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Section 4.1 Quadratic Formula

Subsection 1. Multiply algebraic 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 4.1.

Multiply \(~\dfrac{1}{2}\left(\dfrac{P}{Q}\right)\)

Solution
\(\dfrac{1}{2}\left(\dfrac{P}{Q}\right) = \dfrac{1 \cdot P}{2 \cdot Q} = \dfrac{P}{2Q}\)
Example 4.2.

Multiply \(~4 \left(\dfrac{b}{c}\right)\)

Solution
\(4 \left(\dfrac{b}{c}\right) = \dfrac{4}{1} \left(\dfrac{b}{c}\right) = \dfrac{4 \cdot b}{1 \cdot c} = \dfrac{4b}{c}\)
Example 4.3.

Multiply \(~\dfrac{t}{w} \cdot \dfrac{w}{3t^2}\)

Solution
\(\dfrac{t}{w} \cdot \dfrac{w}{3t^2} = \dfrac{\cancel{t}}{\cancel{w}} \cdot \dfrac{\cancel{w}}{3t \cdot \cancel{t}} = \dfrac{1}{3t}\)
Example 4.4.

Multiply \(~\dfrac{a}{2} \cdot \dfrac{1-b}{b}\)

Solution
\(\dfrac{a}{2} \cdot \dfrac{1-b}{b} = \dfrac{a(1-b)}{2 \cdot b} = \dfrac{a-ab}{2b}\)

Subsubsection Exercises

Multiply \(~\dfrac{3}{2a}\left(\dfrac{a^2}{6}\right)\)

Answer

\(\dfrac{a}{9}\)

Multiply \(~8\left(\dfrac{m}{4x^2}\right)\)

Answer

\(\dfrac{2m}{x^2}\)

Multiply \(~\dfrac{3}{4} \cdot \dfrac{t^2-2}{cw^2}\)

Answer

\(\dfrac{3t^2-6}{4cw^2}\)

Multiply \(~\dfrac{n}{n-1} \cdot \dfrac{p+1}{n^2}\)

Answer

\(\dfrac{p+1}{n^2-n}\)

Subsection 2. Add or subtract algebraic fractions

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.

Subsubsection Examples

Example 4.9.

Subtract \(~\dfrac{2x}{w} - \dfrac{3x}{w}\)

Solution

These are like fractions, so we need only combine their numerators.

\begin{equation*} \dfrac{2x}{w} - \dfrac{3x}{w} = \dfrac{2x-3x}{w} = \dfrac{-x}{w} \end{equation*}
Example 4.10.

Subtract \(~\dfrac{3}{a} - \dfrac{a+2}{a}\)

Solution

These are like fractions, so we need only combine their numerators. Be careful to subtract both terms of the second numerator.

\begin{equation*} \dfrac{3}{a} - \dfrac{a+2}{a} = \dfrac{3-(a+2)}{a} = \dfrac{1-a}{a} \end{equation*}
Example 4.11.

Add \(~2 + \dfrac{1}{x}\)

Solution

We write 2 as a fraction, \(\dfrac{2}{1}\text{,}\) and build it to the LCD, \(x\text{.}\)

\begin{equation*} \dfrac{2}{1} \cdot \alert{\dfrac{x}{x}} + \dfrac{1}{x} = \dfrac{2x}{x} + \dfrac{1}{x} = \dfrac{2x+1}{x} \end{equation*}
Example 4.12.

Subtract \(~\dfrac{a}{2b} - \dfrac{3}{b^2}\)

Solution

We build each fraction to the LCD, \(2b^2\text{.}\)

\begin{align*} \dfrac{a}{2b} - \dfrac{3}{b^2} \amp = \dfrac{a}{2b} \cdot \alert{\dfrac{b}{b}} - \dfrac{3}{b^2} \cdot \alert{\dfrac{2}{2}}\\ \amp = \dfrac{ab}{2b^2} - \dfrac{6}{2b^2} = \dfrac{ab-6}{2b^2} \end{align*}

Subsubsection Exercises

Subtract \(~\dfrac{4a}{b^2} - \dfrac{c}{b^2}\)

Answer
\(\dfrac{4a-c}{b^2}\)

Subtract \(~\dfrac{p+2}{2q} - \dfrac{p-1}{2q}\)

Answer
\(\dfrac{3}{2q}\)

Add \(~N + \dfrac{2}{N}\)

Answer
\(\dfrac{N^2+2}{N}\)

Add \(~\dfrac{2}{xy} - \dfrac{y}{3x}\)

Answer
\(\dfrac{6+y^2}{3xy}\)

Subsection 3. Simplify square roots

Be careful when simplifying radicals after extracting roots.

Subsubsection Example

Example 4.17.

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