If x + y + z = 13, $x^2 + y^2 + z^2 = 91$ and $xz = y^2$, then the difference between z and x is :
Answer & explanation
Correct answer: option 2
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