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Precalculus JumpStart

Section 4.4 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 4.20. 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.

Example 4.22. 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 4.23. Using the Square Root Principle.

Solve
\begin{equation*} 3(x-1)^2 - 6 = 0\text{.} \end{equation*}
Solution.
Isolate the squared expression \((x-1)^2\) first
\begin{align*} 3(x-1)^2 - 6 &= 0\\ 3(x-1)^2 &= 6\\ (x-1)^2 &= \frac{6}{3} = 2\\ x-1 &= \pm \sqrt{2}\\ x &= 1 \pm \sqrt{2} \end{align*}
Estimating with a calculator, \(x = 1 + \sqrt{2} \approx 2.414\) OR \(x=1-\sqrt{2} \approx -0.414\text{.}\)
A third method is to use the Quadratic Formula

Example 4.25. Using the Quadratic Formula.

Use the quadratic formula to solve
\begin{equation*} 4x^2 - 10 x + 2 = 0. \end{equation*}
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*}