SOLUTION: Find three consecutive positive on integers such that the sum of the squares of the first and second integers is equal to the square of the third integer -7

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Question 948636: Find three consecutive positive on integers such that the sum of the squares of the first and second integers is equal to the square of the third integer -7
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
X=first integer; X+2= second integer; X+4=third integer
X%5E2%2B%28X%2B2%29%5E2=%28X%2B4%29%5E2
X%5E2%2BX%5E2%2B4X%2B4=X%5E2%2B8X%2B16 Subtract x^2 from each side
X%5E2%2B4X%2B4=8X%2B16 Subtract 8X from each side.
X%5E2-4X%2B4=16 Subtract 16 from each side.
X%5E2-4X-12=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-12+=+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-12=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+64+%29%29%2F2%5Ca.

X%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+64+%29%29%2F2%5C1+=+6
X%5B2%5D+=+%28-%28-4%29-sqrt%28+64+%29%29%2F2%5C1+=+-2

Quadratic expression 1X%5E2%2B-4X%2B-12 can be factored:
1X%5E2%2B-4X%2B-12+=+1%28X-6%29%2A%28X--2%29
Again, the answer is: 6, -2. 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-12+%29

The positive answer is 6 ANSWER: The first integer is 6
Note if not restricted to positive integers, (-2,0,2) also work.
X+2=8 ANSWER 2: The second integer is 8
X+4=10 ANSWER 3: The third integer is 10
CHECK
Sum of the squares of the first two equal square of the third.
6%5E2%2B8%5E2=10%5E2
36%2B64=100
100=100