In this topic we discuss the usage of the Quadratic Formula. This is probably the one method students have seen before and attempt to use. I write the formula a little differently than most textbooks. This will be discussed more clearly in the example that has complex solutions. The Quadratic Formula is a process that solves the equation: \(ax^2+bx+c=0\) and is given by:
$$x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$
Where did this formula come from? It is simply using the completing the square process on the equation \(ax^2+bx+c=0\) but keeping the \(a,\;b\) and \(c\) in place. To see the proof of the formula, click the button below. However, the focus is just being able to use the quadratic formula correctly to solve equations.
Solve \(6x^2-5x-4=0\) using the Quadratic Formula.
| Equation equals \(0\): | \(a=6,\;b=-5,\;\text{ and }\;c=-4\) |
| Plug into formula: | \(x=\large{\frac{-(-5)}{2(6)}\pm\frac{\sqrt{(-5)^2-4(6)(-4)}}{2(6)}}\) |
| Simplify: | \(x=\large{\frac{5}{12}\pm\frac{\sqrt{25+96}}{12}}\) |
| \(x=\large{\frac{5}{12}\pm\frac{\sqrt{121}}{12}}\) | |
| \(x=\large{\frac{5}{12}\pm\frac{11}{12}}\) | |
| \(x=\large{\frac{5+11}{12}}\)\(\Rightarrow x=\large{\frac{16}{12}}\)\(\Rightarrow x=\large{\frac{4}{3}}\) | |
| \(x=\large{\frac{5-11}{12}}\)\(\Rightarrow x=\large{\frac{-6}{12}}\)\(\Rightarrow x=\large{-\frac{1}{2}}\) |
Solve \(2x^2+19=14x\) using the Quadratic Formula.
| Make equation equal \(0\) with algebra: | \(a=2,\;b=-14,\;\text{ and }\;c=19\) |
| Plug into formula: | \(x=\large{\frac{-(-14)}{2(2)}\pm\frac{\sqrt{(-14)^2-4(2)(19)}}{2(2)}}\) |
| Simplify: | \(x=\large{\frac{14}{4}\pm\frac{\sqrt{196-152}}{4}}\) |
| \(x=\large{\frac{7}{2}\pm\frac{\sqrt{44}}{4}}\) | |
| \(x=\large{\frac{7}{2}\pm\frac{2\sqrt{11}}{4}}\) | |
| \(x=\large{\frac{7}{2}\pm\frac{\sqrt{11}}{2}}\) |
Solve \((x+2)(x-6)=-17\) using the Quadratic Formula.
| Make equation equal \(0\) by algebra: | \(x^2-6x+2x-12+17=0\implies x^2-4x+5=0\) |
| Be careful, looks like it factors, but it doesn't: | \(a=1,\;b=-4,\;\text{ and }\;c=5\) |
| Plug into formula: | \(x=\large{\frac{-(-4)}{2(1)}\pm\frac{\sqrt{(-4)^2-4(1)(5)}}{2(1)}}\) |
| Simplify: | \(x=\large{\frac{4}{2}\pm\frac{\sqrt{16-20}}{2}}\) |
| \(x=2\pm\large{\frac{\sqrt{-4}}{2}}\) | |
| \(x=2\pm\large{\frac{2i}{2}}\) | |
| \(x=2\pm i\) |
This problem has complex numbers as solutions. Complex numbers will be discussed in more detail later on, but just know that \(i=\sqrt{-1}\).
Notice that each example above had solutions of all types of numbers. The first has rational solutions, the second has irrational solutions and the last one has complex solutions. There is a way to determine these types of solutions using the discriminant.
The discriminant is the radicand, or the number under the square root: \(D=b^2-4ac\). The following table describes all possible solutions:
| Discriminant | Types of Solutions |
|---|---|
| Positive & Perfect Square | 2 rational solutions |
| Positive & NOT A Perfect Square | 2 irrational solutions |
| Zero | 1 rational solution |
| Negative | 2 complex solutions |
If you look back at the solutions, you will see that the first example has a discriminant of \(121\). This is a perfect square and positive, hence there are two rational solutions. The second example has a discriminant of \(11\). This is also positive, but NOT a perfect square, leading to the two irrational solutions. Finally, the third example has a discriminant of \(-4\), obviously negative, leading to the two complex solutions.