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Question 939478: I have to find two numbers that equal 60 when multiplyed and equal 4 when added or subtracted. I can't find any though.
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! I have to find two numbers that equal 60 when multiplyed and equal 4 when added or subtracted. I can't find any though.
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x*y = 60
x - y = 4 --> y = x - 4
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Sub for y
x*(x-4) = 60
x^2 - 4x - 60 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
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=256 is greater than zero. That means that there are two solutions: .


Quadratic expression can be factored:

Again, the answer is: 10, -6.
Here's your graph:
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x = 6
y = 10
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x = -6
y = -10
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x*y = 60
x + y = 4 --> y = 4 - x
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Sub for y
x*(4 - x) = 60
-x^2 + 4x - 60 = 0
x^2 - 4x + 60 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
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 -224 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 -224 is + or - .
The solution is , or
Here's your graph:
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x = 2 + sqrt(56)i, y = 2 - sqrt(56)i
x = 2 - sqrt(56)i, y = 2 + sqrt(56)i
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