If x + y + z = 13, $x^2 + y^2 + z^2 = 91$ and $xz = y^2$, then the difference between z and x is :

8

 x + y + z = 13

x² + y² + z² = 91

xz = y²

 We know that,

(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)

(a - b)² = a² - 2ab + b²

So,

x + y + z = 13

= (x + y + z)² = 13²

= x² + y² + z² + 2(xy + yz + xz) = 169

= (xy + yz + xz) = 39

= xy + yz + y² = 39

= y(x + y + z) = 39

= y = 3

= y² = 9

= xz = 9

= 3xz = 27

So,

x² + y² + z² = 91

= x² + xz - 3xz + z² = 91 - 27

= x² - 2xz + z² = 64

= (z - x)² = 64

= (z - x) = 8