SOLUTION: Three consecutive even integers are such that the square of the third is 100 more than the square of the second. Find the three integers.

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Question 178286: Three consecutive even integers are such that the square of the third is 100 more than the square of the second. Find the three integers.
Found 2 solutions by jim_thompson5910, josmiceli:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Consecutive even integers are of the form 2x, 2x%2B2, 2x%2B4, etc...


So the word problem translates to %282x%2B4%29%5E2=%282x%2B2%29%5E2%2B100


%282x%2B4%29%5E2=%282x%2B2%29%5E2%2B100 Start with the given equation.

4x%5E2%2B16x%2B16=4x%5E2%2B8x%2B4%2B100 FOIL

4x%5E2%2B16x%2B16-4x%5E2-8x-4-100=0 Get all terms to the left side.

8x-88=0 Combine like terms.

8x=88 Add 88 to both sides.

x=%2888%29%2F%288%29 Divide both sides by 8 to isolate x.

x=11 Reduce.

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Answer:
So the answer is x=11 which means that the numbers are
2%2811%29=22
2%2811%29%2B2=24
2%2811%29%2B4=26
So the numbers are 22, 24, and 26

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the consecutive even integers
n, n+%2B+2, and n%2B4
given:
%28n+%2B+4%29%5E2+=+%28n+%2B+2%29%5E2+%2B+100
--------
n%5E2+%2B+8n+%2B+16+=+n%5E2+%2B+4n+%2B+4+%2B+100
4n+=+104+-+16
4n+=+88
n+=+22
n+%2B+2+=+24
n+%2B+4+=+26
The numbers are 22,24, and 26
check:
%28n+%2B+4%29%5E2+=+%28n+%2B+2%29%5E2+%2B+100
26%5E2+=+24%5E2+%2B+100
676+=+576+%2B+100
676+=+676
OK