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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    Find the range of \('p\ '\) if roots of the equation \(\large \color{black}{\begin{align} x^2-2x+p^2-3p-4=0 \hspace{.33em}\\~\\ \end{align}}\) are opposite in sign.

  2. SolomonZelman
    • one year ago
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    what do you mean that the roots are opposite in sign, one x solution is negative and the other x solution is positive ?

  3. mathmath333
    • one year ago
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    ya this one

  4. SolomonZelman
    • one year ago
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    And they have to be real number solutions, correct?

  5. SolomonZelman
    • one year ago
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    the x solutions i mean

  6. mathmath333
    • one year ago
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    yes

  7. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle x^2-2x+p^2-3p-4=0 }\) \(\large\color{black}{ \displaystyle x^2-2x=-p^2+3p+4 }\) I am going to treat \(\large\color{black}{ \displaystyle -p^2+3p+4 }\) \(\large\color{black}{ \displaystyle x^2-2x+1=-p^2+3p+4+1 }\) making a perfect square trinomial and factoring \(\large\color{black}{ \displaystyle (x-1)^2=-p^2+3p+5 }\) \(\large\color{black}{ \displaystyle x-1=\pm\sqrt{-p^2+3p+5} }\) \(\large\color{black}{ \displaystyle x=1\pm\sqrt{-p^2+3p+5} }\) this is what I am up to so far

  8. SolomonZelman
    • one year ago
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    now you need to find the p-range that will make one pos. and one neg. solut. for x. I will think how to do it percisely, don't want to make an err anywhere...

  9. SolomonZelman
    • one year ago
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    oh, the 3rd line should say `I am going to treat "-p^2+3p+4 as a constant`

  10. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle -p^2+3p+5 }\) \(\large\color{black}{ \displaystyle -(p^2-3p)+5 }\) \(\large\color{black}{ \displaystyle -(p^2-3p)-\frac{9}{4}+\frac{9}{4}+5 }\) \(\large\color{black}{ \displaystyle -\left(p^2-3p+\frac{9}{4}\right)+\frac{9}{4}+5 }\) \(\large\color{black}{ \displaystyle -\left(p^2-3p+\frac{9}{4}\right)+\frac{29}{4} }\) \(\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2+\frac{29}{4} }\) now, this (the whole expression) has to be greater than 1, so that x is positive or negative. \(\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2+\frac{29}{4}>1 }\) \(\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2>1-\frac{29}{4} }\) \(\large\color{black}{ \displaystyle -\left(p-\frac{3}{2}\right)^2>-\frac{25}{4} }\) \(\large\color{black}{ \displaystyle \left(p-\frac{3}{2}\right)^2<\frac{25}{4} }\) \(\large\color{black}{ \displaystyle \left(p-\frac{3}{2}\right)^2<\frac{25}{4} }\) so this p solution is -1<p<4

  11. SolomonZelman
    • one year ago
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    sorry for long posting. if you prefer a verbal expln. pliz let me know, i will try my best.

  12. mathmath333
    • one year ago
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    excellent

  13. mathmath333
    • one year ago
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    the solutions seems little longer

  14. ganeshie8
    • one year ago
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    \[\large \color{black}{\begin{align} x^2-2x+\color{blue}{p^2-3p-4}=0 \hspace{.33em}\\~\\ \end{align}}\] The constant term here represents the product of roots, so it must be negative for the quadratic equation to have roots that are of different signs : \[\color{blue}{p^2-3p-4}\lt0\\~\\\color{blue}{(p+1)(p-4)}\lt 0\\~\\ \color{blue}{-1\lt p\lt 4}\]

  15. Loser66
    • one year ago
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    @ganeshie8 I have to log in to say: " You are a math wicked"

  16. ganeshie8
    • one year ago
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    lol xD

  17. mathmath333
    • one year ago
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    Perfect!

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