Showing that roots of a quadratic polynomial are of opposite signs. 
Prove that, for all values of $k$, the roots of the quadratic polynomial $x^2 - (2 + k) x - 3$ are real. Show further that the roots are of opposite signs.

For the first part I was able to demonstrate such by using the discriminant of the quadratic, then using the discriminant of the discriminant. 
For the second part I was not able to demonstrate such.
 A: A general quadratic with roots $\alpha$ and $\beta$ can be written
$$
x^2-(\alpha+\beta)x+\alpha\beta=0
$$
The last, constant term is the product of the roots. In your case that equals $-3$. The roots must therefore have opposite signs.
A: In the quadratic $P(x) = x^2 - (2 + k)x - 3 = 0$, the coefficient of $x^2$ is positive. Thus, $P(X)$ and $P(-X)$ is positive for some large enough $X>0$.
But $P(0) = -3$ is negative. So between $-X$ and $0$ there is a $x_1$ where $P(x_1) = 0$ and between $0$ and $X$ there is a $x_2$ where $P(x_2) = 0$: the polynomial has real roots and they are of opposite signs.
A: Well, the solutions to the equation are
$$x=\frac{2+k}{2}\pm\frac12\sqrt{(-2-k)^2-4\cdot1\cdot-3}$$
or, simplified,
$$x=\frac{2+k}2\pm\frac12\sqrt{(2+k)^2+12}$$
and this is equal to
$$x=\frac{2+k}2\pm\sqrt{\left(\frac{2+k}2\right)^2+3}$$
The important step here is
$$\sqrt{\left(\frac{2+k}2\right)^2+3}>\sqrt{\left(\frac{2+k}2\right)^2}=\left|\frac{2+k}2\right|$$
and so if we choose $+$, then we see the root is positive:
$$\frac{2+k}2+\sqrt{\left(\frac{2+k}2\right)^2+3}>\frac{2+k}2+\left|\frac{2+k}2\right|\geq0$$
When we choose $-$, we get something negative:
$$\frac{2+k}2-\sqrt{\left(\frac{2+k}2\right)^2+3}<\frac{2+k}2-\left|\frac{2+k}2\right|\leq0$$
thus, the roots must be of different signs.
