If x + y + z = 19
x2 + y2 + z2 = 133
xz = y2 ,
then find the value of \(\frac{y^2}{3}\) + 26.
Answer & explanation
Correct answer: option 3
x2 + y2 + z2 + 2 (xy + yz + zx) = (x + y + z)2
133 + 2 (xy + yz + y2) = (19)2
y (x + y + z) = \(\frac{361 - 133}{2}\)
⇒ y (19) = \(\frac{228}{2}\)
⇒ y = \(\frac{228}{38}\) = 6
\(\frac{y^2}{3}\) + 26 = \(\frac{6^2}{3}\) + 26 = 38