sqrt(x) + sqrt(y) =11, find x and y.
can only find x as a function of y, and vice versa given you only have one equation
if you square both sides, you are left with: x + y = 121
therefore you can re-arrange your equation to either: x= 121-y or y=121 - x
just square and so me how
... what...?
x and y are positive integer numbers ?
@rahulon i think,the information is not enough
Hint : Square both sides.
Sritama, I think it is not necessary to calculate exact values for x and y, they can be in terms of each other.
x and y are real number
if x and y are real numbers, you will need a 2nd equation to find the value of each, as there are 2 unknowns
\((\sqrt{x} + \sqrt{y} )^2 = 11^2\) \(\implies \sqrt{x}^2 + \sqrt{y}^2 + 2\sqrt{xy} = 121\) \(\implies x + y + 2\sqrt{xy} =121\)
but what about sqrt(xy)
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 \(y= 121-2\sqrt{xy} - x\)
you're right @mathslover, my bad
@rahulon got it?
emm...i think the questions wants us to get numerical values for x and y if im not wrong..lets work on that
and most likely x and y are integers as RadEn said
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...
whats wrong with drawing tool :) ??
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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
Yes!
(x,y)=(121,0),(100,1),(81,4),(64,9),(49,16),(36,25)
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.
^that x and y are real numbers.
ahh, yes...infinite is answer :)
:)
@mukushla , if x and y are integer there are 12 order pairs of (x,y) as solution, right ?
@RaDen yes, thats right...i did'nt write them because of symmetry.
is there any formula to find out all the real values of x and
and y
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