Suppose you can take square roots of negative numbers. Find two numbers with each following sum and product. Sum of 3, product of 5. Sum of 6, product of 15.

Question

I am not sure how to do this as I have not worked with things like this and was hoping for some help.

Answer 1

i is the square root of -1

i^2 = -1

 

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X+Y = 3   and XY=5

 

X(3-x)=5

3x - x^2 - 5 = 0

 

x^2 - 3x + 5 = 0

 

[3 +or- sqrt(  9 - 20) ]/2

 

[ 3 +or- sqrt(-11)]/2

 

[3 +or- i*sqrt(11) ]/2

 

 

[ 3+ i*sqrt(11)]/2 * y = 5

 

y = 10/[3 + i*Sqrt(11)]

 

   = 10 [ 3 - i*Sqrt(11)]/ 20 

 

    = [ 3 - i*Sqrt(11)]/2

 

 

CHECK:

 

  [ 3 + i*Sqrt(11)]/2 * [3 - i*Sqrt(11)]/2

 

 [ 9 + 11]/4 = 20/4 = 5

 

 and the total is 3 because i*sqrt(11) cancels out

 

 

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the second problem is done similarly.

the equations are X+Y = 6 and XY = 15

 

Then Y = 6 - x upon solving the first equation for y.

Next you substitute that into the second equation:

   X(6-x)=15

   6x - x^2 - 15 = 0

 

    x^2 - 6x + 15 = 0

 

    x = [ 6 +or- sqrt(36 - 60)]/2

       = [ 6 +or- sqrt(-24)]/2

 

       = [ 6 +or- 2i*Sqrt(6)]/2

 

        = [ 3 +or- i*Sqrt(6)] <--- reduces by a factor of 2

 

     

 

 Y = 15/[3 + i*Sqrt(6)] = 15 [ 3 - i*Sqrt(6)]/ [ 9 + 6] = 3 - i*sqrt(6)

 

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Answer 2

First problem:

 

x + y = 3

xy = 5

 

Solve by substitution.

 

y = 5/x

x + 5/x = 3

x² + 5 = 3x

x² - 3x + 5 = 0

 

x = (3 ± √-11)/2 = (3 ± i√11)/2,

 

where i = √-1

 

y = 3 - x = 3 - (3 ± i√11)/2 = ?

 

Note you will have 2 solutions for y, just as you have 2 solutions for x.  Now you try working the second problem.