Question

If x+y+z=12 and x^2+y^2+z^2=64 then find xy +yz +zx

Answer 1

Hi friend!!

Given, x+y+z=12

Squaring on both sides, we get

x²+y²+z²+2(xy+yz+zx)=144

It is also given that x²+y²+z²=64

→64+2(xy+yz+zx)=144

2(xy+yz+zx)=80

xy+yz+zx=40

I hope this will help you ;)

Answer 2

x  + y + z  = 12 

Square on both sides,

(x + y + z)^2 = 12^2

x^2 + y^2 + z^2 + 2(xy + yz + zx) = 144

64 + 2(xy + yz + zx) = 144

2(xy + yz + zx) = 144 - 64 

2(xy + yz + zx) = 80 



xy + yz + zx = 80/2 

xy + yz + zx = 40 




i hope this will help you


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