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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
m and n, the consecutive integers.
Let m%3Cn. n=m%2B1.
15%281%2Fm-1%2Fn%29=2------the symbolized translation.
Substitute for n.
15%281%2Fm-1%2F%28m%2B1%29%29=2
-
Solve for m.
15m%28m%2B1%29%281%2Fm-1%2F%28m%2B1%29%29=m%28m%2B1%292
15%28m%2B1%29-15m=2m%5E2%2B2m
15m%2B15-15m=2m%5E2%2B2m
15=2m%5E2%2B2m
2m%5E2%2B2m-15=0
-
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.
15%281%2Fn-1%2F%28n%2B1%29%29=2, exact translation of the description.
n%28n%2B1%2915%281%2Fn-1%2F%28n%2B1%29%29=2%2An%28n%2B1%29
15%28n%2B1%29-15n=2n%5E2%2B2n
15n%2B15-15n=2n%5E2%2B2n
-15=2n%5E2%2B2n
2n%5E2%2B2n%2B15=0
MUST be factorable.
(2n+3)(n+5)? No.
(2n+5)(n+3)? No.
Check the discriminant. discriminant=2%2A2-4%2A2%2815%29%3C0.

The nonfactorability in the second attempt and the negative disciminant mean that the question description is wrong.

Answer by richwmiller(17219) About Me  (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 2x%5E2%2B2x-15, we can see that the first coefficient is 2, the second coefficient is 2, and the last term is -15.



Now multiply the first coefficient 2 by the last term -15 to get %282%29%28-15%29=-30.



Now the question is: what two whole numbers multiply to -30 (the previous product) and add to the second coefficient 2?



To find these two numbers, we need to list all of the factors of -30 (the previous product).



Factors of -30:

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 -30.

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 2:



First NumberSecond NumberSum
1-301+(-30)=-29
2-152+(-15)=-13
3-103+(-10)=-7
5-65+(-6)=-1
-130-1+30=29
-215-2+15=13
-310-3+10=7
-56-5+6=1




From the table, we can see that there are no pairs of numbers which add to 2. So 2x%5E2%2B2x-15 cannot be factored.



===============================================================





Answer:



So 2%2Ax%5E2%2B2%2Ax-15 doesn't factor at all (over the rational numbers).



So 2%2Ax%5E2%2B2%2Ax-15 is prime.


and
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B2x%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=%282%29%5E2-4%2A2%2A-15=124.

Discriminant d=124 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+124+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+124+%29%29%2F2%5C2+=+2.28388218141501
x%5B2%5D+=+%28-%282%29-sqrt%28+124+%29%29%2F2%5C2+=+-3.28388218141501

Quadratic expression 2x%5E2%2B2x%2B-15 can be factored:
2x%5E2%2B2x%2B-15+=+2%28x-2.28388218141501%29%2A%28x--3.28388218141501%29
Again, the answer is: 2.28388218141501, -3.28388218141501. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B2%2Ax%2B-15+%29