Question

Find the two consecutive odd positive integers sum of whose squares is 290

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First odd positive integers be t.

Hence,

Next integer will be t + 2

${\boxed{\sf\:{Situation}}}$

t² + (t + 2)² = 290

t² + t² + 2 × t × 2 + 2² = 290

t² + t² + 4t + 4 - 290 = 0

2t² + 4t - 286 = 0

2(t² + 2t - 143) = 0

t² + 2t - 143 = 0

${\boxed{\sf\:{Solving\;the\;Quadratic\;Equation}}}$

t² + 13t - 11t - 143 = 0

t(t + 13 ) - 11 (t + 13 ) = 0

(t - 11 ) (t + 13 ) = 0

t - 11= 0

t = 11

t + 13 = 0

t = -13

${\boxed{\sf\:{t\;is\;an\;odd\;positive\;integer}}}$

Hence,

t ≠ - 13

t = 11

Therefore,

${\boxed{\sf\:{Two\;consecutive\;odd\;integers}}}$

${\boxed{\sf\:{11\;and\;13}}}$

Answer 2

Answer:

Two consecutive odd integers are 11 and 13.

hope it helps you✌