SOLUTION: "The square of the sum of two consecutive positive even integers is greater than the sum of their squares by 160. Find the larger even integer."

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Question 1043101: "The square of the sum of two consecutive positive even integers is greater than the sum of their squares by 160. Find the larger even integer."
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
2n and 2n+2

%282n%2B%282n%2B2%29%29%5E2-%282n%29%5E2%2B%282n%2B2%29%5E2=160
The hard part was to formulate that equation.

Solving for n is arithmetic according to the rules about numbers you have been studying (algebra, number properties, quadratic equations).

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!

"The square of the sum of two consecutive positive even integers is greater than the sum of their squares by 160. Find the larger even integer."
Let the larger integer be L

Then smaller is: L - 2

We then get: %28L+%2B+L+-+2%29%5E2+=+L%5E2+%2B+%28L+-+2%29%5E2+%2B+160

%282L+-+2%29%5E2+=+L%5E2+%2B+L%5E2+-+4L+%2B+4+%2B+160

4L%5E2+-+8L+%2B+4+=+2L%5E2+-+4L+%2B+164

4L%5E2+-+2L%5E2+-+8L+%2B+4L+%2B+4+-+164+=+0

2L%5E2+-+4L+-+160+=+0

2%28L%5E2+-+2L+-+80%29+=+2%280%29

L%5E2+-+2L+-+80+=+0

Factor this trinomial, set each factor equal to 0, and then obtain the value(s) for L, the larger integer.

Ignore the negative value, as L, or larger integer should be positive (> 0).