How do you find the roots of a quadratic equation?

 
Relevant MM Lessons:This question involves a few major concepts across several lessons - for a deeper dive and for a full gambit of practice and common tricks, make sure to check out each Relevant Mister Math Lesson in the list above.In short, there are three ways we can accomplish this task.

Method 1 - Factor a Quadratic Equation

If a quadratic equation is factorable, this is a very popular way to proceed (and one you'll surely be tested on at some point in your career).Step 1 - Get it Equal to ZeroIf this isn't already done, move all the of terms to one side of the equation via addition and subtraction so that the other side is zero. Preferably, you will have the squared term be positive.Example$$x^2 - 5x = 6 \rightarrow x^2 - 5x -6 = 0$$Step 2 - Begin factored formIf your squared term has a coefficient of 1, like our example above, then you can go ahead and set up the "shell" that we will fill in, like this:$$(x \, \, \, \, \, \, \, \,)(x \, \, \, \, \, \, \, \,) = 0$$Examine whether the constant term is a $+$ of $-$. If it is a $+$, then both symbols in your factored form will be the same sign as the coefficient of your $x$ term. If it is a $-$, then you will have one $+$ and one $-$. In our example, the constant term is $-6$, so we will have one $+$ and one $-$.$$(x \, \, + \, \, \, \,)(x \, \, - \, \, \, \,) = 0$$Step 3 - Find Magic NumbersFinally, find the last numbers in each factor by finding the combination of two numbers which add to the $x$ term coefficient but multiply to the constant term. In this case, we need two numbers, one positive and one negative, that add to $-5$ and multiply to $-6$. I'm thinking $-6$ and $+1$.$$(x + 1)(x - 6) = 0$$Via the zero product property, the $x$ values that will make this product $0$ are $-1$ and $6$. These are the roots of the quadratic, obtained via the factoring method.Let's see another one quickly.$$x^2 - 7x + 12 = 0$$This time the constant term ($+12$) is a positive term, so when we set up our factoring shell, each factor will have the same sign as the $x$ term ($-7x$).$$(x \, \, - \, \, \, \,)(x \, \, - \, \, \, \,) = 0$$Once again, proceed with step 3 and find the numbers that combine to $-7$ and multiply to $+12$. This time it will be $-3$ and $-4$.$$(x - 3)(x - 4) = 0$$Therefore, the roots of $x^2 - 7x + 12$ are $3$ and $4$.NOTE! If the squared term coefficient is not 1, you have to play around with the combinations more. Here's an example.$$6x^2 - 5x - 50 = 0$$There are a few more possibilities for factoring here, since 6 can be factored as $2 \cdot 3$ or $1 \cdot 6$, and 50 can be factored as $1 \cdot 50$, $2 \cdot 25$, or $5 \cdot 10$. AND you have each choice to try in position with the $+$ and the $-$....$$(2x + 10)(3x - 5)$$$$(2x - 10)(3x + 5)$$$$(x + 10)(6x - 5)$$$$...$$However, with some practice, you'll be able to factor these fairly quickly. They are also much less commonly tested by teachers. Finally, if you have a coefficient with the squared term, you might just be better off trying the next method.

Method 2 - Use the Quadratic Formula

For quadratic equations of the form $ax^2 + bx + c = 0$, the quadratic formula will directly tell you the roots of the equation. All you have to do is plug the $a$, $b$, and $c$ coefficients into the formula:$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$Here's an example:$$2x^2 + 7x - 15 = 0$$Using the quadratic formula, the roots are:$$x = \frac{-(7) \pm \sqrt{(7)^2-4(2)(-15)}}{2(2)}$$$$ = \frac{-7 \pm \sqrt{49-(-120)}}{4}$$$$ = \frac{-7 \pm \sqrt{169}}{4}$$$$ = \frac{-7 \pm 13}{4}$$$$ \rightarrow x=-5,\, \frac{3}{2}$$

Method 3 - Graph It

This method is particularly swank if you have a graphing utility on hand, such as a TI graphing calculator, but can also be done by hand. The idea is that you graph the quadratic and use its $x$-intercepts to identify the roots of the equation.First, you must take the equation you're working with and move all the terms to one side, leaving zero on the other side (we had to do this in the other two methods as well). Next, take the quadratic expression you now have, and graph it. With a calculator, using the TI as an example, this means you find the graph equation function and where it says something like $y1=$, you type in the quadratic. Without a calculator, the fastest way to graph is to pick a few $x$ values, plug those values into the quadratic expression, and see what you get for an answer, call it $y$, and write down these pairs in a table. These value pairs represent points on the graph. With 3 or 4 pairs, you can make a decent graph.All that remains is to find the $x$-intercepts. This can be done either by inspection (eyeballing it) or using calculator functionality. With the TI, having graphed the quadratic expression, use the "Calc $\rightarrow$ Zeroes" graph function. Since our teachers are usually nice and give us problems with integer solutions, a good first try if you're working by hand is to eyeball it, see what the $x$-intercepts seem to be based on your sketch, and verify that you're correct by plugging your $x$ values back into the original equation and seeing that the equation is a true statement.Let's see an example:$$x^2 - 2x - 2 = 6$$First, no matter what, we want this equation to be set equal to zero. Let's subtract $6$ from both sides:$$x^2 - 2x - 8 = 0$$Next, if we are using a graphing utility, we would type in $y1=x^2 - 2x - 8$, find the "Calc $\rightarrow$ Zeroes" graph function, and we would have our answers. If we didn't have a calculator, let's proceed by picking some points and putting the results in an $x-y$ table. We'll try $-3$, $0$, $1$, and $3$:We'll try $-3$, $0$, $1$, and $3$:If $x=-3$, we get$$x^2 - 2x - 8 \rightarrow (-3)^2 - 2(-3) - 8 = 7$$If $x=0$, we get$$x^2 - 2x - 8 \rightarrow (0)^2 - 2(0) - 8 = -8$$Similarly, $x=1$ yields $-9$, and $x=3$ yields $-5$. Plot these four order pairs, $(-3,7)$, $(0,-8)$, $(1,-9)$, and $(3,-5)$, then estimate the graph. You might get something close to this:Upon inspection, it looks like $x=-2$ and $x=4$ are our winners! Let's double check.$$x^2 - 2x - 8 \rightarrow (-2)^2 - 2(-2) - 8 = 0$$$$x^2 - 2x - 8 \rightarrow (4)^2 - 2(4) - 8 = 0$$