Question

find to consecutive odd postive integer the sum of squares is 290

Answer 1

Let x = the smaller odd integer
x+2 = the larger odd integer 
x^2 + (x+2)^2 = 290
x^2 + x^2 + 4x + 4 = 290
2x^2 + 4x - 286 = 0 
x^2 + 2x - 143 = 0 
Solve by factoring
(x+13)(x-11) = 0 
x-11 = 0
x = 11
x+2 = 11+2 = 13 
The two consecutive odd integers are 11 and 13.


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Answer 2

Secondary School

Math

5+3 pts

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

 by JeetendraCheema651 10.06.2015

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 .