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!!!!
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.
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
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
To solve this problem, let's break it down step by step.
Let's assume the three consecutive even integers as x, x + 2, and x + 4. Here, x represents the first even integer.
According to the problem statement, twice the product of the first and last numbers is 28 more than the square of the second number, which can be expressed as:
2(x)(x + 4) = (x + 2)^2 + 28
Now, let's solve this equation:
Expanding the square on the right side:
2(x)(x + 4) = x^2 + 4x + 4 + 28
Simplifying further:
2x(x + 4) = x^2 + 4x + 32
Distributing on the left side:
2x^2 + 8x = x^2 + 4x + 32
Moving all terms to one side:
2x^2 + 4x - x^2 - 4x - 32 = 0
Simplifying:
x^2 - 32 = 0
Now, let's solve this quadratic equation by factoring or using the quadratic formula.
Factoring:
(x - 4)(x + 4) = 0
From this, we can see that x can be either 4 or -4. However, we are looking for even integers, so we disregard -4.
Therefore, x = 4.
By substituting x = 4 into our initial assumption, we get the three consecutive even integers as 4, 6, and 8.
So, the three consecutive even integers that satisfy the given conditions are 4, 6, and 8.