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Math Help - Math problem (involves square roots)

  1. #1
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    Math problem (involves square roots)

    Evaluate the expression (SQRT of 3 - SQRT of -4) x (SQRT of 6 - SQRT of -8) and express the result in a + bi form.

    I have the answer:

    - SQRT of 2 - 4 SQRT of 6i

    but have no clue how to get to it. We're not allowed to use calculators for this, so I'm trying to find a way not to. This is my first post, so I'm really sorry if I'm doing something wrong. Also, does anyone know how to type the square root sign? I hope I displayed my problems correctly - any questions, just ask! Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Averee View Post
    Evaluate the expression (SQRT of 3 - SQRT of -4) x (SQRT of 6 - SQRT of -8) and express the result in a + bi form.

    I have the answer:

    - SQRT of 2 - 4 SQRT of 6i

    but have no clue how to get to it. We're not allowed to use calculators for this, so I'm trying to find a way not to. This is my first post, so I'm really sorry if I'm doing something wrong. Also, does anyone know how to type the square root sign? I hope I displayed my problems correctly - any questions, just ask! Thanks.
    so we have
    (sqrt(3) - sqrt(-4))(sqrt(6) - sqrt(-8))

    ok, we will expand these brackets just as we would normally, that is, we take the first term in the first set of brackets and multiply all the terms in the second set of brackets, then we take the second term in the first set of brackets and multiply everything in the second pair of brackets. but first, let's "simplify" the square roots of the negative numbers.

    we know sqrt(-4) = sqrt(4)*sqrt(-1), a law of surds says we can split up square roots like that, and a law of exponents as well, but sqrt(-1) = i

    so sqrt(-4) = sqrt(4)*sqrt(-1) = 2i

    similarly, sqrt(-8) = sqrt(8) i, now back to the problem

    (sqrt(3) - sqrt(-4))(sqrt(6) - sqrt(-8))
    = (sqrt(3) - 2i)(sqrt(6) - sqrt(8)i)
    = sqrt(3)*sqrt(6) - sqrt(3)*sqrt(8)*i - 2i*sqrt(6) + 2*sqrt(8)*i^2 ....but i^2 = -1

    so that is:
    sqrt(3)*sqrt(6) - sqrt(3)*sqrt(8)*i - 2i*sqrt(6) - 2*sqrt(8) .......now to simplify
    = sqrt(18) - 2sqrt(8) - sqrt(24)i - 2sqrt(6)i
    = sqrt(2)*sqrt(9) - 2*sqrt(2)*sqrt(4) - sqrt(4)*sqrt(6)i - 2sqrt(6)i
    = 3sqrt(2) - 4sqrt(2) - 2sqrt(6)i - 2sqrt(6)i
    = -sqrt(2) - 4sqrt(6)i

    now these steps can get a little confusing, you have to keep track of when i'm splitting square roots and when i'm recombining. if you miss a step or don't understand something, just tell me
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  3. #3
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    Thanks, Jhevon. It took me a while, but I think I know what you did. I have to see if I can do it on another problem now...haha. Thanks again.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Averee View Post
    Thanks, Jhevon. It took me a while, but I think I know what you did. I have to see if I can do it on another problem now...haha. Thanks again.
    sure, try a problem, if you run into trouble get back to me.
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