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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  • Fred
    Lv 7
    7 years ago
    Favorite Answer

    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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  • 7 years ago

    Let: x and y = two numbers

    They add to 2, so:

    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

    Quadratic equation:

    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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  • JOS J
    Lv 7
    7 years ago

    The answer is a complex number set

    x = 1 - 3 i Sqrt[11],

    y = 1 + 3 i Sqrt[11]

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  • 7 years ago

    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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  • 7 years ago

    There are no two numbers that fit the parameters, what type of question is this for? I'm assuming splitting the middle term of a quadratic equation?

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