If x + y + z = 19,x^2 +y^2+z^2=133 and x...

If x + y + z = 19,`x^2 +y^2+z^2=133` and `xz=y^2, x>z>0`, what is the value of (x-z) ?

A

-2

B

-5

C

5

D

0

Text Solution

Verified by Experts

The correct Answer is:

C

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