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
Acknowledge(0)
Amend(0)
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
Acknowledge(0)Amend(0)
Answer is X=9 and y=4 but solve it mathemaically.I mean what is the process... Cool Nishi
From India, Jharsuguda
Acknowledge(0)Amend(0)
From India, Jharsuguda
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
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
Acknowledge(0)Amend(0)
[ 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
Acknowledge(0)Amend(0)Engage with peers to discuss and resolve work and business challenges collaboratively - share and document your knowledge. Our AI-powered platform, features real-time fact-checking, peer reviews, and an extensive historical knowledge base. - Join & Be Part Of Our Community.
AMAZING MATHS.ppt
Vedic maths.pdf