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Thread: 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 $\displaystyle (px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $\displaystyle p$ and $\displaystyle p \neq 0$

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

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

    $\displaystyle p^2x^2 +px -4= 0$

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

    $\displaystyle b^2 -4ac = p^2 -4p^2(-4)$

    $\displaystyle = p^2 + 16p^2$

    $\displaystyle = 17p^2$

    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 $\displaystyle (px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $\displaystyle p$ and $\displaystyle p \neq 0$

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

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

    $\displaystyle p^2x^2 +px -4= 0$

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

    $\displaystyle b^2 -4ac = p^2 -4p^2(-4)$

    $\displaystyle = p^2 + 16p^2$

    $\displaystyle = 17p^2$

    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 $\displaystyle 17p^2\ne 0 ~\implies~p\ne 0$
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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 $\displaystyle (px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $\displaystyle p$ and $\displaystyle p \neq 0$

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

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

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

    Suppose $\displaystyle $$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 $\displaystyle (px-1)^2 +3px -5 = 0$ are real and distinct for all real values of $\displaystyle p$ and $\displaystyle p \neq 0$

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

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

    $\displaystyle p^2x^2 +px -4= 0$

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

    $\displaystyle b^2 -4ac = p^2 -4p^2(-4)$

    $\displaystyle = p^2 + 16p^2$

    $\displaystyle = 17p^2$

    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 $\displaystyle p\ne 0$ is a condition of the problem, what can you say about $\displaystyle p^2$?
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