Obviously, $x=\frac{-b\pm \sqrt {b^2-4c}}{2}$ and for real roots we must have $b^2-4c\geq 0$. But for what values of $a,b,c$ will the quadratic have both roots positive?
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asked Jul 3, 2016 at 17:12
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$\begingroup$ I think a sufficient condition should be: $\frac{-b}{2a}>0$ and $\frac{c}{a}>0$. $\endgroup$– StudentCommented Jul 3, 2016 at 17:14
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1$\begingroup$ Well, obviously b must be negative. And $|b| > \sqrt{b^2 - 4c}$ ... So... $\endgroup$– fleabloodCommented Jul 3, 2016 at 17:18
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4 Answers
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- Real roots: $b^2\ge4c$.
- If real roots, same sign: $\;x_1x_2=c>0$.
- If real with same sign, positive: $\;x_1+x_2=-b>0$.
Summing it up, the conditions are $$ b<0,\enspace c>0,\enspace b^2\ge 4c.$$
The solutions can be visualised in a $(b,c)$-plane:
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You are right in that you need $b^2-4c\geq0$. The formula also shows that $x>0$ if and only if $$-b-\sqrt{b^2-4c}>0,$$ or equivalently $-b>\sqrt{b^2-4c}$. In particular $b$ must be negative. Squaring both sides shows that $$b^2>b^2-4c,$$ which means $c$ must be positive. Together with $b^2-4c\geq0$ this also gives $$b\leq-2\sqrt{c}.$$ So in summary, both roots are real positive numbers if and only if $$c>0\qquad\text{ and }\qquad b\leq-2\sqrt{c}.$$
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$\begingroup$ Yes but what should be the precise value of $b$ in terms of $c$? $\endgroup$ Commented Jul 3, 2016 at 17:18
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$\begingroup$ I can see why $b$ needs to be negative from the equation, but how do you derive that $c$ has to be positive from this? Don't you need $(x-r_1)(x-r_2)=...+r_1 r_2$ where the roots are positive? $\endgroup$– OviCommented Jul 3, 2016 at 17:19
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$\begingroup$ c can by any positive. and b can be any negative. They don't have to be related in any sense other than that. $\endgroup$ Commented Jul 3, 2016 at 17:20
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$\begingroup$ @Ovi Squaring the last inequality shows that $b^2>b^2-4c$, hence $c$ must be positive. $\endgroup$– ServaesCommented Jul 3, 2016 at 17:20
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$\begingroup$ @fleablood Well not quite any, you still need $b^2 > 4c$ $\endgroup$– OviCommented Jul 3, 2016 at 17:21
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Hint:
For $x^2+bx+c=0$, we have: $b=-(x_1+x_2)$ and $c=x_1x_2$, where $x_1,x_2$ are the roots.
answered Jul 3, 2016 at 17:15
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$x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$
1)To have any real roots at all $b^2 - 4c \ge 0$. So $b^2 \ge 4c$. So either a) $c < 0$ and $b^2 \ge 0 > 4c$ or b) $c \ge 0$ and $|b| \ge 2\sqrt{c}$.
2) $- b - \sqrt{b^2 - 4c} > 0$ so $b < - \sqrt{b^2 - 4c} \le 0$ so $b < 0$.
3) $-b > \sqrt{b^2 - 4c} \ge 0$ so $b^2 > b^2 -4c$ so $c > 0$
Putting all together we get $b \le -2\sqrt{c}; c > 0$.
