So, I have this page in my agenda that shows how to obtain the solution of a quadratic function.
The solution for a quadratic equation in the form of $ax^2+bx+c=0$ can be found by using the quadratic formula:
$x=\frac {-b\pm\sqrt {b^2-4ac}} {2a}$
I was wondering, what does this formula do and how do I use it? I have already tried to plug in values for a,b and c, but it gives me weird numbers, like, assuming that a=3, b=4 and c=5, then $x=\frac {-2+\sqrt {2^2-4x3x5}} {2x3}= .0913885795...=\frac {\sqrt {56 -2}} {6}$. This number seems random, and I cannot figure out how or why this is useful.
A quadratic function is some function of the form $f(x) = ax^2+bx+c$ where $a \neq 0$. This function only varies with $x$ which is to say that $a, b$ and $c$ are given and fixed. A “solution” would be any $x$ that satisfies $f(x) = 0$ which is to say that $ax^2 + bx+c =0.$ The quadratic formula explicitly gives the $0, 1$ or $2$ possible distinct $x$ values that satisfy (or “solve”) $f(x) = 0$. Again, $a, b$ and $c$ are both given and fixed so the quadratic formula is rather useful since it doesn’t matter what values we have for $a, b$ and $c$ (as long as $a \neq 0$).
For example, suppose you are given the function $f(x) = x^2 + 2x + 1$. Then in this case, $a = c = 1, b = 2$ and if $f(x) = 0$ then the quadratic formula tells us that $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = -1.$$
Indeed, $f(-1) = (-1)^2 + 2(-1) +1 = 1-2+1=0.$