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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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Please include all relevant information in the body, don't rely on the title for content. Thank you. –  Arturo Magidin Apr 7 '11 at 19:02
surely you should solve: $-(m+1) - \sqrt{(m+1)^2 - 4(m+4)} > 0$ for a poisitive x minima? –  yaakov Apr 7 '11 at 19:09
Hint: Note that IF the roots are real, the roots are both positive if and only if the sum and product of the roots are both positive. You can read off these from the coefficients. –  André Nicolas Apr 7 '11 at 19:21
@user6312 then it would be m+1>0 and m+4>0 ? but it's not correct answer –  Templar Apr 7 '11 at 19:37
@Templar: The hint said that IF the roots are real, the roots are both positive iff ..... So we need to use the discriminant condition and the fact that $m+1$ and $m+4$ are positive. Of course it is enough to ask that $m+1$ be positive, since the other follows. And now one can see that the necessary and sufficient condition is that $m \ge 5$. –  André Nicolas Apr 7 '11 at 19:47
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3 Answers

up vote 3 down vote accepted

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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The argument (and answer) are not quite right. For example, if $m=-3$, the equation reads $x^2+2x+1=0$, and the root is not positive. I had mentioned in an earlier comment that IF the roots are real, they are both positive iff their sum and product are positive. This gives an easy way to finish solving the problem. –  André Nicolas Apr 7 '11 at 19:40
@user6312: Ah, yes. I know where the problem is. Thanks for pointing this out. –  Arturo Magidin Apr 7 '11 at 20:01
@Arturo: the above is incorrect. You confused a necessary condition with a sufficient one. See my answer. –  Bill Dubuque Apr 7 '11 at 20:04
@Bill: Yes, I forgot to take out the spurious solutions obtained by the squaring. Your method avoids that issue, of course. –  Arturo Magidin Apr 7 '11 at 20:09
@Arturo: For the record, the above comment by user6312 didn't show in my browser when I added my comment. I say that so you know that I did not mean to emphasize it any further. But I'm sure I refreshed the page in between then to see if you corrected the answer. So something is screwy with the software. –  Bill Dubuque Apr 7 '11 at 20:18
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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 user6312 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. –  Bill Dubuque Apr 7 '11 at 20:32
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Quadratic equation $ax^2+bx+c=0$

has positive roots if


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

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

negative roots if


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

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

two zero roots if

$b=0$ $c=0$

has one positive root and one negative root if


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

and if you use the first statement you should get


Regards MongolGenius

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