Solve It...(ramanujan Maths) - Jpg Download

Dear All,
Please solve it.First of all this equation was solved by great mathematician Ramanujan.
1.Solve this equation :-
√X + Y = 7
X + √Y = 11
This equation is very interesting equation please try it mathimatically.:confused:
With Warm Regards,
Nishi Kant
25th September 2008 From India, Jharsuguda

Answer is X=9 and y=4 but solve it mathemaically.I mean what is the process... Cool Nishi
25th September 2008 From India, Jharsuguda

yar its again mathematically..........
c,,, in d first equation...
if v pt nos frm only 9 wud fit in fr x nd 4 fr y...
nd same in secnd...
yar i knw dere mite b othr ways to solve d equation,,,bt in comptitn.... u need to solve problems quickly............
8th October 2008 From India

Hi Gent,
I know the value of X and Y in this equation.But i want the mathematical proof.
Please note that first of all this equation was mathematically solved by great mathematician Ramanujan after regular work of 17 days on the same equation.
Cheers,
Nishi Kant
14th October 2008 From India, Jharsuguda

Hey ,,,
Nw nishikant,,,y dnt u solve nd let us abt it................it has been quite a long tym since u posted dis equation...nd yes as said in d last post,,,dis is cite hr,,,wherein d users wud hardly find out or devote dere so calld tym on dis..............so y don't u disclose dis for us ....go ahead...:icon10::idea:
17th October 2008 From India

One of my collegues Ms. Varalakshmi has solved the equation. Find your answer below
Let: .u = √x . → . x = u²
. - - . . . . ._
Let: .v = √y . → . y = v²
Substitute: . u + v² .= . 7 -[1]
. . . . . . . . . u² + v .= .11 .[2]
From [1], we have: .u .= .7 - v²
Substitute into [2]: . (7 - v²)² + v .= .11
. . which simplifies to: .v^4 - 14v² + v + 38 .= .0
. . which factors: .(v - 2)(v³ + 2v² - 10v - 19) .= .0
. . and has the rational root: .v = 2
Substitute into [1]: . u + 2² .= .7 . → . u = 3
Therefore: .x = u² = 9, . y = v² = 4
14th May 2009 From India, Hyderabad

My solution🏬⬇⬇⬇⬇
I just got the question and try to solve it and then I got two real values of x and two imaginary value of x and 4 real values of y I just consider the real values
Thank you
18th October 2016

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Let
x=tan^2@
y=sec^2@
1). tan@+sec^2@=7
2) tan^^2@+sec@=11
Take 1)
Tan@+1/cos^2@=7
Solve it u will get
7sin^2@+sin@-6=0.......(A)
Now similarly take equation (2) solve it
U will get
12cos^2@-cos@-1=0......(B)
Take(A)......
Let
Sin@=t
7t^2+t-6=0
Solve it u will get
t=-1. Or. t=6/7
Sin@=-1. Or. Sin@=6/7
Now
Take eq. (B)
Do Similarly
As eq A
12cos^2@-cos@-1=0
Solve it as above
Cos@=1/3 or. Cos@=-1/4
Cos@ value exist in 3 quadrant not -ve
So -1/4 not possible
Cos@=1/3
Sin@ value exist in. 3 quadrant not +ve so
Sin@=-1

Now tan@=sin@/cos@=-1/1/3=-3
x=tan^2@=(-3)^2=9
Put in equation 1
y=4
Therefore
x=9. And y=4 is ans
4th April 2017 From undefined, undefined