Math, asked by 8848224083, 1 year ago

Find 2 consecutive odd positive integers sum of whose squares is 290?

Answers

Answered by Anonymous

Hi there !!

Let the two consecutive odd positive integers be x and x + 2

given ,

x² + (x+2)² = 290
x² + x² + 4x + 4 = 290
2x² + 4x - 286 = 0
dividing by 2 , we get ,

x² + 2x - 143 = 0

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

x = 11 , x = -13

as the integers are positive , x = 11 is the appropriate value

the no:s are :-

x = 11 , x +2 = 11 + 2 = 13

11 and 13 are the no:s

=========================
verification

11² + 13²

= 121 + 169
= 290

verified

Anonymous: u can do it with x , x - 2

Anonymous: (x)^2 - (x-2)^2 =290

Anonymous: x^2 + x^2 - 4x + 4 = 290

Anonymous: 2x^2 - 4x + 4 = 290

Anonymous: 2x^2 - 4x - 286

Anonymous: dividing by 2 , we get ,

Anonymous: x^2 - 2x - 143 = 0

Bhargavi11: i want to say that i had answer it by eq. using x and x-2

Anonymous: hmm fine

Bhargavi11: thats why my answer is not wrong

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