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Section 6.2 Solving Quadratic Equations

Subsection 1. Combine fractions

To solve a quadratic equation by completing the square, we often have to work with 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 6.23.

Multiply.

  1. \(\displaystyle \dfrac{3}{8} \cdot \dfrac{6}{5}\)
  2. \(\displaystyle \dfrac{ab}{6} \cdot \dfrac{3a}{2b}\)
Solution

  1. We can divide out a factor of 2.

    \begin{equation*} \dfrac{3}{8} \cdot \dfrac{6}{5} = \dfrac{3}{\cancel{2} \cdot 4} \cdot \dfrac{\cancel{2} \cdot 3}{5} = \dfrac{9}{20} \end{equation*}
  2. \begin{equation*} \dfrac{ab}{6} \cdot \dfrac{3a}{2b} = \dfrac{a \cancel{b}}{2 \cdot \cancel{3}} \cdot \dfrac{\cancel{3}a}{2 \cancel{b}} = \dfrac{a^2}{4} \end{equation*}
To add or subtract unlike fractions.
  1. Find the LCD for the fractions.
  2. Build each fraction to an equivalent one with the LCD as its denominator.
  3. Add or subtract the numerators. Keep the same denominator.
Example 6.24.

Add.

  1. \(\displaystyle \dfrac{7}{10} + \dfrac{5}{6}\)
  2. \(\displaystyle 6 + \dfrac{4}{9}\)
Solution

  1. 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*}

  2. 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

Multiply.

  1. \(\displaystyle \dfrac{2}{3} \cdot \dfrac{5}{7}\)
  2. \(\displaystyle \dfrac{6}{7} \cdot \dfrac{14}{15}\)
Answer
  1. \(\displaystyle \dfrac{10}{21}\)
  2. \(\displaystyle \dfrac{4}{5}\)

Multiply.

  1. \(\displaystyle \dfrac{12x}{16y} \cdot \dfrac{18}{27xy}\)
  2. \(\displaystyle \dfrac{9c^2}{10c} \cdot \dfrac{25cd}{12d^2}\)
Answer
  1. \(\displaystyle \dfrac{1}{2y^2}\)
  2. \(\displaystyle \dfrac{15c^2}{8d}\)

Add or subtract.

  1. \(\displaystyle \dfrac{5}{8} + \dfrac{1}{12}\)
  2. \(\displaystyle 3 - \dfrac{3}{4}\)
Answer
  1. \(\displaystyle \dfrac{17}{24}\)
  2. \(\displaystyle \dfrac{9}{4}\)

Add or subtract.

  1. \(\displaystyle \dfrac{5}{2} - \dfrac{5}{3}\)
  2. \(\displaystyle 4 + \dfrac{3}{8}\)
Answer
  1. \(\displaystyle \dfrac{5}{6}\)
  2. \(\displaystyle \dfrac{35}{8}\)

Subsection 2. Recognize squares of binomials

To solve a quadratic equation by completint the square, we create the square of a binomial:

\begin{equation*} \blert{(x+p)^2 = x^2 + 2px + p^2} \end{equation*}

Subsubsection Example

Example 6.29.

Write each trinomial as the square of a binomial.

  1. \(\displaystyle x^2+6x+9\)
  2. \(\displaystyle x^2-5x+\dfrac{25}{4}\)
Solution

  1. 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*}

  2. 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

Write \(~x^2+12x+36~\) as the square of a binomial.

Answer
\((x+6)^2\)

Write \(~x^2-26x+169~\) as the square of a binomial.

Answer
\((x-13)^2\)

Write \(~a^2-9a+\dfrac{81}{4}~\) as the square of a binomial.

Answer
\((a-\dfrac{9}{2})^2\)

Write \(~t^2-\dfrac{4}{3}t+\dfrac{4}{9}~\) as the square of a binomial.

Answer
\((t-\dfrac{2}{3})^2\)

Subsection 3. Simplify square roots

Be careful when simplifying radicals after extracting roots. Recall the properties of radicals reviewed in Section 4.4.

Subsubsection Example

Example 6.34.

Can you simplify the first expression into the second expression? (Decide whether the expressions are equivalent.)

  1. Is \(~~\sqrt{4+x^2}~~\) equivalent to \(~~2+x\text{?}\)
  2. Is \(~~\sqrt{\dfrac{x^2}{9}}~~\) equivalent to \(~~\dfrac{x}{3}~~\) for \(~x \ge 0\text{?}\)
  3. Is \(~~\sqrt{w-3}~~\) equivalent to \(~~\sqrt{w} - \sqrt{3}\text{?}\)
Solution

  1. If the expressions are equivalent, they must be equal for every value of the variable. Let's test with \(x=3\text{.}\) Then

    \begin{align*} \sqrt{4+x^2} \amp = \sqrt{4+9} = \sqrt{13} \approx 3.6\\ \text{but}~~~~~~~~~ 2+x \amp = 2+3 = 5 \end{align*}

    No, the expressions are not equivalent.

  2. Because \(\left(\dfrac{x}{3}\right)^2 = \dfrac{x}{3} \cdot \dfrac{x}{3} = \dfrac{x^2}{3^2} = \dfrac{x^2}{9}\text{,}\) it is also true that \(\sqrt{\dfrac{x^2}{9}} = \dfrac{x}{3}\text{.}\) Yes, the expressions are equivalent.

  3. Let \(w=16\text{.}\) Then

    \begin{align*} \sqrt{w-3} \amp = \sqrt{16-3} = \sqrt{13} \approx 3.6\\ \text{but}~~~ \sqrt{w}-\sqrt{3} \amp = \sqrt{16}-\sqrt{3} \approx 4-1.7 = 2.3 \end{align*}

    No, the expressions are not equivalent.

Subsubsection Exercises

Decide whether the expressions are equivalent. Assume all variables are positive.

\(\sqrt{b^2-81}~\) and \(~b-9\)

Answer
No

\(\sqrt{64x^2y^2}~\) and \(~8xy\)

Answer
Yes

\(\sqrt{64+x^2y^2}~\) and \(~8+xy\)

Answer
No

\(\sqrt{\dfrac{c^2+d^2}{4b^2}}~\) and \(~\dfrac{\sqrt{c^2+d^2}}{2b}\)

Answer
Yes