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Math Help - Show that the equation has real and distinct roots

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Exclamation Show that the equation has real and distinct roots

    Question: show that the roots of the equation (px-1)^2 +3px -5 = 0 are real and distinct for all real values of p and p \neq 0

    My workings:
    (px-1)^2 +3px -5 = 0

    p^2x^2 -2px +1 +3px -5 = 0

    p^2x^2 +px -4= 0

    Here a = p^2, b = p, c = -4. Now

    b^2 -4ac = p^2 -4p^2(-4)

    = p^2 + 16p^2

    = 17p^2

    This doesn't prove anything. Can you please help me? I don't have anyone to guide me.
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  2. #2
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    Quote Originally Posted by PythagorasNeophyte View Post
    Question: show that the roots of the equation (px-1)^2 +3px -5 = 0 are real and distinct for all real values of p and p \neq 0

    My workings:
    (px-1)^2 +3px -5 = 0

    p^2x^2 -2px +1 +3px -5 = 0

    p^2x^2 +px -4= 0

    Here a = p^2, b = p, c = -4. Now

    b^2 -4ac = p^2 -4p^2(-4)

    = p^2 + 16p^2

    = 17p^2

    This doesn't prove anything. Can you please help me? I don't have anyone to guide me.
    A square is allways positive or zero. If you multiply a square by a positive number the result is positive too.

    You get two different values if the discriminant unequals zero. Therefore you'll get real and distinct roots of the equation if 17p^2\ne 0 ~\implies~p\ne 0
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by PythagorasNeophyte View Post
    Question: show that the roots of the equation (px-1)^2 +3px -5 = 0 are real and distinct for all real values of p and p \neq 0

    My workings:
    (px-1)^2 +3px -5 = 0

    p^2x^2 -2px +1 +3px -5 = 0

    p^2x^2 +px -4= 0
    Suppose $$p>0 , then by Descartes rule of signs there is exactly one positive roots, and exactly one negative root.

    Suppose $$p<0 , then by Descartes rule of signs there is exactly one positive root and one negative root.

    QED

    CB
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  4. #4
    MHF Contributor

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    Quote Originally Posted by PythagorasNeophyte View Post
    Question: show that the roots of the equation (px-1)^2 +3px -5 = 0 are real and distinct for all real values of p and p \neq 0

    My workings:
    (px-1)^2 +3px -5 = 0

    p^2x^2 -2px +1 +3px -5 = 0

    p^2x^2 +px -4= 0

    Here a = p^2, b = p, c = -4. Now

    b^2 -4ac = p^2 -4p^2(-4)

    = p^2 + 16p^2

    = 17p^2

    This doesn't prove anything. Can you please help me? I don't have anyone to guide me.
    Do you recall why you calculated that? It is the discriminant of the equation and a quadratic equation has distinct real roots if and only if its discriminant is positive. Since p\ne 0 is a condition of the problem, what can you say about p^2?
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