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Math Help - [SOLVED] Algebra Proof

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Algebra Proof

    Consider the statement that may have occurred to early algebraists:
    A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real
    numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.

    Is this statement true or false? If true, give a proof and explain what a proof of a mathematical statement is. If false, give a counterexample and explain what a counterexample
    to a statement is.

    I have no idea how to do this question please help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    Consider the statement that may have occurred to early algebraists:
    A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real
    numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.

    Is this statement true or false? If true, give a proof and explain what a proof of a mathematical statement is. If false, give a counterexample and explain what a counterexample
    to a statement is.

    I have no idea how to do this question please help.
    Hint: look at the quadratic formula. note where the discriminant is. what would happen if it were less than zero?

    when you come up with the reason we can talk about whether or not a counter-example exists and how to find one if it does
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  3. #3
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    Quote Originally Posted by ronaldo_07 View Post
    Consider the statement that may have occurred to early algebraists: A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.
    What do you know about a, b, & c?
    Consider the quadratic equation ix^2  - x + i = 0.
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  4. #4
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    Quote Originally Posted by ronaldo_07 View Post
    Consider the statement that may have occurred to early algebraists:
    A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real
    numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.

    Is this statement true or false? If true, give a proof and explain what a proof of a mathematical statement is. If false, give a counterexample and explain what a counterexample
    to a statement is.

    I have no idea how to do this question please help.
     ax^2+bx+c = 0

     = a(x^2+\frac{b}{a}x)+c = 0

     = a(x^2+\frac{b}{a}x + (\frac{b}{2a})^2-(\frac{b}{2a})^2)+c = 0


     = a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2)+c = 0

     = a(x+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c = 0

     = a(x+\frac{b}{2a})^2-(\frac{b^2}{4a})+c = 0


     = a(x+\frac{b}{2a})^2 =(\frac{b^2}{4a})^2-c

     = (x+\frac{b}{2a})^2=(\frac{b^2}{4a^2})-\frac{c}{a}

     = (x+\frac{b}{2a})^2 =(\frac{b^2}{4a^2})-\frac{4ac}{4a^2}

     = (x+\frac{b}{2a}) =\pm \sqrt{(\frac{b^2-4ac}{4a^2})}


     = (x+\frac{b}{2a}) =\pm (\frac{\sqrt{b^2-4ac}}{2a})

     = x = -\frac{b}{2a} \pm (\frac{\sqrt{b^2-4ac}}{2a})

     = x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    And you can see from this that there are only real solutions if  b^2-4ac \geq 0.
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  5. #5
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    Quote Originally Posted by Mush View Post
    And you can see from this that there are only real solutions if  b^2-4ac \geq 0.
    Mush, have you considered my example?
    ix^2  - x + i = 0
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  6. #6
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    Quote Originally Posted by Plato View Post
    Mush, have you considered my example?
    ix^2  - x + i = 0
    Ah. I didn't realise there wasn't a restriction for a, b and c to be real.
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  7. #7
    Member ronaldo_07's Avatar
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    Thanks for the help I just used an example of a quadratic equation and showed that the negative can not be square rooted. is that enough for the proof?
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  8. #8
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    ronaldo, a polynomial with two real roots implies that the discriminant is non-negative. However, a non-negative discriminant does not imply that the solutions are real. The proposition is then false. Plato gave the couterexample that proves this by contradiction:

    Quote Originally Posted by Plato View Post
    Mush, have you considered my example?
    ix^2  - x + i = 0
    b^2-4ac = 1 - 4i^2 = 1 + 4 = 5 but the solutions are not real.
    Last edited by mr fantastic; January 7th 2009 at 07:53 PM. Reason: Fixed up the post
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  9. #9
    Member ronaldo_07's Avatar
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    what are these examples please explain
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  10. #10
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    Quote Originally Posted by ronaldo_07 View Post
    what are these examples please explain
    A simple counter-example has been given several times in this thread.

    Read post #8 again.
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  11. #11
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    What if for example A,B,C, are real numbers where A can not equal zero, whould this statement then hold.
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  12. #12
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    Quote Originally Posted by 1234567 View Post
    What if for example A,B,C, are real numbers where A can not equal zero, whould this statement then hold.
    That had been assumed and discussed early in the thread.
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