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Mathematics 20 Online
OpenStudy (anonymous):

sqrt(x) + sqrt(y) =11, find x and y.

OpenStudy (jack1):

can only find x as a function of y, and vice versa given you only have one equation

OpenStudy (jack1):

if you square both sides, you are left with: x + y = 121

OpenStudy (jack1):

therefore you can re-arrange your equation to either: x= 121-y or y=121 - x

OpenStudy (anonymous):

just square and so me how

OpenStudy (jack1):

... what...?

OpenStudy (raden):

x and y are positive integer numbers ?

OpenStudy (anonymous):

@rahulon i think,the information is not enough

mathslover (mathslover):

Hint : Square both sides.

mathslover (mathslover):

Sritama, I think it is not necessary to calculate exact values for x and y, they can be in terms of each other.

OpenStudy (anonymous):

x and y are real number

OpenStudy (jack1):

if x and y are real numbers, you will need a 2nd equation to find the value of each, as there are 2 unknowns

mathslover (mathslover):

\((\sqrt{x} + \sqrt{y} )^2 = 11^2\) \(\implies \sqrt{x}^2 + \sqrt{y}^2 + 2\sqrt{xy} = 121\) \(\implies x + y + 2\sqrt{xy} =121\)

OpenStudy (anonymous):

but what about sqrt(xy)

mathslover (mathslover):

Sritama and Jack1 are right if you want to get the values for the 2 unknowns , though I can say that : \(x = 121 - y - 2\sqrt{xy}\) or \(y= 121-2\sqrt{xy} - x\)

OpenStudy (jack1):

you're right @mathslover, my bad

mathslover (mathslover):

@rahulon got it?

OpenStudy (anonymous):

emm...i think the questions wants us to get numerical values for x and y if im not wrong..lets work on that

OpenStudy (anonymous):

and most likely x and y are integers as RadEn said

mathslover (mathslover):

I don't think that there is ANY way to calculate the numerical values for x and y as it is an equation of two variables. if x = 0 then y = 121 if x = 1 then y = 100 .. and so on...

OpenStudy (anonymous):

whats wrong with drawing tool :) ??

OpenStudy (anonymous):

|dw:1365589853647:dw|

OpenStudy (anonymous):

there are infinite real pairs of x and y from real numbers satisfying the equation and we can find integer ones because we know that x,y<=121

mathslover (mathslover):

Yes!

OpenStudy (anonymous):

(x,y)=(121,0),(100,1),(81,4),(64,9),(49,16),(36,25)

mathslover (mathslover):

But as the user said they x and y are real numbers : \((x,y) \in \mathbb{R}\) So , x and y can have infinite values.

mathslover (mathslover):

^that x and y are real numbers.

OpenStudy (anonymous):

ahh, yes...infinite is answer :)

mathslover (mathslover):

:)

OpenStudy (raden):

@mukushla , if x and y are integer there are 12 order pairs of (x,y) as solution, right ?

OpenStudy (anonymous):

@RaDen yes, thats right...i did'nt write them because of symmetry.

OpenStudy (anonymous):

is there any formula to find out all the real values of x and

OpenStudy (anonymous):

and y

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