Putting the fancy name aside, a discriminant is the \(b^2-4ac\) in \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) (i.e. the quadratic equation).
It turns out that the value of this expression can tell us about the number of intersections a quadratic equation will have with a straight line on a graph.
If, for whatever reason, we feel the urge to know how many intersections the line \(y=3x^2-7x+9\) will have with \(y=0\) (the \(x\) axis), then we can simply calculate the value of its discriminant.
if \(a=3, b=-7, c=9\), then the discriminant is \((-7)^2-4(3)(9)\) which yields -59. Looking at the rules above, we can satisfy our urge to know the number of intersections this line has with \(y=0\): zero.
Similarly, we can use this information to find any missing values, or a possible value, in a quadratic if we know the number of intersections. Take \(5x^2-3x+k=0\), which supposedly has a tangent that is the \(x\) axis. We need the value of \(k\).
We know that for a tangent to exist, our discriminant has to be 0. Hence,
\((-3)^2-4(5)(k)=0\)
Solve as normal,
\(k=\frac{9}{20}\)
Now, let's find the value of \(k\) if \(x^2+y^2=50\) and \(x=2y+k\) have two points of intersection.
\(x\) is already the subject in the second equation, so let's put that into the first and solve from there.
\((2y+k)^2+y^2=50\)
\(4y^2+4ky+k^2+y^2=50\)
\(5y^2+4ky+k^2-50=0\) (relate this to \(ax^2+bx+c=0\))
Now let's work on the discriminant. We know it's greater than zero, since there are two solutions. Hence, we'll start working with inequalities.
\((4k)^2-4(5)(k^2-50)>0\)
\(16k^2-20k^2+1000>0\)
\(-4k^2>-1000\)
\(k^2\gt250\)
Let \(k^2=250\)
\(k=\pm5\sqrt{10}\)
This tells us that \(k\) lies somewhere in between \(-5\sqrt{10}\lt k\lt5\sqrt{10}\).
That's enough.
This article was written on 27/09/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!