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Having $px^3+(p-3)x^2+(2-p)x=0$ how to find p that this equality have at least 1 positive root ? How can we solve this and similiar things? Because i'm stuck... i did it for quaratic, but i can't solve it.

Any help will be appreciated.

P.S this is not homework, but it was long time ago, now i feel stupid and decided to revise it.

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In your case, you can factor by $x$ and find a quadratic. – Joel Cohen Nov 4 '11 at 12:03
up vote 1 down vote accepted

Firstly, $px^3+(p-3)x^2+(2-p)x=x[px^2+(p-3)x+(2-p)]$. Solution $x=0$ is not a positive root and can, therefore, be safely discarded. We are left with a quadratic in x. This quadratic can only have real solutions when its discriminant $D(p)=(p-3)^2-4p(2-p)=(p-1)(5p-9)$ is non-negative, hence $$ \mbox{either}\quad p\le1 \quad\mbox{or}\quad p\ge 9/5.\qquad (1) $$ The quadratic has two solutions $$ x_{1,2}=\frac{3}{2p}-\frac{1}{2}\pm\frac{1}{2p}\sqrt{(p-1)(5p-9)}.\qquad (2_{1,2}) $$ For $p>0$ solution ($2_1$) is the greater of two. When $p=9/5$, ($2_1$) gives $x=1/3>0$, and since $D(p)$ is an increasing function for $p\ge 9/5$, we conclude that $p\ge 9/5$ would satisfy the problem requirements.

The case when $p\le1$ is a bit more complicated. There are several cases to consider.

  1. When $0<p\le 1$, solution ($2_1$) gives $x=1>0$, and since $\sqrt{D(p)}/p$ is an increasing function for $0<p\le 1$, we conclude that $0<p\le 1$ satisfies the problem requirements.
  2. Solution (2) is invalid when $p=0$. Direct substitution of $p=0$ into the original equation gives $x=2/3$, hence, $p=0$ satisfies the problem requirements (the same result can be obtain by taking a limit of ($2_2$) as $p\to +0$).
  3. When $p<0$, solution ($2_2$) is the greater of two. Analysing whether it is positive directly is a bit difficult. Instead, we can do a simpler argument, similar to pharmine's. We know about the constant term of the quadratic that $(2-p)/p=x_1x_2$. When $p<0$ this term is always negative. Overall, (1) says that quadratic has two roots and $(2-p)/p=x_1x_2<0$ says that these roots are of the opposite sign, hence, one of them will always be positive.

We can now combine these observations into a full answer. Given equation has at least one positive solution provided $$ p\le 1 \quad\mbox{or}\quad p\ge 9/5. $$

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Seems like a good solution, but i can't find p for which this equation has none positive roots, i tried something like $p=-0.15$… – Spinach Nov 4 '11 at 13:34
(2) has a typo: $\frac{1}{2p}\sqrt{\cdots}$, not $\frac12\sqrt{\cdots}$ – pharmine Nov 4 '11 at 13:48
@Chris: Sorry, I made a typo while copying the quadratic and solved wrong (harder) problem. Will fix in a moment. – Aleksey Pichugin Nov 4 '11 at 14:36
@pharmine: Sorry, made a typo. Will fix now. – Aleksey Pichugin Nov 4 '11 at 14:37
Thanks, everything's clear now :) – Spinach Nov 4 '11 at 15:06

Following Aleksey Pichugin's argument, the question is to find the condition in which $px^2+(p-3)x+(2-p)=0$ ($\ast$) has at least one positive root.

First, its discriminant must be non-negative: $(p-3)^2-4p(2-p)=(p-1)(5p-9)\ge 0$, which means $p\le 1$ or $p\ge 9/5$. (1)

Provided (1) is satisfied, the negation of 'the equation ($\ast$) has at least one positive root' is 'the equation ($\ast$) has two nonpositive (=negative or zero) roots'. Let's consider the latter condition, which seems a bit easier.

If $p=0$, ($\ast$) reduces to $-3x+2=0$ and it has a positive root; we assume $p\ne 0$ below. Let the two roots of the equation be $\alpha$ and $\beta$ (provided (1) is satisfied). Then '$\alpha\le 0$ and $\beta\le 0$' is equivalent to '$\alpha\beta\ge 0$ and $\alpha+\beta\le 0$'. Here $\alpha\beta=\frac{2-p}{p}$, while $\alpha+\beta=-\frac{p-3}{p}$, so this means '$p(2-p)\ge 0$ and $p(3-p)\le 0$'. There are no $p\ne 0$ satisfying both inequalities.

Therefore the condition we want to have is (1), which is '$p\le 1$ or $p\ge 9/5$'.

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$\Rightarrow x_1=0$ and: $$x_{2,3}=\frac{3-p \pm \sqrt {5p^2-14p+9}}{2p}$$ so you have that:

$$\frac{3-p - \sqrt {5p^2-14p+9}}{2p}>0 \lor \frac{3-p + \sqrt {5p^2-14p+9}}{2p}>0$$

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