Principle 5.4.1. Zero Product Theorem.
Suppose a product of two expressions \(A\) and \(B\) is zero,
\begin{equation*}
A\cdot B = 0.
\end{equation*}
Then \(A=0\) OR \(B=0\text{.}\)
Section 5.4 Quadratic Equations
A quadratic equation is an equation of the form
\begin{equation*}
ax^2+bx+c=0,
\end{equation*}
where \(a\neq 0\text{.}\) Let’s discuss some ways we might solve them. The first method is to take advantage of the so-called Zero Product Principle
Example 5.4.2. Solving a quadratic by factoring.
To solve
\begin{equation*}
x^2-2 x-3=0\text{,}
\end{equation*}
we attempt to factor
\begin{align*}
x^2-2 x-3 &=0\\
(x+1)(x-3) &=0
\end{align*}
Thus, by the Zero Product Theorem, \(x+1 = 0\) OR \(x-3=0\text{.}\) Solving each, we find that \(x=-1\) OR \(x=3\text{.}\)
The second method is to take advantage of the Square Root Principle.
Principle 5.4.3. Square Root Principle.
Suppose \(A^2 = B\text{,}\) where \(B\geq 0\text{.}\) Then \(A= \sqrt{B}\) OR \(A=-\sqrt{B}\text{.}\) That is, \(A= \pm \sqrt{B}\text{.}\)
Example 5.4.4. Using the Square Root Principle.
To solve
\begin{equation*}
4x^2 - 9 = 0\text{,}
\end{equation*}
we first note that there is no degree one term. So
\begin{align*}
4x^2 - 9 &= 0\\
4x^2 &= 9\\
x^2 &= \frac{9}{4}\\
x &= \pm \sqrt{\frac{9}{4}} = \pm \frac{\sqrt{9}}{\sqrt{4}}= \pm \frac{3}{2}
\end{align*}
That is, \(x=3/2\) OR \(x=-3/2\text{.}\)
Example 5.4.5. Using the Square Root Principle.
Solution.
A third method is to use the Quadratic Formula
Theorem 5.4.6. Quadratic Formula.
The solutions to
\begin{equation*}
ax^2+bx+c=0, \quad a\neq 0
\end{equation*}
are given by
\begin{equation*}
x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}.
\end{equation*}
Example 5.4.7. Using the Quadratic Formula.
Solution.
\begin{align*}
x &= \frac{10\pm \sqrt{100-32}}{8}\\
&= \frac{10\pm \sqrt{68}}{8}\\
&= \frac{10\pm 2\sqrt{17}}{8}\\
&= \frac{2(5\pm \sqrt{17})}{8}\\
&= \frac{5\pm \sqrt{17}}{4}
\end{align*}
