Section 3.4 Completing the Square
To solve a quadratic equation by completing the square, we often have to work with fractions.
Subsection 1. Multiply fractions
To multiply two fractions together, we multiply their numerators together, and multiply their denominators together. We can divide out any common factors in numerator and denominator before we multiply.
Subsubsection Examples
Example 3.58.
Multiply \(~\dfrac{3}{8} \cdot \dfrac{6}{5}\)
We can divide out a factor of 2.
Example 3.59.
Multiply \(~\dfrac{ab}{6} \cdot \dfrac{3a}{2b}\)
We divide out common factors before multiplying.
Subsubsection Exercises
Checkpoint 3.60.
Multiply.
- \(\displaystyle \dfrac{2}{3} \cdot \dfrac{5}{7}\)
- \(\displaystyle \dfrac{6}{7} \cdot \dfrac{14}{15}\)
- \(\displaystyle \dfrac{10}{21}\)
- \(\displaystyle \dfrac{4}{5}\)
Checkpoint 3.61.
Multiply.
- \(\displaystyle \dfrac{12x}{16y} \cdot \dfrac{18}{27xy}\)
- \(\displaystyle \dfrac{9c^2}{10c} \cdot \dfrac{25cd}{12d^2}\)
- \(\displaystyle \dfrac{1}{2y^2}\)
- \(\displaystyle \dfrac{15c^2}{8d}\)
Subsection 1. Add fractions
To add or subtract unlike fractions.
- Find the LCD for the fractions.
- Build each fraction to an equivalent one with the LCD as its denominator.
- Add or subtract the numerators. Keep the same denominator.
Subsubsection Examples
Example 3.62.
Add.
- \(\displaystyle \dfrac{7}{10} + \dfrac{5}{6}\)
- \(\displaystyle 6 + \dfrac{4}{9}\)
-
Step1: Find the LCD. Factor each denominator.
\begin{gather*} 10=2\cdot 5\\ 6 = 2\cdot 3 \end{gather*}The LCD is \(~2 \cdot 3 \cdot 5 = 30\text{.}\)
Step2: Build each fraction to a denominator of 30. The building factor for the first fraction is \(\blert{3}\text{,}\) and \(\blert{5}\) for the second fraction.
\begin{equation*} \dfrac{7}{10} \cdot \blert{\dfrac{3}{3}} = \dfrac{21}{30}~~~~\text{and}~~~~\dfrac{5}{6} \cdot \blert{\dfrac{5}{5}} = \dfrac{25}{30} \end{equation*}Step 3: Add the two like fractions, and reduce.
\begin{align*} \dfrac{7}{10} + \dfrac{5}{6} \amp = \dfrac{21}{30} + \dfrac{25}{30} = \dfrac{46}{30}\\ \amp = \dfrac{\cancel{2} \cdot 23}{\cancel{2} \cdot 15} = \dfrac{23}{15} \end{align*} -
Step1: The LCD is 9.
Step 2: Build the whole number to a denominator of 9.
\begin{equation*} \dfrac{6}{1} \cdot \blert{\dfrac{9}{9}} = \dfrac{54}{9} \end{equation*}Step 3: Add the two like fractions.
\begin{equation*} 6 + \dfrac{4}{9} = \dfrac{54}{9} + \dfrac{4}{9} = \dfrac{58}{9} \end{equation*}
Subsubsection Exercises
Checkpoint 3.63.
Add or subtract.
- \(\displaystyle \dfrac{5}{8} + \dfrac{1}{12}\)
- \(\displaystyle 3 - \dfrac{3}{4}\)
- \(\displaystyle \dfrac{17}{24}\)
- \(\displaystyle \dfrac{9}{4}\)
Checkpoint 3.64.
Add or subtract.
- \(\displaystyle \dfrac{5}{2} - \dfrac{5}{3}\)
- \(\displaystyle 4 + \dfrac{3}{8}\)
- \(\displaystyle \dfrac{5}{6}\)
- \(\displaystyle \dfrac{35}{8}\)
Subsection 3. Recognize squares of binomials
To solve a quadratic equation by completint the square, we create the square of a binomial:
Subsubsection Example
Example 3.65.
Write each trinomial as the square of a binomial.
- \(\displaystyle x^2+6x+9\)
- \(\displaystyle x^2-5x+\dfrac{25}{4}\)
-
In the formula above, note that the coefficient of \(x\) is \(2p\) and the constant term is \(p^2\text{.}\) In this example, \(2p=6\) and \(p^2=9\text{,}\) so \(p=\alert{3}\text{,}\) and
\begin{equation*} x^2+6x+9 = (x+\alert{3})^2 \end{equation*} -
The coefficient of \(x\) is \(2p=-5\) and the constant terms is \(p^2=\dfrac{25}{4}\text{,}\) so \(p=\alert{\dfrac{-5}{2}}\text{,}\) and
\begin{equation*} x^2-5x+\dfrac{25}{4} = \left(x-\alert{\dfrac{5}{2}}\right)^2 \end{equation*}
Subsubsection Exercises
Checkpoint 3.66.
Write \(~x^2+12x+36~\) as the square of a binomial.
Checkpoint 3.67.
Write \(~x^2-26x+169~\) as the square of a binomial.
Checkpoint 3.68.
Write \(~a^2-9a+\dfrac{81}{4}~\) as the square of a binomial.
Checkpoint 3.69.
Write \(~t^2-\dfrac{4}{3}t+\dfrac{4}{9}~\) as the square of a binomial.