Question 130543
I'll do the long (and correct) version of the solution:
Let the first odd integer be (2x+1) and the next consecutive odd integer be (2x+3)
Why (2x+1) and (2x+3)?
This insures that the integers will be odd, no matter what value of integer x you may substitute.
{{{(2x+1)^2+(2x+3)^2 = 34}}} "The sum of the squares of two consecutive odd integers is 35"
Simplify and solve for x.
{{{(2x+1)^2+(2x+3)^2 = 34}}} Square the terms on the left side.
{{{(4x^2+4x+1)+(4x^2+12x+9) = 34}}} Simplify.
{{{8x^2+16x+10 = 34}}} Subtract 34 from both sides.
{{{8x^2+16x-24 = 0}}} Divide through by 8 to simplify.
{{{x^2+2x-3 = 0}}} Factor this quadratic equation.
{{{(x-1)(x+3) = 0}}} Apply the zero products rule.
{{{x-1 = 0}}} or {{{x+3 = 0}}}, so then...
{{{x = 1}}} or {{{x = -3}}}
The integers are:
For x = 1:
2x+1 = 2(1)+1 = 3
2x+3 = 2(1)+3 = 5 and the other pair of integers is:
For x = -3:
2x+1 = 2(-3)+1 = -6+1 = -5
2x+3 = 2(-3)+3 = -3+3 = -3
So there are two solutions to this problem:
Consecutive odd integers: 3 and 5
Consecutive odd integers -5 and -3