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)?

5

Given,

 x + y + z = 19

 x² + y² + z² = 133

 xz = y²

 (a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

Now,

 x + z = 19 – y

 x² + z² = 133 – y²

Square on both side

x² + z² + 2xz = 361 + y² – 38y  

= 133 – y² + 2y² = 361 + y² – 38y [x² + z² = 133 – y² and xz = y²]

= 38y = 361 – 133 = 228

= y = 6

(x – z)² = x² + z² – 2xz

= (x – z)² = 133 – y² – 2y²

= (x – z)² = 133 – 3y²

= (x – z)² = 133 – 3 × 6²

= (x – z)² = 25

= (x – z)  = 5