SOLUTION: please help me solve this 5n(squared+2n+6=0

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Question 278949: please help me solve this 5n(squared+2n+6=0
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
5n^2+2n+6=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 :

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

-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 NumberSecond NumberSum
1301+30=31
2152+15=17
3103+10=13
565+6=11
-1-30-1+(-30)=-31
-2-15-2+(-15)=-17
-3-10-3+(-10)=-13
-5-6-5+(-6)=-11




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



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





Answer:



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



So is prime.


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 : .

The discriminant -116 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -116 is + or - .

The solution is

Here's your graph:

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