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)?
Answer & explanation
Correct answer: option 1
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