SOLUTION: what two numbers multiply to negative 14 and add to 1??

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Question 1024905: what two numbers multiply to negative 14 and add to 1??
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52784) About Me  (Show Source):
You can put this solution on YOUR website!
.
what two numbers multiply to negative 14 and add to 1??
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Let x and y be these two numbers. Then

x + y = 1,   (1)
xy = -14.    (2)

From (1) x = 1-y. Substitute it into (2). You will get

y*(1-y) = 14,   or

y%5E2+%2B+y+%2B+14 = 0.

Solve this quadratic equation by using the quadratic formula (factoring doesn't work):

y%5B1%2C2%5D = %28-1+%2B-+sqrt%281%5E2+-+4%2A14%29%29%2F2 = %28-1+%2B-+sqrt%28-55%29%29%2F2.

The roots are  y%5B1%5D = %28-1+%2B+i%2Asqrt%2855%29%29%2F2  and  y%5B2%5D = %28-1+-+i%2Asqrt%2855%29%29%2F2.

Thus the solutions to the problem are these two numbers 

%281+%2B+i%2Asqrt%2855%29%29%2F2, %281+-+i%2Asqrt%2855%29%29%2F2.


Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Use x and y
x + y = 1 --> y = 1-x
xy = -14
------
x*(1-x) = 14
x^2 - x + 14 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B14+=+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=%28-1%29%5E2-4%2A1%2A14=-55.

The discriminant -55 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 -55 is + or - sqrt%28+55%29+=+7.41619848709566.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B14+%29

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The numbers are 1%2F2%2B-sqrt%2855%29i%2F2