# Math Help - Discriminants

1. ## Discriminants

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

2. Originally Posted by xwrathbringerx
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
Yoo should be able to use the quadratic formula to get $x = p \pm \sqrt{p^2 - p + 2}$.

So you require the values of p that satisfy the inequality $p < \sqrt{p^2 - p + 2}$ (why?)

3. Originally Posted by mr fantastic
So you require the values of p that satisfy the inequality $p < \sqrt{p^2 - p + 2}$ (why?)
Hmmmm I have no clue. Why exactly does it have to satisfy that specific inequality?

4. Hi

The product of the roots (real or complex) of equation $ax^2 + bx + c = 0$ is $\frac{c}{a}$

Therefore $ax^2 + bx + c = 0$ has 2 real roots with one positive and one negative if both conditions are realised :
(i) $b^2 - 4ac \geq 0$
(ii) $\frac{c}{a} < 0$

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 $x^2 - 2px + p - 2 = 0$ has 2 real roots with one positive and one negative if $p-2 < 0$ then $p < 2$

5. Originally Posted by xwrathbringerx
Hmmmm I have no clue. Why exactly does it have to satisfy that specific inequality?
Look at the solution for x. It has the form $x = p \pm D$, that is, $x = p + D$ and $x = p - D$ where $D = \sqrt{p^2 - p + 2}$.

$x = p + D$ is always positive so that's the positive solution.

The negative solution therefore has to come from $x = p - D$. So you require $p - D < 0 \Rightarrow p < D \Rightarrow p < \sqrt{p^2 - p + 2}$.