If x + y + z = 19, x2 + y2 + z2 = 133, and xz = y2, x > z > 0, what is the value of (x - z)?
A. 0
B. 5
C. -2
D. -5
Answer: Option B
Solution(By Examveda Team)
x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2(x - z) = ?
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
361 = 133 + 2(xy + yz + y2)
228 = 2(x + y + z)y
$$y = \frac{{114}}{{19}} = 6$$
x + z = 13
xz = 36
x - z = ?
(x + z)2 - (x - z)2 = 4xz
169 - (x - z)2 = 144
x - z = 5
This Question Belongs to Arithmetic Ability >> Algebra
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