Question 130543: The sum of the squares of two consecutive odd integers is 34. Find the two integers. Found 2 solutions by Earlsdon, checkley71:Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! 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. "The sum of the squares of two consecutive odd integers is 35"
Simplify and solve for x. Square the terms on the left side. Simplify. Subtract 34 from both sides. Divide through by 8 to simplify. Factor this quadratic equation. Apply the zero products rule. or , so then... or
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
You can put this solution on YOUR website! x^2+(x+2)^2=34
x^2+x^2+4x+4=34
2x^2+4x+4-34=0
2x^2+4x-30
2(x^2+2x-15)=0
2(x+5)(x-3)=0
x+5=0
x=-5 answer.
x-3=0
x=3 answer.
thus the answers are (3,-3,5&-5)
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