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Section A.8 Chapter 8 Polynomial and Rational Functions
Subsection A.8.1 Polynomial Functions
Subsubsection A.8.1.1 Compute sums and products
Compare the rules for simplilfying products to the rules for simplifying sums.
Example A.8.1 .
Simplify each expression if possible.
\(\displaystyle 3x^2-5x^3\)
\(\displaystyle 3x^2(-5x^3)\)
Solution .
This expression is a difference of terms, but they are not like terms (because the variable has different exponents), so we cannot combine them.
This expression is a product, and the powers have the same base, so we can apply the first law of exponents to get \(~3x^2(-5x^3)=-15x^5\text{.}\)
Example A.8.2 .
Simplify each expression if possible.
\(\displaystyle -6t^4-8t^4\)
\(\displaystyle -6t^4(-8t^4)\)
Solution .
This expression is a difference of like terms, so we can combine their coefficients to get \(~-6t^4-8t^4=-14t^4\)
This expression is a product, and the powers have the same base, so we can apply the first law of exponents to get \(~-6t^4(-8t^4)=48t^8\)
Checkpoint A.8.3 .
Simplify each expression if possible.
\(\displaystyle 2a^2-9a^3+a^2\)
\(\displaystyle 2a^2(-9a^3+a^2)\)
Answer .
\(\displaystyle 3a^2-9a^3\)
\(\displaystyle -18a^5+2a^4\)
Checkpoint A.8.4 .
Simplify each expression if possible.
\(\displaystyle 7-4a^3+2a^3\)
\(\displaystyle 7-4a^2(2a^3)\)
Answer .
\(\displaystyle 7-2a^3\)
\(\displaystyle 7-8a^5\)
Subsubsection A.8.1.3 Square binomials
Sometimes it is easier to use formulas to square binomials.
Special Products of Binomials.
\begin{align*}
\amp(a + b)^2 = (a + b) (a + b) = a^2 + 2ab + b^2\\
\amp(a - b)^2 = (a - b) (a - b) = a^2 - 2ab + b^2\\
\amp(a + b) (a - b)= a^2 -b^2
\end{align*}
Example A.8.9 .
Use the identity
\(~(a+b)^2=a^2+2ab+b^2~\) to expand
\(~(3h^2+4k^3)^2\text{.}\)
Solution .
We substitute \(3h^2\) for \(a\) and \(4k^3\) for \(b\) into the identity.
\begin{align*}
(3h^2+4k^3)^2 \amp =(3h^2)^2 +2(3h^2)(4k^3) +(4k^3)^2\\
\amp = 9h^4+24h^2k^3+16k^6
\end{align*}
Example A.8.10 .
Use the identity
\(~(a-b)^2=a^2-2ab+b^2~\) to expand
\(~(2xy^2-5)^2\text{.}\)
Solution .
We substitute \(2xy^2\) for \(a\) and \(5\) for \(b\) into the identity.
\begin{align*}
(2xy^2-5)^2 \amp =(2xy^2)^2 -2(2xy^2)(5) +5^2\\
\amp = 4x^2y^4 - 20xy^2+25
\end{align*}
Checkpoint A.8.11 .
Expand
\(~(8w^4-3w^3)^2\)
Checkpoint A.8.12 .
Expand
\(~(a^3b+9ab^3)^2\)
Answer .
\(a^6b^2+18a^4b^4+81a^2b^6\)
Subsection A.8.2 Algebraic Fractions
Subsubsection A.8.2.1 Factor a polynomial
To reduce an algebraic fraction, we must factor its numerator and denominator.
Example A.8.13 .
Factor.
\(\displaystyle 4x^2-4x\)
\(\displaystyle 4x^2-1\)
\(\displaystyle 4x^2-4x+2\)
Solution .
We factor out a common factor of
\(4x\) to get
\(4x(x-1)\text{.}\)
This is a difference of two squares that factors as
\((2x-1)(2x+1)\text{.}\)
This is the square of a binomial,
\((2x-1)^2\text{.}\)
Example A.8.14 .
Factor.
\(\displaystyle 27a^2-3\)
\(\displaystyle 27a^3-1\)
\(\displaystyle 81a^3-a\)
Solution .
We first factor out 3 to find
\(3(9a^2-1)\text{,}\) then factor the difference of two squares to get
\(3(3a-1)(3a+1)\text{.}\)
This is a differece of two cubes, which factors as
\((3a-1)(9a^2+3a+1)\text{.}\)
We first factor out
\(a\) to get
\(a(81a^2-1)\text{,}\) then factor the difference of two squares to get
\(a(9a-1)(9a+1)\) .
Checkpoint A.8.15 .
Factor completely
\(2x^3+16y^3\)
Answer .
\(2(x+2y)(x^2-2xy+4y^2)\)
Checkpoint A.8.16 .
Factor completely
\(4x^2y-36y^3\)
Checkpoint A.8.17 .
Factor completely
\(2b^3-6b^2-36b\)
Checkpoint A.8.18 .
Factor completely
\(9b^4+9b^2\)
Subsubsection A.8.2.2 Find the opposite of a binomial
To find the opposite or negative of a binomial we multiply by
\(-1\text{.}\)
Example A.8.19 .
Which of these is the opposite of \(m^2-p~\text{?}\)
\(\displaystyle m^2+p\)
\(\displaystyle m-p^2\)
\(\displaystyle p-m^2\)
Solution .
The opposite of
\(m^2-p\) is
\(-(m^2-p) = -m^2+p\text{,}\) or
\(p-m^2\text{.}\)
Example A.8.20 .
Which of these pairs of binomials are opposites?
\(3c-5\) and \(5+3c\)
\(5-3c\) and \(3-5c\)
\(5c-3\) and \(3-5c\)
Solution .
The opposite of
\(5c-3\) is
\(-(5c-3)=-5c+3\text{,}\) or
\(3-5c\text{,}\) so (c) is correct.
Checkpoint A.8.21 .
Find the opposite of the binomial
\(~2x+1\)
Checkpoint A.8.22 .
Find the opposite of the binomial
\(~b^2-b\)
Checkpoint A.8.23 .
Find the opposite of the binomial
\(~-4n+8\)
Checkpoint A.8.24 .
Find the opposite of the binomial
\(~-3z^2-2\)
Subsubsection A.8.2.3 Asymptotes
The asymptotes of rational functions are horizontal and vertical lines.
Example A.8.25 .
Give the equation and slope of the line.
Solution .
This line is a vertical line. All the points on the line have
\(x\) -coordinate
\(-15\text{,}\) so the equation of the line is
\(x=-15\text{.}\) Because
\(\Delta x=0\) between any two points on the line, its slope is undefined.
Example A.8.26 .
Give the equation and slope of the line.
Solution .
This line is a horizontal line. All the points on the line have
\(y\) -coordinate
\(04\text{,}\) so the equation of the line is
\(y=40\text{.}\) Because
\(\Delta y=0\) between any two points on the line, its slope is
\(0\text{.}\)
Checkpoint A.8.27 .
Give the equation and slope of the line.
Checkpoint A.8.28 .
Give the equation and slope of the line.
Answer .
\(x=30\text{;}\) \(~ m~\) is undefined
Checkpoint A.8.29 .
Give the equation and slope of the line.
Checkpoint A.8.30 .
Give the equation and slope of the line.
Answer .
\(x=-24\text{;}\) \(~ m~\) is undefined
Subsection A.8.3 Operations on Algebraic Fractions
Subsubsection A.8.3.1 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.
Example A.8.31 .
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 A.8.32 .
Multiply
\(~4 \left(\dfrac{b}{c}\right)\)
Solution .
\(4 \left(\dfrac{b}{c}\right) = \dfrac{4}{1} \left(\dfrac{b}{c}\right) = \dfrac{4 \cdot b}{1 \cdot c} = \dfrac{4b}{c}\)
Checkpoint A.8.33 .
Multiply
\(~\dfrac{3}{2a}\left(\dfrac{a^2}{6}\right)\)
Checkpoint A.8.34 .
Multiply
\(~8\left(\dfrac{m}{4x^2}\right)\)
Subsubsection A.8.3.2 Divide fractions
To divide one fraction by another, we first take the reciprocal of the second fraction, then proceed as in multiplication.
Example A.8.35 .
Divide
\(~\dfrac{t}{w} \div \dfrac{3t^2}{w}\)
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 A.8.36 .
Divide
\(~\dfrac{a}{2} \div \dfrac{b}{1-b}\)
Solution .
\(\dfrac{a}{2} \cdot \dfrac{1-b}{b} = \dfrac{a(1-b)}{2 \cdot b} = \dfrac{a-ab}{2b}\)
Checkpoint A.8.37 .
Divide
\(~\dfrac{3}{4} \div \dfrac{cw^2}{t^2-2}\)
Answer .
\(\dfrac{3t^2-6}{4cw^2}\)
Checkpoint A.8.38 .
Divide
\(~\dfrac{n}{n-1} \div \dfrac{n^2}{p+1}\)
Subsubsection A.8.3.3 Use factoring
Example A.8.39 .
Multiply
\(~\dfrac{4y^2-1}{4-y^2} \cdot \dfrac{y^2-2y}{4y+2}\)
Solution .
We factor each numerator and denominator, and look for common factors.
\begin{align*}
\amp \dfrac{4y^2-1}{4-y^2} \cdot \dfrac{y^2-2y}{4y+2} \amp \\
\amp\qquad=\dfrac{(2y-1)\cancel{(2y+1)}}{\cancel{(2-y)}(2+y)} \cdot \dfrac{y(-1)\cancel{(y-2)}}{2\cancel{(2y+1)}} \amp \amp \blert{\text{Divide out common factors.}}\\
\amp\qquad = \dfrac{-y(2y-1)}{2(y+2)} \amp \amp \blert{\text{Note:}~~y-2=-(2-y)}
\end{align*}
Example A.8.40 .
Divide
\(~\dfrac{6ab}{2a+b} \div (4a^2b)\)
Solution .
We multiply the first fraction by the reciprocal of the second fraction.
\begin{align*}
\dfrac{6ab}{2a+b} \div (4a^2b) \amp = \dfrac{\blert{\cancel{2} \cdot 3}\cancel{ab}}{2a+b} \cdot \dfrac{1}{\blert{\cancel{2} \cdot 2}a \cdot \cancel{ab}} \amp \amp \blert{\text{Divide out common factors.}}\\
\amp = \dfrac{3}{2a(2a+b)}
\end{align*}
Checkpoint A.8.41 .
Multiply
\(~\dfrac{3xy}{4xy-6y^2} \cdot \dfrac{2x-3y}{12x}\)
Checkpoint A.8.42 .
Multiply
\(~\dfrac{9x^2-25}{2x-2} \cdot \dfrac{x^2-1}{6x-10}\)
Answer .
\(\dfrac{(3x+5)(x+1)}{4}\)
Checkpoint A.8.43 .
Divide
\(~(x^2-9) \div \dfrac{x^2-6x+9}{3x}\)
Checkpoint A.8.44 .
Divide
\(~\dfrac{x^2-1}{x+3} \div \dfrac{x^2-x-2}{x^2+5x+6}\)
Answer .
\(\dfrac{(x-1)(x+2)}{x-2}\)
Subsubsection A.8.3.4 Find an LCD
The first step in adding unlike fractions is to find the lowest common denominator, or LCD.
Example A.8.45 .
Find the LCD for the fractions
\(~\dfrac{2}{5} + \dfrac{3}{10}\)
Solution .
We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.
\begin{align*}
15 \amp = \blert{3} ~~~\cdot~~~ 5\\
10 \amp = ~~~~~~~~~~~~\blert{5}~~~\cdot~~~\blert{2}
\end{align*}
The LCD is \(3 \cdot 5\cdot 2\text{,}\) or 30.
Example A.8.46 .
Find the LCD for the fractions
\(~\dfrac{5}{6a} - \dfrac{2}{9a^2}\)
Solution .
We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.
\begin{align*}
6a \amp = \blert{2}~~~\cdot~~~ 3~~~\cdot~~~ a\\
9a^2 \amp = ~~~~~~~~~~~~\blert{3}~~~\cdot~~~\blert{3}~~~\cdot~~~ \blert{a}~~~\cdot~~~ \blert{a}
\end{align*}
The LCd is \(2 \cdot 3 \cdot 3 \cdot a \cdot a\text{,}\) or \(18a^2\text{.}\)
Checkpoint A.8.47 .
Find the LCD for the fractions.
\(\displaystyle \dfrac{7}{8} - \dfrac{1}{6}\)
\(\displaystyle \dfrac{52}{75} + \dfrac{13}{24}\)
Answer .
\(\displaystyle 24\)
\(\displaystyle 600\)
Checkpoint A.8.48 .
Find the LCD for the fractions.
\(\displaystyle \dfrac{3}{4a^2b^2} + \dfrac{7}{10ab^3}\)
\(\displaystyle \dfrac{3}{8t^4} - \dfrac{3}{5t^2}\)
Answer .
\(\displaystyle 20a^2b^3\)
\(\displaystyle 40t^4\)
Subsubsection A.8.3.5 Build fractions
Before we can add unlike fractions, we must build each fraction to an equivalent one with the LCD as denominator.
Example A.8.49 .
Write an equivalent fraction with the new denominator:
\begin{equation*}
~\dfrac{3}{s} = \dfrac{?}{2s^2}
\end{equation*}
Solution .
We first find the building factor for the fraction: what must we mulitply the old denominator by to get the new denominator? We factor the new denominator to see what factors are missing.
The new denominator is \(2 \cdot s \cdot 3\text{,}\) so we need to multiply the old denominator by \(2s\text{.}\) This is the building factor. We multiply top and bottom of the old fraction by the building factor:
\begin{equation*}
\dfrac{3}{s} \cdot \blert{\dfrac{2s}{2s}} = \dfrac{6}{2s^2}
\end{equation*}
Example A.8.50 .
Write an equivalent fraction with the new denominator:
\begin{equation*}
~\dfrac{2}{t+1} = \dfrac{?}{t^2+t}
\end{equation*}
Solution .
The new denominator factors as \(t(t+1)\text{,}\) so the building factor is \(t\text{.}\) We multiply top and bottom of the old fraction by \(t\) to obtain:
\begin{equation*}
\dfrac{2}{t+1} \cdot \blert{\dfrac{t}{t}} = \dfrac{2t}{t^2+t}
\end{equation*}
Checkpoint A.8.51 .
Write an equivalent fraction with the new denominator.
\(\displaystyle \dfrac{1}{n} = \dfrac{?}{n^2-n}\)
\(\displaystyle \dfrac{4}{a-1} = \dfrac{?}{a^2-1}\)
Answer .
\(\displaystyle \dfrac{n-1}{n^2-n}\)
\(\displaystyle \dfrac{4a+4}{a^2-1}\)
Checkpoint A.8.52 .
Write an equivalent fraction with the new denominator.
\(\displaystyle \dfrac{b}{b-2} = \dfrac{?}{(b+3)(b-2)}\)
\(\displaystyle \dfrac{x-1}{x^2+2x} = \dfrac{?}{(x^2-x)(x+2)}\)
Answer .
\(\displaystyle \dfrac{b^2+3b}{(b+3)(b-2)}\)
\(\displaystyle \dfrac{x^2-2x+1}{(x^2-x)(x+2)}\)
Subsubsection A.8.3.6 Add or subtract 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.
Example A.8.53 .
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 A.8.54 .
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 A.8.55 .
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 A.8.56 .
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*}
Checkpoint A.8.57 .
Subtract
\(~\dfrac{4a}{b^2} - \dfrac{c}{b^2}\)
Checkpoint A.8.58 .
Subtract
\(~\dfrac{p+2}{2q} - \dfrac{p-1}{2q}\)
Checkpoint A.8.59 .
Add
\(~N + \dfrac{2}{N}\)
Checkpoint A.8.60 .
Add
\(~\dfrac{2}{xy} - \dfrac{y}{3x}\)
Subsection A.8.4 More Operations on Fractions
Subsubsection A.8.4.1 Use negative exponents
Recall that a negative exponent indicates a reciprocal.
Example A.8.61 .
Write each expression without negative exponents.
\(\displaystyle \dfrac{1}{4}xy^{-2}\)
\(\displaystyle a^{-3}b^{-2}\)
\(\displaystyle \dfrac{2a^{-1}}{{bc}^{-2}}\)
\(\displaystyle \dfrac{x}{y^{-2}}+\dfrac{x^{-2}}{y}\)
Solution .
We use the fact that \(~a^{-n} = \dfrac{1}{a^n}\text{,}\) and consequently that \(~\dfrac{1}{a^{-n}} = a^n\text{.}\)
\(\displaystyle \dfrac{x}{4y^2}\)
\(\displaystyle \dfrac{1}{a^{3}b^{2}}\)
\(\displaystyle \dfrac{2c^2}{ab}\)
\(\displaystyle xy^{2}+\dfrac{1}{x^2y}\)
Example A.8.62 .
Simplify where possible using the laws of exponents.
\(\displaystyle 3x^{-3}x^5\)
\(\displaystyle \dfrac{4a^{-4}}{8a^{-8}}\)
\(\displaystyle (2bc^{-3})^{-2}\)
\(\displaystyle 3x^{-4}-2x^{-3}\)
Solution .
Add the exponents: \(~3x^{-3}x^5 = 3x^2\)
Subtract the exponents: \(~\dfrac{4a^{-4}}{8a^{-8}} = \dfrac{1}{2}a^{-4-(-8)} = \dfrac{a^4}{2}\)
Raise each factor to the power \(-2\text{.}\) Multiply exponents:
\begin{equation*}
(2bc^{-3})^{-2} = 2^{-2}b^{-2}(c^{-3})^{-2} = \dfrac{c^6}{4b^2}
\end{equation*}
We cannot add or subtract powers with different exponents.
\begin{equation*}
3x^{-4}-2x^{-3} = \dfrac{3}{x^4} - \dfrac{2}{x^3}
\end{equation*}
Checkpoint A.8.63 .
Simplify where possible. Write your answer without negative exponents.
\begin{equation*}
(2x^3y^{-4})(\dfrac{3}{4}x^{-2}y^2)
\end{equation*}
Checkpoint A.8.64 .
Simplify where possible. Write your answer without negative exponents.
\begin{equation*}
~\dfrac{ab^{-3}}{(3ab)^{-2}}
\end{equation*}
Checkpoint A.8.65 .
Simplify where possible. Write your answer without negative exponents.
\begin{equation*}
2x^{-2}-(2x)^{-2}
\end{equation*}
Checkpoint A.8.66 .
Simplify where possible. Write your answer without negative exponents.
\begin{equation*}
2x^{-2}(-2x)^{-2}
\end{equation*}
Subsubsection A.8.4.2 Improper fractions
An improper fraction is one in which the numerator is larger than the denominator.
Example A.8.67 .
Write an improper fraction for the sum
\(~2 + \dfrac{3}{8}\)
Solution .
We write the whole number with the same denominator as the fraction. Thus,
\begin{equation*}
\blert{2} + \dfrac{3}{8} = \blert{\dfrac{16}{8}}+\dfrac{3}{8} = \dfrac{16+3}{8} = \dfrac{19}{8}
\end{equation*}
Example A.8.68 .
Write an improper fraction for the difference
\(~3 - \dfrac{1}{5}\)
Solution .
We write the whole number with the same denominator as the fraction. Thus,
\begin{equation*}
\blert{3} - \dfrac{1}{5} = \blert{\dfrac{15}{5}}-\dfrac{1}{5} = \dfrac{15-1}{5} = \dfrac{14}{5}
\end{equation*}
Checkpoint A.8.69 .
Write an improper fraction for each sum or difference.
\(\displaystyle 1 + \dfrac{3}{4}\)
\(\displaystyle 2 - \dfrac{3}{10}\)
Answer .
\(\displaystyle \dfrac{7}{4}\)
\(\displaystyle \dfrac{17}{10}\)
Checkpoint A.8.70 .
Write an improper fraction for each sum or difference.
\(\displaystyle 1 - \dfrac{23}{100}\)
\(\displaystyle 3 + \dfrac{3}{25}\)
Answer .
\(\displaystyle \dfrac{77}{100}\)
\(\displaystyle \dfrac{78}{25}\)
Subsubsection A.8.4.3 Check a division
Remember that division is the inverse operation for multiplication.
Example A.8.71 .
Check that the division is correct:
\(~536\div 15 = 35\dfrac{11}{15}\)
Solution .
The quotient tells us that 15 divides into 536 thirty-five times, with a remainder of 11. This in turn means that if we multiply 15 by 35, and then add 11, we should get 536 back again.
\begin{equation*}
15 \times 35 + 11 = 525 + 11 = 536
\end{equation*}
Note the pattern: divisor \(\times\) quotient \(+\) remainder \(=\) starting number
Example A.8.72 .
Check that the division is correct:
\begin{equation*}
(3n^2+n-6)\div (n+2) = 3n-5 + \dfrac{4}{n+2}
\end{equation*}
Solution .
The answer tells us that \(n+2\) divides into \(3n^2+n-6\) to give a quotient of \(3n-5\text{,}\) with a remainder of 4. If we multiply \(n+2\) by \(3n-5\text{,}\) and then add 4, we should get \(3n^2+n-6\) back again.
\begin{equation*}
(n+2)(3n-5)+4 = (3n^2+n-10)+4 = 3n^2-n-6
\end{equation*}
Checkpoint A.8.73 .
Check the division.
\begin{equation*}
25 \div 4 = 6\dfrac{1}{4}
\end{equation*}
Checkpoint A.8.74 .
Check the division.
\begin{equation*}
1331 \div 28 = 47\dfrac{15}{28}
\end{equation*}
Checkpoint A.8.75 .
Check the division.
\begin{equation*}
(n^2+3n+6) \div (n+1) = n+2 +\dfrac{4}{n+1}
\end{equation*}
Answer .
\((n+1)(n+2)+4 = n^2+3n+6\)
Checkpoint A.8.76 .
Check the division.
\begin{equation*}
(2x^3+7x^2+9x+40) \div (2x-3) = x^2+5x+12 + \dfrac{40}{2x-3}
\end{equation*}
Answer .
\((2x-3)(x^2+5x+12)+40 = 2x^3+7x^2+9x+40\)
Subsection A.8.5 Equations with Fractions
Subsubsection A.8.5.1 Solve quadratic equations
Once we have cleared the fractions from an equation, we may have a quadratic equation to solve. We can choose the easiest method to solve: factoring, extracting roots, or the quadratic formula.
Example A.8.77 .
Solve each quadratic equation by the easiest method.
\(\displaystyle 2x^2-2x=3\)
\(\displaystyle (2x-1)^2=3\)
\(\displaystyle 2x^2-x=3\)
Solution .
Because \(~2x^2-2x-3~\) does not factor, we use the quadratic formula.
\begin{equation*}
x = \dfrac{2 \pm \sqrt{(-2)^2-4(2)(-3)}}{2(2)} = \dfrac{2\pm \sqrt{28}}{4} = \dfrac{1 \pm \sqrt{7}}{2}
\end{equation*}
We use extraction of roots.
\begin{align*}
2x-1 \amp = \pm \sqrt{3}\\
x \amp = \dfrac{1 \pm \sqrt{3}}{2}
\end{align*}
We write the equation in standard form and factor the left side.
\begin{align*}
2x^2-x-3 \amp = 0\\
(2x-3)(x+1) \amp = 0\\
2x-3=0~~~~x+1 \amp = 0\\
x = \dfrac{3}{2}~~~~x \amp = -1
\end{align*}
Checkpoint A.8.78 .
Solve each equation by the easiest method.
\(\displaystyle 3x^2+10x=8\)
\(\displaystyle x^2+6x+9=8\)
\(\displaystyle 81x^2-18x+1=0\)
\(\displaystyle 9x^2+18x=27\)
Answer .
\(\displaystyle x=-4,~\dfrac{2}{3}\)
\(\displaystyle x=-2\pm2\sqrt{2}\)
\(\displaystyle x=\dfrac{1}{9},~\dfrac{1}{9}\)
\(\displaystyle x=-3,~1\)
Subsubsection A.8.5.2 Solve equations graphically
If we can’t solve an equation algebraically, we may be able use a graph to find at least an approximation for the solution.
Example A.8.79 .
Use a graph to solve the equation
\(~2x^3+9x^2-8x+36=0\)
Solution .
We graph the equation
\(~y=2x^3+9x^2-8x-36~\) and look for the points where
\(~y=0~\) (the
\(x\) -intercepts).
From the graph, we estimate the solutions at
\(~x=-4.5,~x=-2,\) and
\(~x=2\text{.}\) By substituting each of these values into the original equation, you can verify that they are indeed solutions.
Example A.8.80 .
Use a graph to solve the equation
\(~x^2+2x+3 = 15-2x\)
Solution .
We graph the equations
\(~y_1=x^2+2x+3~\) and
\(~y_2=15-2x~\) and look for points on the two graphs where the coordinates are equal (intersubsection points).
From the graph, we see that the points with
\(~x=-6~\) and
\(~x=2~\) have the same
\(y\) -coordinate on both graphs. In other words,
\(~y_1=y_2~\) when
\(~x=-6~\) or
\(~x=2~\text{,}\) so
\(~x=-6~\) and
\(~x=2~\) are the solutions.
Checkpoint A.8.81 .
Use a graph to solve the equation
\(~2x^3+7x^2-7x-12=0\)
Answer .
\(x = -4,~ -1,~ \dfrac{3}{2}\)
Checkpoint A.8.82 .
Use a graph to solve the equation
\(~\dfrac{24}{x+4}=11+2x-x^2\)
Subsubsection A.8.5.3 Choose the correct technique
Example A.8.83 .
Choose the appropriate technique for each problem.
Cross-multiply
Multiply each term by the LCD
Multiply top and bottom by the LCD
Find building factors
Combine \(~\dfrac{8}{x+2} + \dfrac{x}{x-3}\)
Solve \(~\dfrac{8}{x+2} = \dfrac{x}{x-3}\)
Solve \(~\dfrac{8}{x+2} + 1 = \dfrac{x}{x-3}\)
Simplify \(~\dfrac{\dfrac{8}{x} + 1}{\dfrac{x}{x-3}+\dfrac{2}{x}}\)
Solution .
To add fractions, we find an LCD and build each fraction, so choice IV is correct.
To solve a proportion, we can cross-multiply, so choice I is correct.
To clear fractions from an equation, we multiply by the LCD, so choice II is correct.
To simplify a complex fraction, we apply the fundamental pricnicple of fractions, so choice III is correct.
Checkpoint A.8.84 .
Write the first step for the problem.
Solve
\(~\dfrac{3}{x} + 3 = \dfrac{1}{x+3}\)
Checkpoint A.8.85 .
Write the first step for the problem.
Combine
\(~\dfrac{3}{x} + 3 - \dfrac{1}{x+3}\)
Answer .
\(\dfrac{3(x+3)}{x(x+3)} + \dfrac{3x(x+3)}{x(x+3)} - \dfrac{x}{x(x+3)}\)
Checkpoint A.8.86 .
Write the first step for the problem.
Simplify
\(~\dfrac{\dfrac{3}{x} + 1} {3- \dfrac{1}{x+3}}\)
Answer .
\(\dfrac{3(x+3)-x(x+3)}{3x(x+3)-x}\)
Checkpoint A.8.87 .
Write the first step for the problem.
Solve
\(~\dfrac{3}{x} = \dfrac{1}{x+3}\)
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