Q376
If x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2, then the difference between z and x is:
A.
5
AnswerB.
3
C.
6
D.
4
Answer: Option A
Solution
Answer: Option A
Solution:
Given:x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2,
Formula used:
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
Calculation:
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
⇒ (19)2 = 133 + 2(xy + yz + y2)
⇒ 133 + 2[y(x + y + z)] = 361
⇒ 2y(19) = 361 - 133
⇒ y = 6
x + y + z = 19
⇒ x + z = 13
The possible value of x and z is 9 and 4
x - 4
⇒ 9 - 4
⇒ 5
∴ The value is 5