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Sum of the squares of three consecutive natural numbers is 110. |
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Answer» If the numbers are x, x + 1, x + 2, then x2 + ( x + 1)2 + (x + 2)2 = 110 x2 + x2 + 2x + 1 + x2 + 4x + 4 = 110 3x2 + 6x – 105 = 0 \(x=\frac{-6\pm\sqrt{6^2-4\times3\times-105}}{2\times3}\) = \(\frac{-6\pm\sqrt{1296}}{6}=\frac{-6\pm36}{6}=5.7\) |
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