An equation of the form \$ax^2 + bx + c = 0\$, where \$a, b, c\$ are constants and \$a \neq 0\$.
NoteDon't forget the \$\pm\$ sign! Both positive and negative roots are solutions.
Self reviewSolve the equation \$x^2 = 25\$. What is the solution set?
Solving Factored Polynomial Equations
If a polynomial equation is already factored, like \$(x - 2)(x + 3) = 0\$, you can use the Zero Product Property to solve it.
If \$ab = 0\$, then \$a = 0\$ or \$b = 0\$.
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}\$.
Self reviewSolve the equation \$(x + 2)(x - 5) = 0\$. What is the solution set?
Solving Quadratic Equations by Factoring
Not all quadratic polynomials factor. If one does in an equation where there is a zero on the right-hand side of the equal sign, the solution set can be found very quickly.
\$\$x^2 - 5x + 6 = 0\$\$
If possible, start by factoring the left-hand side:
\$\$ \begin{aligned} (x - 2) (x - 3) &= 0\ (x - 2 = 0) \quad &\text{or} \quad (x - 3 = 0)\ \frac{+2 = +2}{x = 2} \quad &\text{or} \quad \frac{+3 = +3}{x = 3} \end{aligned} \$\$
The solution set is \${2, 3}\$.
NoteThe two solutions, \$4\$ and \$-2\$, are the opposites of the two constants in the factors, \$(x - 4)\$ and \$(x + 2)\$. In general, if the factorsof a quadratic expression are \$(x - a)\$ and \$(x - b)\$, then the zerosor rootsof the quadratic expression are \$x = a\$ and \$x = b\$. This also works in reverse. If the zeros of a quadratic expression are \$x = a\$ and \$x = b\$, then the factors of the expression are \$(x - a)\$ and \$(x - b)\$.
Self reviewSolve the equation \$x^2 + 3x - 10 = 0\$ by factoring. What is the solution set?
Solving Quadratic Equations with the Quadratic Formula
When the quadratic expression does not factor, the equation has irrational roots and can be solved with the quadratic formula.
The solutions of the quadratic equation \$ax^2 + bx + c = 0\$ are given by:
\$\$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\$\$
NoteSimplifying radicals is explained in more detail in Chapter 4.
Self reviewUse the quadratic formula to solve the equation \$2x^2 - 3x + 1 = 0\$. What is the solution set?
Theory of KnowledgeHow do we know that the quadratic formula always works? What is the relationship between algebraic methods and geometric interpretations of quadratic equations?
