Hi
For the question: Find all values of p so that x^2 - 2px + p - 2 = 0 has one positive root and a negative root.
What is the condition of the discriminant for a positive and a negative root?
Thanx
Hi
Maybe I can add something
The product of the roots (real or complex) of equation $\displaystyle ax^2 + bx + c = 0$ is $\displaystyle \frac{c}{a}$
Therefore $\displaystyle ax^2 + bx + c = 0$ has 2 real roots with one positive and one negative if both conditions are realised :
(i) $\displaystyle b^2 - 4ac \geq 0$
(ii) $\displaystyle \frac{c}{a} < 0$
Condition (ii) implies that a and c have different sign ; then their product is negative
It means that condition (ii) implies condition (i) and therefore only condition (ii) is sufficient
In your specific case $\displaystyle x^2 - 2px + p - 2 = 0$ has 2 real roots with one positive and one negative if $\displaystyle p-2 < 0$ then $\displaystyle p < 2$
Look at the solution for x. It has the form $\displaystyle x = p \pm D$, that is, $\displaystyle x = p + D$ and $\displaystyle x = p - D$ where $\displaystyle D = \sqrt{p^2 - p + 2}$.
$\displaystyle x = p + D$ is always positive so that's the positive solution.
The negative solution therefore has to come from $\displaystyle x = p - D$. So you require $\displaystyle p - D < 0 \Rightarrow p < D \Rightarrow p < \sqrt{p^2 - p + 2}$.