Question

Let x, y, z, w satisfy \[\frac{x^2}{4^2-1^2}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1\]\[\frac{x^2}{6^2-1^2}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1\]\[\frac{x^2}{8^2-1^2}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1\]\[\frac{x^2}{10^2-1^2}+\frac{y^2}{10^2-3^2}+\frac{z^2}{10^2-5^2}+\frac{w^2}{10^2-7^2}=1\] Then the value of \(\Large{x^2+y^2+z^2+w^2}\) is

Answer 1

@HolsterEmission @ParthKohli

Answer 2

132

Answer 3

Agree with 132

Answer 4

Change the variables for x1=x^2, y1=y^2,z1=z^2,w1=w^2,
You will obtain 4 linear equations in 4 unknown.
Solve and find the sum

Answer 5

There must be an easier way than the one I described above

Answer 6

Just a heads up :
Can't use modular arithmetic as x, y, z, w are not integers.


http://www.wolframalpha.com/input/?i=solve+x%2F(3*5)+%2B+y%2F(1*7)%2Bz%2F(-1*9)%2Bw%2F(-3*11)%3D1,+x%2F(5*7)+%2B+y%2F(3*9)%2Bz%2F(1*11)%2Bw%2F(-1*13)%3D1,++x%2F(7*9)+%2B+y%2F(5*11)%2Bz%2F(3*13)%2Bw%2F(1*15)%3D1,+x%2F(9*11)+%2B+y%2F(7*13)%2Bz%2F(5*15)%2Bw%2F(3*17)%3D1

Answer 7

That's a long method which i won't try to use during examination.

Answer 8

I'd plug it into my TI-84 and invert the matrix to get the answer on my calculator in less than a minute if I was allowed.

Answer 9

We are not given more than 2 minutes for solving questions in jee advanced examination

Answer 10

There is a unique set of scalars a,b,c,d to make the
a* eqn1 + b* eqn2 + c*eqn3 +d*eqn4= x+y+z+w=a+b+c+d=132
This also requires solving 4 equations in 4 unknowns.
@jjiteshmeghwal9 , if you know an easier way, please post it. Many people would love to see it

Answer 11

U r right there is no shortcut method for this question but a less time consuming approach is by assuming 4^2 as t in equation 1 and then u will get a quadratic equation