A quadratic equation is an equation formed by putting a quadratic polynomial equal to zero.


Standard Form

The standard form of the quadratic equation is a polynomial of the second degree, written as ax²+bx+c=0, where x is unknown and a, b, c are real numbers with a ≠ 0. The equation is not quadratic if a=0, because the term ax² is reduced to zero leaving only bx+c=0, which is the equation of a line.


The Quadratic Formula

The easiest way to solve a quadratic equation is probably the quadratic formula. The quadratic formula can be derived by completing the square on the quadratic equation ax²+bx+c=0. The quadratic formula is:

\(\textrm{x=}\)\(\frac{-\textrm{b}\pm\sqrt{\textrm{b}^2-\textrm{4ac}}}{\textrm{2a}}\)(a ≠ 0)

Using the quadratic formula to solve a quadratic equation is straight-forward, as shown by the following example:

Given the equation x²+10x+21=0, we get:

\(\textrm{x=}\frac{-10\pm\sqrt{10^2-4(1)(21)}}{2(1)}\)

\(\,\,\textrm{=}\frac{-10\pm\sqrt{100-84}}{2}\)

\(\,\,\textrm{=}\frac{-10\pm\sqrt{16}}{2}\)

\(\,\,\textrm{=}\frac{-10\pm 4}{2}\)

\(\,\,\textrm{=}\{\frac{-10-4}{2}, \frac{-10+4}{2}\}\)

\(\,\,\textrm{=}\{\frac{-14}{2}, \frac{-6}{2}\}\)

\(\,\,\textrm{=}\{\)−7, −3\(\}\)


Solution by Factoring

Solving quadratic equations can be done using various methods. One way to solve a quadratic equation is by factoring. It is important to keep in mind, however, that not every quadratic equation can be factored. Factoring a trinomial expression is part of another topic (please refer to Factoring Polynomials for a detailed step-by-step explanation.)


Once the quadratic equation is factored, the roots of the equation can be easily found. For example, the quadratic expression x²+7x+12 factors to (x+4)(x+3). Since assigning a value of either -3 or -4 to x satisifies the equation x²+7x+12=0, the solution set is x={-4,-3}.


A slightly more complicated example is one such as 6x²+6x-12. After factoring, it becomes (2x+4)(3x-3)=0. We then need to solve each of the roots algebraically:

2x+4=0  →  2x=-4  →  x=-2;
3x-3=0  →  3x=3  →  x=1;

The solution set is therefore x={-2, 1}.


Completing the Square

Since it's not possible to factor every quadratic expression, we often solve quadratic equations by the method known as completing the square. In order to solve a quadratic equation by completing the square, we perform the following steps:

1) Write the equation on the standard form ax²+bx+c=0.
2) Divide both sides of the equation by the coefficient of if not 1 (i.e, divide by a.)
3) Subtract the constant term from both sides (i.e, subtract c.)
4) Divide the coefficient of x by 2, square the result, then add the resulting value to both sides (i.e, add (½b)² or ¼b².)
5) Factor the left side of the equation.
6) Apply the square root property ±√…
7) Check the results if required. (i.e, plug the results into the equation to test it.)
8) Write the solution set.


For example, given the quadratic equation 3x²-10x-2, we follow the steps given above:

1) Write the equation in standard form:
3x²-10x-2=0

2) Divide both sides of the equation by the coefficient of , which is 3:
x²-\(\frac{10}{3}\)x-\(\frac{2}{3}\)=0

3) Subtract the constant term from both sides:
x²-\(\frac{10}{3}\)x=\(\frac{2}{3}\)

4) Divide the coefficient of x by 2, square the result, then add the resulting value to both sides:
(-\(\frac{10}{3}\)⋅\(\frac{1}{2}\))²=-(\(\frac{5}{3}\))²=\(\frac{25}{9}\)

x²-\(\frac{10}{3}\)+x+\(\frac{25}{9}\)=\(\frac{2}{3}\)+\(\frac{25}{9}\)=\(\frac{31}{9}\)

5) Factor the left side of the equation:
(x-\(\frac{5}{3}\))²=\(\frac{31}{9}\)

6) Apply the square root property:
x-\(\frac{5}{3}\)= ±\(\sqrt{\frac{31}{9}}\)

x-\(\frac{5}{3}\)= ±\(\frac{\sqrt{31}}{3}\)

x=\(\frac{5}{3}\)±\(\frac{\sqrt{31}}{3}\)

7) Check the results if required:
3(\(\frac{5}{3}+\frac{\sqrt{31}}{3}\))²−10(\(\frac{5}{3}+\frac{\sqrt{31}}{3}\))−2=0

3(\(\frac{5}{3}-\frac{\sqrt{31}}{3}\))²−10(\(\frac{5}{3}-\frac{\sqrt{31}}{3}\))−2=0

8) Write the solution set:
x = {\(\frac{5-\sqrt{31}}{3}\), \(\frac{5+\sqrt{31}}{3}\)}  (or)  x = {\(\frac{5±\sqrt{31}}{3}\)}

Since completing the square is more complex than factoring or using the quadratic formula, it's always a good idea to try to use those methods before using this one (that is, unless you're doing homework and the assignment is to solve the problem by completing the square!)



Algebra: Factoring Polynomials Algebra: Factorials