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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1Find the range of \('p\ '\) if roots of the equation \(\large \color{black}{\begin{align} x^22x+p^23p4=0 \hspace{.33em}\\~\\ \end{align}}\) are opposite in sign.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4what do you mean that the roots are opposite in sign, one x solution is negative and the other x solution is positive ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4And they have to be real number solutions, correct?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4the x solutions i mean

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle x^22x+p^23p4=0 }\) \(\large\color{black}{ \displaystyle x^22x=p^2+3p+4 }\) I am going to treat \(\large\color{black}{ \displaystyle p^2+3p+4 }\) \(\large\color{black}{ \displaystyle x^22x+1=p^2+3p+4+1 }\) making a perfect square trinomial and factoring \(\large\color{black}{ \displaystyle (x1)^2=p^2+3p+5 }\) \(\large\color{black}{ \displaystyle x1=\pm\sqrt{p^2+3p+5} }\) \(\large\color{black}{ \displaystyle x=1\pm\sqrt{p^2+3p+5} }\) this is what I am up to so far

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4now you need to find the prange 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...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4oh, the 3rd line should say `I am going to treat "p^2+3p+4 as a constant`

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle p^2+3p+5 }\) \(\large\color{black}{ \displaystyle (p^23p)+5 }\) \(\large\color{black}{ \displaystyle (p^23p)\frac{9}{4}+\frac{9}{4}+5 }\) \(\large\color{black}{ \displaystyle \left(p^23p+\frac{9}{4}\right)+\frac{9}{4}+5 }\) \(\large\color{black}{ \displaystyle \left(p^23p+\frac{9}{4}\right)+\frac{29}{4} }\) \(\large\color{black}{ \displaystyle \left(p\frac{3}{2}\right)^2+\frac{29}{4} }\) 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

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4sorry for long posting. if you prefer a verbal expln. pliz let me know, i will try my best.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1the solutions seems little longer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\large \color{black}{\begin{align} x^22x+\color{blue}{p^23p4}=0 \hspace{.33em}\\~\\ \end{align}}\] 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 : \[\color{blue}{p^23p4}\lt0\\~\\\color{blue}{(p+1)(p4)}\lt 0\\~\\ \color{blue}{1\lt p\lt 4}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 I have to log in to say: " You are a math wicked"
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