The sum of the squares of three consecutive positive integers is 50. Find the integers.
Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).
∵X2 + (X + 1)2 + (X + 2)2 = 50
X2 + X2 + 1 + 2X + X2 + 4 + 4X = 50
On simplifying further,
3X2 + 6X– 45 = 0
X2 + 2X – 15 = 0
On applying Sreedhracharya formula
X = – 5 or 3
∴ X = 3 (Only Positive values)
∴ X + 1 = 4
∴ X +2 = 5
∴ The numbers are 3, 4 & 5.
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