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.

  • $\begingroup$ Also, we have Citardauq's variation of the quadratic formula: $$x = \frac{-2c}{-b\pm \sqrt{b^2-4ac}}.\tag{$c\neq 0$}$$ Go here to find out more about it $\longrightarrow$ math.stackexchange.com/questions/2105240/… For more about the standard quadratic formula and why it works, go here $\longrightarrow$ math.stackexchange.com/questions/49229/… $\endgroup$ – Mr Pie May 13 '18 at 2:02
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    $\begingroup$ "What does this formula do?" It gives you the two values of $x$ which are solutions to the equation $ax^2+bx+c = 0$. "How do I use it?" You take the values of $a$, $b$, and $c$ and plug them in... $\endgroup$ – Alex Kruckman May 13 '18 at 2:09
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    $\begingroup$ Do you want to know what it means to have a "solution" of an equation, do you want to know why the "solution" of this equation has $x$ alone on the left side of another equation, or do you want real life applications of this formula? Or something else? $\endgroup$ – David K May 13 '18 at 2:10
  • $\begingroup$ @user477343 In citardauq's formula there is no $-b$ in the denominator, just $b$ $\endgroup$ – imranfat May 13 '18 at 2:30
  • $\begingroup$ @imranfat oh yes of course, because the numerator is already negative. Hahahah, thanks for picking that up. This just shows how often I don't use the formula :) $\endgroup$ – Mr Pie May 13 '18 at 5:30

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, $f(-1) = (-1)^2 + 2(-1) +1 = 1-2+1=0.$

  • $\begingroup$ There are 0 or 2 real solutions, always 2 complex solutions, but never 1 in either case $\endgroup$ – AEngineer May 13 '18 at 2:02
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    $\begingroup$ @Skip find 0 or 2 distinct real or complex numbers that satisfy $x^2 + 2x + 1 = 0.$ $\endgroup$ – user328442 May 13 '18 at 2:06
  • $\begingroup$ @Skip because by the Fundamental Theorem of Algebra, any polynomial function equal to $0$; with a highest degree $n$; and a single indeterminate $x$, has exactly $n$ solutions for $x$ at most. This theorem applies to all polynomials, even with such functions that have a highest degree $n > 4$, and those kinds do not have a specific formula (Abel's Theorem)! $\endgroup$ – Mr Pie May 13 '18 at 2:06

The quadratic formula tells you the $x$ intercepts (where your graph crosses the $x$ axis) of your equation. You use the quadratic equation by putting the coefficients from your quadratic function $(ax^2+bx+c)$ into the appropriate spots $(a,b,c)$ in the quadratic formula. Simply solve and you will know all potential $(x,0)$ points.

This is particularly useful when your $x's$ are not easily factorable.


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