SOLUTION: Please help me solve this word problem { { { Let "x" be the first of three consecutive even integers. If the sum of the squares of the first and second is eighty-four more than the

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Question 587484: Please help me solve this word problem { { { Let "x" be the first of three consecutive even integers. If the sum of the squares of the first and second is eighty-four more than the square of the largest, find all three integers.
Answer by dfrazzetto(283) About Me  (Show Source):
You can put this solution on YOUR website!
x, x+2, x+4
x^2 + (x+2)^2 = (x+4)^2 + 84
x^2 + x^2 + 4x + 4 = x^2 + 8x + 16 + 84
2x^2 + 4x + 4 = x^2 + 8x + 100
x^2 - 4x - 96 = 0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%2B-96+=+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=%28-4%29%5E2-4%2A1%2A-96=400.

Discriminant d=400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+400+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+400+%29%29%2F2%5C1+=+12
x%5B2%5D+=+%28-%28-4%29-sqrt%28+400+%29%29%2F2%5C1+=+-8

Quadratic expression 1x%5E2%2B-4x%2B-96 can be factored:
1x%5E2%2B-4x%2B-96+=+1%28x-12%29%2A%28x--8%29
Again, the answer is: 12, -8. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B-96+%29


x = 12, -8
technically both {-8, -6, -4} AND {12, 14, 16} satisfy