Question 893361: Help? Im kinda confused on this one:
Find three consecutive integers such that the product of the first and the third is 20 more than the second.(Application of Quadratic function.)
I've tried with
xz=y+20
or
(a sub 1)(a sub 3)= (a sub 2)+ 20
but cant derive any equations for the three variables
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39617)   (Show Source):
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! x^2+2x-21=0
| 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 :
1,3,7,21
-1,-3,-7,-21
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to .
1*(-21) = -21
3*(-7) = -21
(-1)*(21) = -21
(-3)*(7) = -21
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 | -21 | 1+(-21)=-20 | | 3 | -7 | 3+(-7)=-4 | | -1 | 21 | -1+21=20 | | -3 | 7 | -3+7=4 |
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.
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