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5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z
Asked by Maradana Rajesh kumar Last Modified
Answer 
Vivek Sharma
Atma Nand
Experienced (10+ Y) & Certified Mathematics Assistant Professor from Uttaranchal University, Dehradun
Susanta Kumar Mallick
Computer Professional Teacher with more than 15 years of experience in Computer Science
In this condition, there is no exact value given for the equation. Hence we cannot solve it without a given state. We cannot find a specific answer for each variable as there are no values given. I also tried the trial and error method, but the answer keeps fluctuating. We might have to create assumptions, and then we can solve it.
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5(x2 + y2 + z2) = 4(xy + yz + zx)
or, 5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx
or, 4x2 + 4y2 + 4z2 + x2 + y2 + z2 = 4xy + 4yz + 4zx
or, (4x2 – 4xy + y2) + (4y2 – 4yz + z2) + (4z2 – 4zx + x2) = 0
or, (2x - y)2 + (2y - z)2 + (2z - x)2 = 0
We know that, since sum of squares is 0, each square must be equal to 0.
Therefore,
(2x - y)2 = (2y - z)2 = (2z - x)2 = 0
or, (2x - y) = (2y - z) = (2z - x) = 0
Therefore,
2x = y, 2y = z, 2z = x.
Thus converting all terms in respect to ‘x’ we get,
y = 2x
z = x/2
Therefore,
x : y : z = x : 2x : x/2
= 1 : 2 : ½
= 2 : 4 : 1
Answer: x : y : z = 2 : 4 : 1
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