Originally Posted by

**MarkFL2** When dealing with quadratic equations, we analyze the discriminant to find whether we have two distinct real roots, one real repeated root, or complex conjugate roots.

If the discriminant is positive, then we have two distinct real roots. The discriminant is

$\displaystyle D=1-4k$ and with $\displaystyle k<0$ we know:

$\displaystyle D>0$

The quadratic formula tells us the roots will be:

$\displaystyle x=\frac{1\pm\sqrt{1-4k}}{2k}$

Let's examine the root:

$\displaystyle x=\frac{1-\sqrt{1-4k}}{2k}$

$\displaystyle 0<\frac{1-\sqrt{1-4k}}{2k}<1$

Multiply through by $\displaystyle 0<-2k$

$\displaystyle 0<\sqrt{1-4k}-1<-2k$

$\displaystyle 1<\sqrt{1-4k}<1-2k$

$\displaystyle 1<1-4k<1-4k+4k^2$

$\displaystyle 0<-4k<-4k+4k^2$

$\displaystyle 4k<0<4k^2$

$\displaystyle k<0<k^2$

Since this must be true, then the root we began with must be on the required interval.