An equation that involves a polynomial expression on one or both sides of the equal sign.
The solution set of a polynomial equation is the set of numbers that make the left side of the equation equal to the right side of the equation.
Polynomial equations usually have more than one solution.
Solving Quadratic Equations that Have No \$x\$-Term
A polynomial equation where the highest exponent is \$2\$.
The simplest type of quadratic equation is when there is no \$x\$-term, such as the quadratic equation \$x^2 = 9\$.
To solve \$x^2 = 9\$, take the square root of each side. Be sure to put a \$\pm\$ on the square root on the right-hand side of the equal sign to account for the fact that \$(+3)^2\$ and \$(−3)^2\$ both equal \$+9\$.
$$\begin{aligned} x^2 &= 9 \ \sqrt{x^2} &= \pm\sqrt{9} \ x &= \pm3 \end{aligned}$$
The solution set is \${3, −3}\$.
Solving Factored Polynomial Equations
The equation \$(x − 2)(x + 3) = 0\$ is a quadratic equation. If the left side was simplified, the highest exponent would be \$2\$.
Equations like this with two or more factors on the left side of the equation and a zero on the right side can be solved very quickly.
The only way \$(x − 2)(x + 3)\$ can equal zero is if one of the two factors is also equal to zero.
So \$(x − 2)(x + 3)\$ will equal zero if \$x − 2 = 0\$ or if \$x + 3 = 0\$.
$$\begin{aligned} (x-2)(x-3)&=0\ (x-2=0) &\text{or}&x-3=0\ \frac{+2=+2}{x=2} &\text{or}&\frac{+3=+3}{x=3} \end{aligned}$$
The solution set is \${2, 3}\$.
| x - 1 = 0 | or | x - 3 = 0 | |||
|---|---|---|---|---|---|
| x = 1 | or | x = 3 |
Simplifying radicals is explained in more detail in Chapter 4.
An example of a quadratic equation where the quadratic expression cannot factor is \$x^2 − 4x + 1 = 0\$, where \$a = 1\$, \$b = −4\$, and \$c = 1\$.
According to the quadratic formula:
$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$
The solution set is \${2 + \sqrt{3}, 2 - \sqrt{3}}\$.
1. Solve the equation \$x^2 - 4x + 4 = 0\$ by factoring. 2. Use the quadratic formula to solve \$x^2 + 2x - 3 = 0\$. 3. Find the roots of the cubic equation \$(x + 1)(x - 2)(x + 3) = 0\$.
How do different methods for solving polynomial equations (factoring, quadratic formula, etc.) reflect the nature of mathematical problem-solving? Are some methods more "elegant" or "beautiful" than others?
