Question

find two consecutive positive integers sum of whose square is 265​

Answer 1

Hi

Let the 2 consecutive positive integers be (x) and (x+1)

$({x})^{2} + ({x + 1})^{2} = 265$

${x}^{2} +( {x}^{2} + 1 + 2x) = 265$

(since (a+b)^2 = a^2 + b^2 + 2ab)

$\implies \: 2 {x}^{2} + 2x +1 = 265$

$\implies \: 2 {x}^{2} + 2x - 264 = 0$

Now, take out 2 as a common factor

${x}^{2} + x - 132 = 0$

On factorising above equation,

${x}^{2} + 12x - 11x - 132 = 0$

$\implies \: x(x + 12) - 11(x + 12)$

$\implies \: (x - 11) (x + 12)$

$\implies \red{x = 11} \: or \: \orange{x = - 12}$

According to the question, we are asked about positive integer

$\huge{ \boxed{ \therefore \: x = 11}}$

Hope it helps you

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