find the value of (x+y/z) if x2+y2+z2-4x-8y-6z+29=0
Answers
Answered by
12
Solution for the question is-
x²+y²+z²-4x-8y-6z+29=0
(x²-4x+4)+(y²-8y+16)+(z²-6z+9) = 0
(x-2)²+(y-4)²+(z-3)² = 0
Each term here is positive
So each term must be zero to get the sum as zero.
so, x-2 = 0
y-4 = 0
z-3 = 0
we get
x =2, y=4, z=3
x+y/z = 2+4/3 = 2
x²+y²+z²-4x-8y-6z+29=0
(x²-4x+4)+(y²-8y+16)+(z²-6z+9) = 0
(x-2)²+(y-4)²+(z-3)² = 0
Each term here is positive
So each term must be zero to get the sum as zero.
so, x-2 = 0
y-4 = 0
z-3 = 0
we get
x =2, y=4, z=3
x+y/z = 2+4/3 = 2
Similar questions