SOLUTION: the product of two consecutive even integers is 126 greater than three times the larger number. what are the two numbers?

Question 945475: the product of two consecutive even integers is 126 greater than three times the larger number. what are the two numbers?
Found 2 solutions by Fombitz, macston:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!

Only solving for even solution.

So the two integers are 12 and 14.

Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
let A=the first integer
(A)(A+2)=3(A+2)+126
(A)(A+2)=3A+6+126
A^2+2A-3A-6-126=0
A^2-A-132=0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=529 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 12, -11. Here's your graph:

Answers 12,-11 Even result=12
ANSWER 1: The first integer is 12
A+2=12+2=14 ANSWER 2: The second integer is 14
CHECK
(12)(14)=3(14)+126
168=42+126
168=168

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