SOLUTION: The sum of the squares of two consecutive positive even integers is 340. Find the integers. Can't for the life of me figure this one out. Please help!! Joanne

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: The sum of the squares of two consecutive positive even integers is 340. Find the integers. Can't for the life of me figure this one out. Please help!! Joanne      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 110810: The sum of the squares of two consecutive positive even integers is 340. Find the integers.
Can't for the life of me figure this one out. Please help!!
Joanne

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Consecutive even integers follow the form 2x, 2x+2, etc

So the sum of their squares is

%282x%29%5E2%2B%282x%2B2%29%5E2=340


4x%5E2%2B%282x%2B2%29%5E2=340 Square 2x to get 4x^2


4x%5E2%2B4x%5E2%2B8x%2B4=340 Foil


4x%5E2%2B4x%5E2%2B8x%2B4-340=0 Subtract 340 from both sides


8x%5E2%2B8x-336=0 Subtract 340 from both sides



Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve 8%2Ax%5E2%2B8%2Ax-336=0 ( notice a=8, b=8, and c=-336)




x+=+%28-8+%2B-+sqrt%28+%288%29%5E2-4%2A8%2A-336+%29%29%2F%282%2A8%29 Plug in a=8, b=8, and c=-336



x+=+%28-8+%2B-+sqrt%28+64-4%2A8%2A-336+%29%29%2F%282%2A8%29 Square 8 to get 64



x+=+%28-8+%2B-+sqrt%28+64%2B10752+%29%29%2F%282%2A8%29 Multiply -4%2A-336%2A8 to get 10752



x+=+%28-8+%2B-+sqrt%28+10816+%29%29%2F%282%2A8%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-8+%2B-+104%29%2F%282%2A8%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-8+%2B-+104%29%2F16 Multiply 2 and 8 to get 16

So now the expression breaks down into two parts

x+=+%28-8+%2B+104%29%2F16 or x+=+%28-8+-+104%29%2F16

Lets look at the first part:

x=%28-8+%2B+104%29%2F16

x=96%2F16 Add the terms in the numerator
x=6 Divide

So one answer is
x=6



Now lets look at the second part:

x=%28-8+-+104%29%2F16

x=-112%2F16 Subtract the terms in the numerator
x=-7 Divide

So another answer is
x=-7

So our solutions are:
x=6 or x=-7




Since the values of x are x=6 and x=-7, this means our two numbers are

2%286%29=12, 12%2B2=14

or

2%28-7%29=-14, -14%2B2=-12