Asked by Stupid
Find three consecutive even integers such that twice the product of the first and last numbers is 28 more than the square of the second number.
I know the answers i just need to know how to get to them and ive spent 30 min on it HELP ME!!!!
I know the answers i just need to know how to get to them and ive spent 30 min on it HELP ME!!!!
Answers
Answered by
bobpursley
let the integers be n, n+1, and n+2
2n*(n+2)-28=(n+1)^2
solve for n, notice it is a quadratic.
2n*(n+2)-28=(n+1)^2
solve for n, notice it is a quadratic.
Answered by
Stupid
Buy why wouldn't it be n, n+2, and N+4 because it is even ingergers,
Answered by
Not answered
this question is still not answered will some one walk through it step by step
Answered by
Reiny
you are right, it should have been
n, n+2 and n+4 so let's adjust bobpursley's equation to look like
2n(n+4) - 28 = (n+2)^2
n, n+2 and n+4 so let's adjust bobpursley's equation to look like
2n(n+4) - 28 = (n+2)^2
Answered by
Adithya
lets start with the equation
2n(n+4) - 28 = (n+2)^2
using formula (a+b)^2=a^2+2ab+b^2
2n^2+8n-28=n^2+4n+32
simplyfying,
n2+4n=32
Solving for n, we get n=4
So the even consecutive numbers are 4,6,and 8.....
checking...
2(4*8)-28=36
2n(n+4) - 28 = (n+2)^2
using formula (a+b)^2=a^2+2ab+b^2
2n^2+8n-28=n^2+4n+32
simplyfying,
n2+4n=32
Solving for n, we get n=4
So the even consecutive numbers are 4,6,and 8.....
checking...
2(4*8)-28=36
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