SOLUTION: Therr consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1,3, and 5.Find three other

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Therr consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1,3, and 5.Find three other      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 660287: Therr consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1,3, and 5.Find three other consecutive odd integers that also satisfy the given conditions
Answer by kevwill(135) About Me  (Show Source):
You can put this solution on YOUR website!
Let x, x+2, and x+4 be the three consecutive odd integers. We are looking for x such that
x%5E2+%2B+%28x%2B2%29%5E2+%2B+15+=+%28x%2B4%29%5E2
Expanding
x%5E2+%2B+x%5E2+%2B+4x+%2B+4+%2B+15+=+x%5E2+%2B+8x+%2B+16
2x%5E2+%2B+4x+%2B+19+=+x%5E2+%2B+8x+%2B+16
2x%5E2+%2B+4x+%2B+19+-+x%5E2+-+8x+-+16+=+0
x%5E2+-+4x+%2B+3+=+0
We can solve for x using the quadratic equation x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ with a=1, b=-4, and c=3
x+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%2A1%2A3+%29%29%2F%282%2A1%29+
x+=+%284+%2B-+sqrt%28+16-12+%29%29%2F2+
x+=+%284+%2B-+sqrt%28+4+%29%29%2F2+
x+=+%284+%2B-+2%29%2F2+ so we have
x+=+%284+-+2%29%2F2+=+2%2F2+=+1 or
x+=+%284+%2B+2%29%2F2+=+6%2F2+=+3
For x = 1, we have the solution 1, 3, 5 which was given in the problem statement.
For x = 3, we have the second solution: 3, 5, 7.