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

Question 202691: Three consecutive even integers are such that the square of the third is 100 more than the suare of the second. Find the three integers. Solve involving applications of polynomials.
Answer by rajagopalan(174) (Show Source): You can put this solution on YOUR website!
Let the 3 numbers be (n-2), n , (n+2)
then
Square of third number=(n+2)*((n+2)=n^2+4n+4
Square of second number=n^2
Subtracting we get...4n+4 =100
4n=100-4=96
n=96/4=24
Numbers are 22, 24 and 26.
***
Verification
Square of third number=26*26=676
Square of Second number=24*24=576
Difference=100
OK
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