SOLUTION: given two consecutive positive even integers, it turns out that the square of the larger minus 10 times the smaller is 76. Find the two integers.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: given two consecutive positive even integers, it turns out that the square of the larger minus 10 times the smaller is 76. Find the two integers.      Log On


   

Question 962495: given two consecutive positive even integers, it turns out that the square of the larger minus 10 times the smaller is 76. Find the two integers.
Answer by hkwu(60) About Me  (Show Source):
You can put this solution on YOUR website!
Let a be the larger integer and b be the smaller integer. We know that
a = b + 2
since they're consecutive even integers and a is the larger one. Also,
a%5E2-10b=76
as we are told. Substituting, we get
%28b%2B2%29%5E2-10b=76
b%5E2%2B4b%2B4-10b=76
b%5E2-6b%2B4=76
b%5E2-6b-72=0
We use the quadratic formula to get
b+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A1%2A-72+%29%29%2F%282%2A1%29+
b=%286%2B-sqrt%28324%29%29%2F2
b=%286+%2B-+18%29%2F2
b=3+%2B-+9
So b could be 12 or -6, but since the question says the numbers are positive, b must be 12. We conclude that a is 14.
Check:
14%5E2-10%2812%29=196-120=76