# what two numbers multippiles to 100 and adds to 2?

Update:

i meant -2

Update 2:

llaffer your answer does not work -99-1=100 but does not multiple to 100 hundred

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llaffer's method will work with the new sum, -2, in place of the old, +2, from the start of his answer.

I like to do these like this. Manipulate the sum, S and product, P, to give you the difference, D:

S = x + y = -2

P = xy = 100

S² - 4P = -396 = (x² + 2xy + y²) - 4xy = x² - 2xy + y² = (x - y)²

D = x - y = ±6i√11

Then use the sum and difference to find the numbers:

{x,y} = ½(S±D)

= -1 ± 3i√11

Check:

(-1 + 3i√11)(-1 - 3i√11) = 1 - 9(-1)(11) = 1 + 99 = 100

(-1 + 3i√11) + (-1 - 3i√11) = -2

Another handy way to work these is to know that the solutions of the quadratic

x² - Sx + P = 0

add up to S, and multiply out to P, because, if you call the roots, a and b, then

(x - a)(x - b) = x² - (a+b)x + ab

So for your problem, you will have

x² + 2x + 100 = 0

to solve. Its two roots are your two numbers.

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• Let: x and y = two numbers

x + y = 2

and multiply to 100, so ..

xy = 100

We now have a system of equations. I'll set equation 1 to x in terms of y and substitute it into the second equation.

x + y = 2

x = 2 - y

xy = 100

(2 - y)y = 100

2y - y² = 100

0 = y² - 2y + 100

y = [ -b ± √(b² - 4ac)] / (2a)

y = [ -(-2) ± √((-2)² - 4(1)(100))] / (2(1))

y = [ 2 ± √(4 - 400)] / 2

y = [ 2 ± √(-396)] / 2

y = [ 2 ± √(36 * -11)] / 2

y = [ 2 ± 6√(-11)] / 2

y = 1 ± 3√(-11)

y = 1 ± 3i√(11)

This ends up being both of your numbers:

1 + 3i√11 and 1 - 3i√11

We can test it to make sure the rules work. First, I'll add them to make sure they add up to 2.

The 3i√11 and -3i√11 cancels out, leaving: 1 + 1 = 2.

Now to multiply:

(1 + 3i√11)(1 - 3i√11)

FOIL ...

1 - 3i√11 + 3i√11 - 9i²(11)

1 - 99i²

1 - 99(-1)

1 + 99

100

Both tests work. Your solution is:

1 ± 3i√11

EDIT: Now that you changed the answer to -2, the steps above are still the same, you just have different numbers to work with. So you can use my work as an example and try x + y = -2 and xy = 100 on your own.

EDIT2: I didn't say that my numbers were 1 and 99. I said that they are 1 + 3i√11 and 1 - 3i√11

I showed above how you multiply those two irrational numbers together and it gives you 100, satisfying one of the conditions of the problem.

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• The answer is a complex number set

x = 1 - 3 i Sqrt,

y = 1 + 3 i Sqrt

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• x*y=100

x+y=2

x=2-y

-y^2-2y=100 y^2+2y+100=0

a=1 b=2 c=100

-2+/- 2^2-4(1)(100)/ 2

x=-2 +/- sq rt 4-400/ 2

x=-2+/- sq rt -396 / 2

x=-2+/- 2i sq rt 99/2

x=-1+i sq rt 99 -1-i sq rt 99 no solution

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