Question

the two consecutive positive odd integers such that the sum of their squares is 290 are

Answer 1

hey mate here is yr answer


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. 

Answer 2

x,x+2
x^2+(x+2)^2=290
x^2+x^2+4x+4=290
2x^2+4x+4=290
2(x^2+2x+2)=290
x^2+2x+2=145
x^2+2x-143=0
x^2+13x-11x-143=0
x(x+13)-11(x+13)=0
(x-11)(x+13)=0
x-11=0
x=11
x+13=0
x=-13(that is neglected)


11,13
your answer