Hello all,

I have never really thought about the following question:

-when does the quadratic equation have only positive solutions?
-when does the quadratic equation have only negative solutions?
-When are the roots in a biquadratic equation positive/negative?

Thanks

You use the discriminant: $\displaystyle \Delta = b^2-4ac$

If $\displaystyle \Delta > 0$ then there are two real solutions
If $\displaystyle \Delta = 0$ then there are two, equal real roots[/tex]
If $\displaystyle \Delta < 0$ then there are no real solutions (2 complex ones)[/tex]

"only positive solutions" essentially means that any solution is to the right of 0 on the x-axis.

"only positive solutions" essentially means that any solution is to the right of 0 on the x-axis
that is exactly what I wanted to know. When is this the case, i.e what needs to holds in order for the roots to be only positive/negative. Could one set up some type of criteria?

I'm not entirely sure. We can rearrange the general form of the quadratic into $\displaystyle (x-h)^2+k$ form. From that we could link "only positive solutions" to h.

$\displaystyle ax^2+bx+c = 0 \Longleftrightarrow x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$

$\displaystyle \left(x+\dfrac{b}{2a}\right)^2 - \dfrac{b^2-4ac}{4a^2} = 0$

Since the "+k" part is a vertical shift it's value does not make a difference on where the roots lie so we can forget about it for now

Spoiler:
If you want to try this out for yourself you're more than welcome to - test $\displaystyle y = (x-1)^2 + 4 \text{ and } y = (x-1)^2 +1$ and see what happens. It's always good to check for yourself anyway

Due the properties of squaring real expressions we know that $\displaystyle \left(x+\dfrac{b}{2a}\right)^2 \geq 0$

For positive solutions it follows that $\displaystyle \displaystyle | x + \dfrac{b}{2a}| \geq 0$

$\displaystyle x \geq -\dfrac{b}{2a}$

I'm not 100% this is right you may want to try with some examples