Let's have the following polynomial: $0.5x^2 + bx +c$
I want to check what the conditions are for having a positive root, and a negative root.
But I'm encountering a mistake which I cannot find.
I mark: $s = \sqrt{b^2-4ac}$
So the roots are: $R_{1,2}=-b \pm s $. (Since a=0.5)
The positive one will be $-b+s > 0$
The negative one will be $-b-s < 0$.
So I multiply the 2nd inequality by $-1$ and we get $b+s > 0$
I sum both inequalities and we get $2s>0$.
Therefore, $s>0$ is the condition for having 1 positive root, and 1 negative root, which means $b < -\sqrt{2c} $ and $b > \sqrt{2c}$ are the conditions.
But from checks I've performed, this is incorrect.
What algebraic mistake am I making here?
Thanks.
$\endgroup$1 Answer
$\begingroup$Since your roots are real, the discriminant is positive, and $s$ is positive always, because it is the square root of a positivie number.
Take a look to the following. You need $-b+s>0$ and $-b-s<0$. This is $s>b$ and $b>-s$, in other words, $|b|<s$. Squaring, this is equivalent to $b^{2} < s^{2}$.
Now, remember that $s^{2}$ is just the discriminant, so $b^{2} < b^{2} - 4ac$, which is $4ac<0$. This is $2c<0$, or $c<0$.
$\endgroup$ 2
