solve

find x,y,z

4x²+25y²+9z²-10xy-15yz-6zx=0

and

x+y+z=5

4 Answers

The first equaiton is of the form (2x)^2+(5y)^2+(3z)^2-(2x)(5y)-(2x)(3z)-(5y)(3z)=0

Hence: (2x-5y)^2+(5y-3z)^2+(3z-2x)^2=0

Now It is a piece of cake..

as 2x=5y=3z=k

Then k/2+k/3+k/5=5

So k(31/30)=5

K= 150/31

x=75/31, y=30/31 and z=50/31

that is a simple one
the eqn will be of d form
1/2[(a-b)² +(b-c)²+(c-a)²]
n then it can b easily solved

nishant sir, how did u get the second line, is it some standard formula

yes...

but you should try to expand (a-b)²+(b-c)²+(c-a)² and see what happens..

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