If the square of the sum of two successive natural numbers exceeds the sum of their square by 112, then the larger of the two is :

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Solution

The correct option is C 8
Let two successive natural numbers are x and (x+1).
Then, $(x + x + 1)^{2} - [x^{2} + (x + 1)^{2}] = 112$
$⟹ 4 x^{2} + 4 x + 1 - x^{2} - x^{2} - 2 x - 1 = 112$
$⟹ x^{2} + x - 56 = 0$
Solving we get x = 7
Hence the lareger of the two is
$x + 1 = 7 + 1 = 8$

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