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    hi can someone help me with this proof.

    prove that the equation

    x^2 + px + q = 0

    has distinct, real roots. and only when p^2 > 4q


    thank you if you can help

    x
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    A quadratic has distinct real roots if the discriminant is greater than zero; in other words, if b^2 - 4ac > 0.
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    (Original post by rachio)
    hi can someone help me with this proof.

    prove that the equation

    x^2 + px + q = 0

    has distinct, real roots. and only when p^2 > 4q


    thank you if you can help

    x
    p^2- 4q > 0

    p^2 > 4q
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    (Original post by TheWolf)
    p^2- 4q > 0

    p^2 > 4q
    is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

    thank you!
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    (Original post by rachio)
    is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

    thank you!
    b^2 - 4ac > 0 if there are real roots

    so you just do it
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    (Original post by rachio)
    is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

    thank you!
    They probably show why if b^2 > 4ac there are two distinct real roots. However, you may just quote this in your actual exam.
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    I think it was looking for

    x^2 + px + q = 0

    (x + p/2)^2 - p^2/4 + q = 0

    (x+p/2)^2 = p^2/4 - q

    x + p/2 = +/- (p^2/4 - q)^1/2

    x = -p/2 +/- (p^2/4 - q)^1/2

    you can see that that will only be real when p^2/4 - q > 0

    or p^2 > 4q

    QED
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    (Original post by fishpaste)
    I think it was looking for

    x^2 + px + q = 0

    (x + p/2)^2 - p^2/4 + q = 0

    (x+p/2)^2 = p^2/4 - q

    x + p/2 = +/- (p^2/4 - q)^1/2

    x = -p/2 +/- (p^2/4 - q)^1/2

    you can see that that will only be real when p^2/4 - q > 0

    or p^2 > 4q

    QED

    what the hells this is this in the syllabus?
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    (Original post by TheWolf)
    what the hells this is this in the syllabus?
    Making x the subject and obtaining the quadratic formula, duuh.
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    (Original post by rachio)
    is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

    thank you!
    Well, the longer method is to complete the square

    x^2 + px + q = 0
    ( x^2 + px + p^2/4 ) - p^2/4 + q = 0
    (x + p/2)^2 = p^2/4 - q
    x + p/2 = sqrt( p^2/4 - q )

    Now we consider the square root. Square root does not have a real value if p^2/4 - q is negative. The square root has only one value if p^2/4 - q = 0. We require the equation to have distinct real roots, so that means that the square root must be real and have more than one value. This means that p^2/4 - q > 0, therefore p^2 > 4q.
 
 
 
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