA5m_52cdcee.jpg



And the 4 answers to it are:


f_RA5m_17bd5fa.jpg



Also, I didn't understand your post :(, I though I had to get the values for "x" and "y" first using ( If x+y=5, xy = 1 and x>y ),
and replace those values on (sqrx + sqry) / (sqrx - sqry) = ???

Thanks
 
No you don't need to get values for x and y - however you'll need to calculate x-y

Use

(x-y)^2 = (x+y)^2 - 4xy

Then use this value and other known value to the expression I had derived.

(x-y)^2 = 5^2 - 4*1 = 21

x-y = (21)^(1/2)

Then

[(x)^(1/2) + (y)^(1/2)] / [(x)^(1/2) - (y)^(1/2)]

= [(x)^(1/2) + (y)^(1/2)] / [(x)^(1/2) - (y)^(1/2)] * [(x)^(1/2) + (y)^(1/2)] / [(x)^(1/2) + (y)^(1/2)]

= [(x)^(1/2) + (y)^(1/2)] (^2)/ [(x) - (y)]

= [x + y + 2(xy)^(1/2)] / (x-y)

= (5 + 2) / (21)^(1/2)

Now finish it....
 
Sorry but I still don't get it

What do you mean with "calculate x-y"?

Also, is (x+y)^2 - 4xy some sort of the discriminant? (what is found under the square root part of the cuadratic formula)

What would you recommend me to read to understand how to solve this type of problems?

Thanks
 
\(\displaystyle \L\ \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}}\ = \frac{(\sqrt{x} + \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})}\\)

\(\displaystyle \L\ = \frac{x + y + 2\sqrt{xy}}{x - y}\\)

\(\displaystyle \L\ = \frac{5 + 2(1)}{x - y}\\)

You were told what x - y was by SK.
 
Also, is (x+y)^2 - 4xy some sort of the discriminant? (what is found under the square root part of the cuadratic formula)
Click to expand...

No - it is just an extension of the square formulae

(x+y)^2 = x^2 + y^2 + 2xy

(x-y)^2

= x^2 + y^2 - 2xy

= x^2 + y^2 + 2xy - 4xy

= (x+y)^2 - 4xy

Now you can calculate (x-y) knowing (x+y) and xy
 
You can also be sure to take the positive square root only of what SK posted above, as it is given that x > y (implying x - y > 0). So there is one distinct value. Also, since x and y are given an "order" on the reals, then this implies they are both real and therefore must be non-negative.
 
I hope you do realize that all the given choices of answer are numerically equivalent.
 
Who, me? Yes of course. I was highlighting the importance of the information "x is greater than y."
 
Thank you guys for all you help, there is something though I have still not understood:

morson said:
\(\displaystyle \L\ \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}}\ = \frac{(\sqrt{x} + \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})}\\)

\(\displaystyle \L\ = \frac{x + y + 2\sqrt{xy}}{x - y}\\)

\(\displaystyle \L\ = \frac{5 + 2(1)}{x - y}\\)
Click to expand...

Why is \(\displaystyle \L\ \frac{\sqrt{x} + \sqrt{y}}{sqrt{x} - \sqrt{y}}\\) multiplied by \(\displaystyle \L\ (\sqrt{x} + \sqrt{y})\\) ?
 
It's multiplied by \(\displaystyle \L\ \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}}\\), which is 1. It's just a means of simplifying.
 
Patrick,

Another way you could do is:

x - y = (21)^(1/2)
x + y = 5

Then

x = 1/2 * [5 + (21)^(1/2)]

and

y = 1/2 * [5 - (21)^(1/2)]

Then use these into your given equation and simplify to get the required form. But that would be very tedious.
 
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