SOLUTION: Find two consectuive Integers such that 15 times the difference of their reciprocols gives 2

Question 862530: Find two consectuive Integers such that 15 times the difference of their reciprocols gives 2
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
m and n, the consecutive integers.
Let . .
------the symbolized translation.
Substitute for n.

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Solve for m.

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That quadratic is expected to be factorable because you know m and n are integers.
(2m+3)(m-5)????? No.
(2m+5)(m-3)? No.
(2m+15)(m-1)? No.
(2m+1)(m+15)?

Try again. Integers n and n+1.
, exact translation of the description.

MUST be factorable.
(2n+3)(n+5)? No.
(2n+5)(n+3)? No.
Check the discriminant. .

The nonfactorability in the second attempt and the negative disciminant mean that the question description is wrong.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
Here are two ways to show there is something wrong with the problem.

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?

To find these two numbers, we need to list all of the factors of (the previous product).

Factors of :

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .

1*(-30) = -30
2*(-15) = -30
3*(-10) = -30
5*(-6) = -30
(-1)*(30) = -30
(-2)*(15) = -30
(-3)*(10) = -30
(-5)*(6) = -30

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number Second Number Sum
1 -30 1+(-30)=-29
2 -15 2+(-15)=-13
3 -10 3+(-10)=-7
5 -6 5+(-6)=-1
-1 30 -1+30=29
-2 15 -2+15=13
-3 10 -3+10=7
-5 6 -5+6=1

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

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Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

and

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=124 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2.28388218141501, -3.28388218141501. Here's your graph:

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