Question

Find two consecutive positive integers. Sum of whose square is 365

Answer 1

182 and 183 is your answer.

Answer 2

Let the two consecutive no. to be

x and x+1.

${x}^{2} + (x + 1) {}^{2} = 365$

${x}^{2} + ( {x}^{2} + 2x + 1) = 365$

$2 {x }^{2} + 2x + 1 = 365$

$2 {x}^{2} + 2x = 364$

${x }^{2} + x = 182$

${x}^{2} + x - 182 = 0$

${x}^{2} + 14x - 13x - 182 = 0$

$x(x + 14) - 13(x + 14) = 0$

$(x -13)(x + 14) = 0$

$x - 13 = 0 \: then \: x = 13$

$x + 1 = 13 + 1 = 14$

The two consecutive no. is 13 and 14.

I hope this answer will help u...