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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
N%28N%2B2%29=126%2B3%28N%2B2%29
N%28N%2B2%29-3%28N%2B2%29=126
%28N-3%29%28N%2B2%29=126
N%5E2-N-6=126
N%5E2-N-132=0
%28N%2B11%29%28N-12%29=0
Only solving for even solution.
N-12=0
N=12
So the two integers are 12 and 14.

Answer by macston(5194) About Me  (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 ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-132+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-132=529.

Discriminant d=529 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+529+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+529+%29%29%2F2%5C1+=+12
x%5B2%5D+=+%28-%28-1%29-sqrt%28+529+%29%29%2F2%5C1+=+-11

Quadratic expression 1x%5E2%2B-1x%2B-132 can be factored:
1x%5E2%2B-1x%2B-132+=+1%28x-12%29%2A%28x--11%29
Again, the answer is: 12, -11. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-132+%29

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