SOLUTION: I have asked this question within the past two days without a response. I am having a tough time figuring out how to set this problem up. Once I know how to set it up and why, I ca

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I have asked this question within the past two days without a response. I am having a tough time figuring out how to set this problem up. Once I know how to set it up and why, I ca      Log On


   

Question 68069: I have asked this question within the past two days without a response. I am having a tough time figuring out how to set this problem up. Once I know how to set it up and why, I can probably solve the equation. I hope someone out there can help me with this one, please.
Find two consecutive positive integers such that the sum of their squares is 85. [You must use algebra to solve this, setting up an equation with a variable, and using algebra to solve for the variable.]
Found 2 solutions by Earlsdon, checkley71:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the first positive integer. The next consecutive positive integer would then be x+1.
The sum of the squares of these is 85.
x%5E2+%2B+%28x%2B1%29%5E2+=+85 Simplify and solve for x.
x%5E2+%2B+%28x%5E2+%2B+2x+%2B+1%29+=+85 Simplify.
2x%5E2+%2B+2x+%2B+1+=+85 Subtract 85 from both sides.
2x%5E2+%2B+2x+-+84+=+0 Divide through by 2.
x%5E1+%2B+x+-+42+=+0 Factor.
%28x%2B7%29%28x-6%29+=+0 Apply the zero product principle.
x%2B7+=+0 and/or x-6+=+0
So, x = -7 and x = 6 Discard the negative value as the problem requires positive integers.
The two integers are: 6 and 7
Check:
6%5E2+%2B+7%5E2+=+36+%2B+49 = 85

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
let x^2 be the first number & (x+1)^2 the next number
x^2+(x+1)^2=85
x^2+x^2+2x+1=85
2x^2+2x+1-85
2x^2+2x-84=0
x^2+x^2-42=0
(x+7)(x-6)=0
x-6=0
x=6 answer for the first number
6+1=7 for the next number.
proof
6^2+7^2=84
36+49=84
84=84