what are 3 consecutive positive even integers that the product of the second and third integers is twebnty more than ten times the first integer?
n,n+2, n+4 are the integers.
(n+2)(n+4)=20+10n
solve first for n.
thank you
can you please explain?
To find the three consecutive positive even integers, let's break down the problem step by step.
Let's assume the first even integer is "x". Since we are looking for consecutive positive even integers, the second even integer would be "x + 2" and the third even integer would be "x + 4".
According to the given information, the product of the second and third integers [(x + 2) * (x + 4)] is twenty more than ten times the first integer (10x + 20).
So, we can set up an equation:
(x + 2) * (x + 4) = 10x + 20
Now, let's solve this equation to find the value of "x" and subsequently the three consecutive positive even integers.
Expanding the equation:
x^2 + 6x + 8 = 10x + 20
Simplifying:
x^2 - 4x - 12 = 0
Now, we need to solve this quadratic equation. We can factor it:
(x - 6)(x + 2) = 0
Setting each factor equal to zero:
x - 6 = 0 or x + 2 = 0
Solving for "x":
x = 6 or x = -2
Since we are looking for positive integers, we discard the negative value, so x = 6.
Now, substituting the value of "x" back into the equation, we can find the three consecutive even integers:
First integer (x): 6
Second integer (x + 2): 6 + 2 = 8
Third integer (x + 4): 6 + 4 = 10
Therefore, the three consecutive positive even integers that satisfy the given condition are 6, 8, and 10.