Find three consecutive integers such that four times the sum of all three is 2 times more than the product of the larger two?

Question

It should have two possible answers and it is a quadratic equation.

Answer 1

If n is an integer, then the three consecutive integers would be n, n+1, and n+2.

"Four times the sum of all three" becomes 4 (n + n+1 + n+2) = 4(3n+3) = 12n + 12

"is" means EQUALS in math-speak

So now we have 12n + 12 =

"two times more than the product of the larger two" --

"the product of the larger two" becomes (n+1)(n+2)

So "two times" then makes this part of the sentence say 2*(n+1)(n+2) = 2(n² + 3n + 2) = 2n² + 6n + 4

Putting all of this together, we have

12n + 12 = 2n² + 6n + 4

put everythign on the right side

0 = 2n² + 6n + 4 - 12n - 12

or

0 = 2n² - 6n - 8

Factor out a 2

0 = 2(n² - 3n - 4) (divide both sides by 2 and get 0 = n² -3n -4)

and solve n² - 3n - 4 = 0

factor

(n-4)(n+1) = 0

n-4 = 0 or n+1 = 0

so n=4 or n = -1

Now let's check:

if n=4, then our 3 consecutive integers are 4, 5, 6. 4(4+5+6) = 60 and 2n² + 6n + 4 = 2(16)+6(4)+4 = 60

So 4,5,6 works!

if n= -1, then our 3 consecutive integers are -1, 0, 1 4(-1`+0+1) = 0 2(-1)² + 6(-1) + 4 = 2-6+4 = 0

So -1, 0, 1 works!

There are the two answers! :-)