<s>[INST] 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.
 [/INST] 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.
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