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I need to find which values of $m$ will cause $x^2-mx-x+m+4=0$ to have two positive solutions.

So what I know that discriminant should be positive too $D>0$ $(m+1)^2-4(m+4)>0$ $(m+3)(m-5)>0$ $m\in (-\infty;-3)\cup(5;\infty)$ I also know that x minima should be positive, so $-\frac{-(m+1)}{2}>0$ $m>-1$

But unfortunately it's not enough, could you help me to find solution?

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    @Bill: It was not correct. I've since fixed it. Thank you and thanks to user6312 for pointing out the error.2011-04-07

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Assuming that when you say "two positive solutions" you mean "two positive distinct solutions" you are correct that you need the discriminant to be positive, and you've correctly computed this.

But I don't think you figured out the correct condition for the roots to be positive. You can have a quadratic with vertex in the positive numbers, but one root positive and one root negative. For example, take $(x+1)(x-4)=x^2-3x-4$. The minimum occurs at $x=\frac{3}{2}$, but one root is negative.

So while having the vertex in the positive numbers is necessary, it is not sufficient.

Instead, think about the quadratic formulas for the roots. Since the roots are $\frac{(m+1) \pm \sqrt{D}}{2}.$

The smaller root will occur when we take $\frac{m+1-\sqrt{D}}{2}$, so you want $m+1-\sqrt{D}\gt 0$, which means you want $m+1\gt \sqrt{D}$.

Edit and fix. If you square both sides, we get $(m+1)^2 \gt D = (m+1)^2 - 4(m+4)$, which means you want $0\gt -4(m+4)$, which means you want $-4(m+4)\lt 0$, or $m+4 \gt 0$. But this may add spurious solutions (because of the squaring), which occur when $m+1 \lt - \sqrt{D}\lt 0$. So we need to take out these possibilities. These spurious solutions are introduced example when $|m+1|\geq \sqrt{D}$ and $m+1\lt 0$; that is, $(m+1)^2\geq D$ and $m+1\lt 0$. As above, these occur when $m+4\gt 0$ and $m+1\lt 0$, that is, when $-4\lt m\lt -1$.

So we need $m\gt -4$, and not to have $m\lt -1$. This gives $m\geq -1$ in summary.

So the conditions are: $m\in (-\infty,-3)$ and $m\geq -1$ (impossible); or $m\in (5,\infty)$ and $m\geq -1$, which yield $m\in (5,\infty)$.

If you are okay with a single double root which is positive, then you also allow $D=0$, which means you need to include $x=5$ into your solution set.

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    @Bill: Again: I'm not *complaining* that you added the comment. I'm only explaining why added my coda. Of course, insofar as the answer is no longer incorrect, you could remove the comment; or not. For the very same reason that you wated to notify the OP (and other readers) that the answer was incorrect at the time, I wanted to notify readers that this was no longer the case.2011-04-07
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If both roots are $> 0$ then so is their sum $\rm\:m+1\:,\:$ hence $\rm\: m> -1\:,\:$ which excludes $\rm\:m\in (-4,-3)\:.$ In the remaining "real" interval $\rm\:I = (5,\infty)$ the root sum $\rm\:m+1 > 0\:,\:$ thus at least one root is $ > 0\:.\:$ But since also in $\rm\:I\:$ the root product $\rm\:m+4\: > 0\:,\:$ it follows that both roots are $> 0$ for $\rm\:m\in (5,\infty)$.

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    I just noticed way down at the bottom of the page (off my screen) that Fabian mentioned a similar way in a deleted answer, and user63$1$2 hinted similarly in the comments (alas, I rarely have time to read comments before answering). Please do post answers (not comments), and don't delete correct answers! Not everybody reads comments, nor does everybody see deleted answers.2011-04-07
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Quadratic equation $ax^2+bx+c=0$

has positive roots if

$b^2-4ac>0$

$x_1+x_2={-b\over a}>0$

$x_1*x_2={c\over a}>0$


negative roots if

$b^2-4ac>0$

$x_1+x_2={-b\over a}<0$

$x_1*x_2={c\over a}>0$


two zero roots if

$b=0$ $c=0$


has one positive root and one negative root if

$b^2-4ac>0$

$x_1*x_2={c\over a}<0$


and if you use the first statement you should get

$m\in(5,\infty)$


Regards MongolGenius

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    Why is this true?2018-05-21