SOLUTION: Find four consecutive even integers such that the sum of the squares of the first and the second is 12 more than the last.

Question 228683: Find four consecutive even integers such that the sum of the squares of the first and the second is 12 more than the last.
Found 2 solutions by rapaljer, solver91311:
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let x = first
x+2= second
x+4= third
x+6= fourth

x^2 + (x+2)^2 = (x+6) + 12
x^2 + x^2 + 4x + 4 = x+ 18

Set the equation equal to zero:
2x^2 + 4x-x + 4 - 18=0
2x^2 + 3x -14= 0
(2x____ )(x______)=0
(2x + 7)(x-2)=0

Two possible solutions:

First solution:
2x+7=0
2x=-7
x=-7/2 Not an integer, so reject the solution.

Second Solution:
x-2=0
x=2
x+2=4
x+4=6
x+6=8

The numbers are 2, 4, 6, and 8.

Check: 2^2 + 4^2 = 8+12
4 + 16=20 IT CHECKS!!!

Dr. Robert J. Rapalje, Retired
Seminole State College
Altamonte Springs Campus
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Let represent the smallest integer.

Then is the next consecutive even integer.

The next one after that is

And the last one is

The square of the first one is:

The square of the second one is:

The sum of the squares of the first two is:

Twelve more than the last is

So:

Solve the quadratic for to find the smallest integer, and then count by 2s to get the next three.

Hint: This quadratic factors. You will get two roots, but one of them will not be an integer. The positive integer root is your answer.

John

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