find 2 consecutive odd positive integer whose squares have sum 290
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Hey mate here is the answer..
Let one of the odd positive integer be x
then the other odd positive integer is x+2
their sum of squares = x² +(x+2)²
= x² + x² + 4x +4
= 2x² + 4x + 4
Given that their sum of squares = 290
⇒ 2x² +4x + 4 = 290
⇒ 2x² +4x = 290-4 = 286
⇒ 2x² + 4x -286 = 0
⇒ 2(x² + 2x - 143) = 0
⇒ x² + 2x - 143 = 0
⇒ x² + 13x - 11x -143 = 0
⇒ x(x+13) - 11(x+13) = 0
⇒ (x-11) = 0 , (x+13) = 0
Therfore x = 11 or -13
We always take positive value of x
So , x = 11 and (x+2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13 .
Hope you got it:-)
Let one of the odd positive integer be x
then the other odd positive integer is x+2
their sum of squares = x² +(x+2)²
= x² + x² + 4x +4
= 2x² + 4x + 4
Given that their sum of squares = 290
⇒ 2x² +4x + 4 = 290
⇒ 2x² +4x = 290-4 = 286
⇒ 2x² + 4x -286 = 0
⇒ 2(x² + 2x - 143) = 0
⇒ x² + 2x - 143 = 0
⇒ x² + 13x - 11x -143 = 0
⇒ x(x+13) - 11(x+13) = 0
⇒ (x-11) = 0 , (x+13) = 0
Therfore x = 11 or -13
We always take positive value of x
So , x = 11 and (x+2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13 .
Hope you got it:-)
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