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Q4. Find two consecutive positive integers, sum of whose squares is 365.
Answer :

Let the first number be \(x\)
The second number can be represented by \(x + 1\).
From the given condition,
\(x^2 + (x + 1)^2 = 365\)
\(x^2 + x^2 + 2x + 1 - 365 = 0\)
\(2x^2 + 2x -364 = 0\)
\(x^2 + x - 182 = 0\)
(Splitting x as 14x - 13x )
\(x^2 + 14x - 13x - 182 = 0\)
\(x(x + 14) - 13(x + 14) = 0\)
\((x - 13)(x + 14) = 0\)
The roots of this equation are the values of x for which \( (x + 14)(x - 13) = 0 \)
which are,
\( x + 14 = 0 \) or \( x - 13 = 0\)
Thus, \(x = -14\) or \(x = 13\)

Since the integer is supposed to be positive, \(x = -14\) is eliminated.
Thus, the two consecutive positive integers are 13 and 13 + 1 = 14.