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If x + y + z = 19 x2 + y2 + z2 = 133 xz = y2 , then find the value of \(\frac{y^2}{3}\) + 26. |
4 6 38 36 |
38 |
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 |
