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
From India, Jharsuguda
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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
From India, Jharsuguda
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Answer is X=9 and y=4 but solve it mathemaically.I mean what is the process... Cool Nishi
From India, Jharsuguda
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From India, Jharsuguda
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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............
From India
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2c,,, 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............
From India
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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
From India, Jharsuguda
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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
From India, Jharsuguda
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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:
From India
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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:
From India
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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
From India, Hyderabad
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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
From India, Hyderabad
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My solution: 🧐🔗🔗🔗🔗
I just received the question and attempted to solve it. I obtained two real values of x, two imaginary values of x, and four real values of y. I only considered the real values.
Thank you
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I just received the question and attempted to solve it. I obtained two real values of x, two imaginary values of x, and four real values of y. I only considered the real values.
Thank you
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Let:
[ x = tan^2@ ]
[ y = sec^2@ ]
1) ( tan@ + sec^2@ = 7 )
2) ( tan^2@ + sec@ = 11 )
Take 1):
[ tan@ + frac{1}{cos^2@} = 7 ]
Solve it, and you will get:
[ 7sin^2@ + sin@ - 6 = 0 quad (A) ]
Now, similarly, take equation (2) and solve it.
You will get:
[ 12cos^2@ - cos@ - 1 = 0 quad (B) ]
Taking equation (A):
Let:
[ sin@ = t ]
[ 7t^2 + t - 6 = 0 ]
Solve it, and you will get:
[ t = -1 text{ or } t = frac{6}{7} ]
[ sin@ = -1 text{ or } sin@ = frac{6}{7} ]
Now, take equation (B) and do similarly as equation A:
[ 12cos^2@ - cos@ - 1 = 0 ]
Solve it as above:
[ cos@ = frac{1}{3} text{ or } cos@ = -frac{1}{4} ]
Since the cosine value exists in the 3rd quadrant, not negative, (-frac{1}{4}) is not possible. Therefore:
[ cos@ = frac{1}{3} ]
The sine value exists in the 3rd quadrant, not positive, so:
[ sin@ = -1 ]
Now,
[ tan@ = frac{sin@}{cos@} = -1/frac{1}{3} = -3 ]
[ x = tan^2@ = (-3)^2 = 9 ]
Put in equation 1:
[ y = 4 ]
Therefore:
[ x = 9 text{ and } y = 4 text{ is the answer.} ]
From undefined, undefined
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[ x = tan^2@ ]
[ y = sec^2@ ]
1) ( tan@ + sec^2@ = 7 )
2) ( tan^2@ + sec@ = 11 )
Take 1):
[ tan@ + frac{1}{cos^2@} = 7 ]
Solve it, and you will get:
[ 7sin^2@ + sin@ - 6 = 0 quad (A) ]
Now, similarly, take equation (2) and solve it.
You will get:
[ 12cos^2@ - cos@ - 1 = 0 quad (B) ]
Taking equation (A):
Let:
[ sin@ = t ]
[ 7t^2 + t - 6 = 0 ]
Solve it, and you will get:
[ t = -1 text{ or } t = frac{6}{7} ]
[ sin@ = -1 text{ or } sin@ = frac{6}{7} ]
Now, take equation (B) and do similarly as equation A:
[ 12cos^2@ - cos@ - 1 = 0 ]
Solve it as above:
[ cos@ = frac{1}{3} text{ or } cos@ = -frac{1}{4} ]
Since the cosine value exists in the 3rd quadrant, not negative, (-frac{1}{4}) is not possible. Therefore:
[ cos@ = frac{1}{3} ]
The sine value exists in the 3rd quadrant, not positive, so:
[ sin@ = -1 ]
Now,
[ tan@ = frac{sin@}{cos@} = -1/frac{1}{3} = -3 ]
[ x = tan^2@ = (-3)^2 = 9 ]
Put in equation 1:
[ y = 4 ]
Therefore:
[ x = 9 text{ and } y = 4 text{ is the answer.} ]
From undefined, undefined
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