SOLUTION: I am thinking of three consecutive integers. The square of the largest is 12 less than the sum of the squares of the other two. Find the integers.

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Question 446568: I am thinking of three consecutive integers. The square of the largest is 12 less than the sum of the squares of the other two. Find the integers.
Answer by chriswen(106) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the first integer.
Let x+1 be the second integer.
Let x+2 be the third integer.
...
x^2+(x+1)^2=(x+2)^2+12
x^2+x^2+2x+1=x^2+4x+4+12
2x^2+2x+1=x^2+4x+16
2x^2-x^2+2x-4x+1-16=0
x^2-2x-15=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-15+=+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=%28-2%29%5E2-4%2A1%2A-15=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+64+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+64+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-2%29-sqrt%28+64+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-2x%2B-15 can be factored:
1x%5E2%2B-2x%2B-15+=+1%28x-5%29%2A%28x--3%29
Again, the answer is: 5, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-15+%29

(x-5)(x+3)=0
x=5 .... or ... x= -3
x+1=6 or x+1=-2
x+2=7 or x+2=-1
Therefore, the three consecutive integers are 5,6 and 7.
Or, -3, -2, and -1.