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Math Help - Factoring to find a solution to what x is

  1. #1
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    Factoring to find a solution to what x is

    Find all solutions of the equation x^2+3x+8=0 and express them in the form a+bi: (I have no idea what 'i' stands for)

    I have to find 2 sets of numbers in which the first set b < 0, then another set with b > 0. A and B are both real numbers.

    I can't find an easy way to factor it, and when I put it into the quadratic formula it isn't possible because you cant squareroot -26.
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    Quote Originally Posted by Lolcats View Post
    Find all solutions of the equation x^2+3x+8=0 and express them in the form a+bi: (I have no idea what 'i' stands for)

    I have to find 2 sets of numbers in which the first set b < 0, then another set with b > 0. A and B are both real numbers.

    I can't find an easy way to factor it, and when I put it into the quadratic formula it isn't possible because you cant squareroot -26.
    (actually, it's \sqrt {-23} that you have to deal with) well, that's the point of the i. you have complex roots here. i = \sqrt{-1}. so, \sqrt {-23} = \sqrt{(-1)(23)} = \sqrt {-1} \sqrt {23}= i \sqrt {23}
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    Quote Originally Posted by Jhevon View Post
    (actually, it's \sqrt {-23} that you have to deal with) well, that's the point of the i. you have complex roots here. i = \sqrt{-1}. so, \sqrt {-23} = \sqrt{(-1)(23)} = \sqrt {-1} \sqrt {23}= i \sqrt {23}
    Ohay, so that means I know what my i value is now? So should i just multiply my i value by my b value in the quadratic formula to get the values?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lolcats View Post
    Ohay, so that means I know what my i value is now? So should i just multiply my i value by my b value in the quadratic formula to get the values?
    uh...no. do the quadratic formula, simplify the negative square root as i showed you, and split the answer in two terms, one with i in it, one without i in it.
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    Quote Originally Posted by Jhevon View Post
    uh...no. do the quadratic formula, simplify the negative square root as i showed you, and split the answer in two terms, one with i in it, one without i in it.

    If one solution is complex, then the other solution is complex. This is a major theorem.

    Both solutions will have  i in it if one does (the roots are complex conjugates)
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    Ahhh, alright I'll try that. Thank you.
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    One last thing... when their asking me to express it in a + bi... is that like in terms of the variables from the quadric formula? Or am I suppose to let a + bi = the two x's found, and then solve for a and b?
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    Quote Originally Posted by Lolcats View Post
    One last thing... when their asking me to express it in a + bi... is that like in terms of the variables from the quadric formula? Or am I suppose to let a + bi = the two x's found, and then solve for a and b?
    a+bi is the general form a complex number, i know it might be confusing, but has nothing to do with the a,b from the quadratic formula, it just means that when you run through the quadratic formula you will get an answer, part of it will have i and part of it wont

    write it as (real number)+(real number)i
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by emathinstruction View Post
    a+bi is the general form a complex number, i know it might be confusing, but has nothing to do with the a,b from the quadratic formula, it just means that when you run through the quadratic formula you will get an answer, part of it will have i and part of it wont

    write it as (real number)+(real number)i
    haha, oh, now i see where the confusion came from i was wondering why the OP said to multiply i by b. yes, the b in a + ib is not the same b as the one in the quadratic formula.
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    Uhh k..

    First input the solution with b< 0 here:
    the real number a equals_______ and the real number b equals_______
    Then input the solution with b0 here:
    the real number a equals_______ and the real number b equals _______

    Thats the format I'm suppose to answer it in, I'm just really confused now how to get through this...

    Sorry for dragging this out, I've never had a question like this or delt with 'complex conjugates'
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lolcats View Post
    Uhh k..

    First input the solution with b< 0 here:
    the real number a equals_______ and the real number b equals_______
    Then input the solution with b>0 here:
    the real number a equals_______ and the real number b equals _______

    Thats the format I'm suppose to answer it in, I'm just really confused now how to get through this...

    Sorry for dragging this out, I've never had a question like this or delt with 'complex conjugates'
    example: lets say you did all i said and you ended up with something like \frac {1 \pm i \sqrt 2}2

    first step: \frac 12 \pm i \frac { \sqrt 2}2

    so a = \frac 12, and b = \pm \frac {\sqrt 2}2. got it?

    so answering your questions with these, it would be:

    (1) _____1/2___ and ____-_sqrt(2)/2_________

    (2) _____1/2___ and _____sqrt(2)/2_________
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  12. #12
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    OOooooooh, yeah now I see it. Thank you for taking your time to answer this!
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