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.
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!!!!
5 answers
Buy why wouldn't it be n, n+2, and N+4 because it is even ingergers,
this question is still not answered will some one walk through it step by step
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
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
2 answers