Under what conditions will $x^2+bx+c=0$ have both roots real and positive?

Question

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?

Answer 1

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 $$c>0\qquad\text{ and }\qquad b\leq-2\sqrt{c}.$$

Answer 2

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.

Answer 3

• 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:

Answer 4

$x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$

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 $b \le -2\sqrt{c}; c > 0$.