Question

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

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Answer 1

let a number be x
and it's cosecutuve number be (x+2)
$in \: above \: question \\ {x}^{2} + {(x + 2)}^{2} = 290 \\ {x }^{2} + {x}^{2} + 4 + 4x = 290 \\ 2 {x}^{2} + 4x - 286 = 0 \\ (divide \: by \: 2) \\ {x}^{2} + 2x - 143 = 0 \\ {x}^{2} + 13x - 11x - 143 = 0 \\ x(x + 13) - 11(x + 13) \\ (x + 13)(x - 11) = 0$
positive number is 11
and its consecutive number is 13
so,
Two numbers are 11 and 13
Please mark it as brainliest answer.

Answer 2

Answer:yes

Step-by-step explanation: