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Worked-Out Exercises: Preliminaries
The following set of exercises is given together with hints, solutions, and solution paths. They are designed such that
you can first try to solve them and in case you need help, please, open the "hints" section. For checking your
answer, please, open the "solution" section and for getting more details or checking your way of solving the
exercise, please, open the "solution path" section.
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Exercise 1: Evaluating expressions with
exponents |
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Evaluate these expressions (without using your calculator):
\[
{\bf{a)}} \quad 8^{-1/3} \, , \qquad {\bf{b)}} \quad \left( \tfrac{1}{100} \right)^{-3/2} \, , \qquad \text{and} \qquad {\bf{c)}} \quad 5^0 \, ,
\]
as well as
\[
{\bf{d)}} \quad (2^{-2})^3 \, , \qquad {\bf{e)}} \quad \frac{3^3}{3^{1/3} \cdot 3^{2/3}} \, , \qquad \text{and} \qquad
{\bf{f)}} \quad 2^{7/4} \cdot 8^{-1/4} \, .
\]
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 Hint
(please, click on the "+" sign to read more)
Apply the rules of exponents and recall that \( 8 = 2^3 \), \( a^{-1} = \frac{1}{a} \), and \( 100 = 10^2 \).
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Solution
(please, click on the "+" sign to read more)
a) \( \frac{1}{2} \), b) \( 1000 \), c) \( 1 \), d) \( \frac{1}{64} \), e) \( 9 \), and f) \( 2 \).
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Solution Path
(please, click on the "+" sign to read more)
a) We have
\[
8^{-1/3} \, \, = \, \, \frac{1}{8^{1/3}} \, \, = \, \, \frac{1}{\sqrt[3]{8}} \, \, = \, \, \frac{1}{2}
\]
b) We have
\[
\left( \tfrac{1}{100} \right)^{-3/2} \, \, = \, \, 100^{3/2} \, \, = \, \, \left( \sqrt{100} \right)^3 \, \, = \, \, 10^3 \, \, = \, \, 1000
\]
c) We have
\[
5^0 \, \, = \, \, 1
\]
d) We have
\[
(2^{-2})^3 \, \, = \, \, 2^{-6} \, \, = \, \, \frac{1}{2^6} \, \, = \, \, \frac{1}{64}
\]
e) We have
\[
\frac{3^3}{3^{1/3} \cdot 3^{2/3}} \, \, = \, \, \frac{3^3}{3^{1/3 + 2/3}} \, \, = \, \, \frac{3^3}{3^1} \, \, = \, \, 3^2 \, \, = \, \, 9
\]
f) We have
\[
2^{7/4} \cdot 8^{-1/4} \, \, = \, \, 2^{7/4} \cdot (2^3)^{-1/4} \, \, = \, \, 2^{7/4} \cdot 2^{-3/4} \, \, = \, \,
2^{7/4 - 3/4} \, \, = \, \, 2^1 \, \, = \, \, 2
\]
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Exercise 2: Simplifying an algebraic
equation |
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Simplify the expression:
\[
\frac{4 (x+3)^4 (x-2)^2 - 6 (x+3)^2 (x-2)^3}{(x+3) (x-2)^3}
\]
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Hint
(please, click on the "+" sign to read more)
Apply the rules of exponents, factor linear terms and divide equal terms.
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Solution
(please, click on the "+" sign to read more)
\[ \frac{2 (x+3) \left( 2x^2 + 9x + 24 \right)}{x-2} \]
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Solution Path
(please, click on the "+" sign to read more)
We have
\begin{eqnarray*}
\frac{4 (x+3)^4 (x-2)^2 - 6 (x+3)^2 (x-2)^3}{(x+3) (x-2)^3} & = &
\frac{4 (x+3)^3 - 6 (x+3) (x-2)}{x-2} \\[1mm]
& = &
\frac{2 (x+3) \left( 2 (x+3)^2 - 3 (x-2) \right)}{x-2} \\[1mm]
& = &
\frac{2 (x+3) \left( 2x^2 + 9x + 24 \right)}{x-2}
\end{eqnarray*}
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Exercise 3: Using the root formula for
quadratic polynomials |
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Solve the quadratic equations (where possible):
\[
{\bf{a)}} \quad x^2 + 3x + 1 \, \, = \, \, 0 \, , \qquad \text{and} \qquad
{\bf{b)}} \quad x^2 + x + 1 \, \, = \, \, 0 \, .
\]
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Hint
(please, click on the "+" sign to read more)
Apply the solution formula for quadratic equations and take care of the sign of the discriminat.
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Solution
(please, click on the "+" sign to read more)
a) approx. \( -0.381966 \) and \( -2.618034 \), b) there are no real solutions.
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Solution Path
(please, click on the "+" sign to read more)
a) We have
\[
x_{1/2} \, \, = \, \, \frac{-3 \pm \sqrt{9-4}}{2} \, \, = \, \, \frac{-3 \pm \sqrt{5}}{2} \, \, = \, \left\{ \begin{array}{c}
-0.381966 \\[1mm] -2.618034
\end{array} \right.
\]
b) We have
\[
x_{1/2} \, \, = \, \, \frac{-1 \pm \sqrt{1-4}}{2} \qquad \text{as $-3 < 0$ there are no real solutions}
\]
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Exercise 4: Factoring
polynomials |
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Factor the polynomials
- \( p(x) = 12 x^2 - 11 x - 15 \) by using the solution formula for quadratic polynomials.
- \( p(x) = 4(x-2)^3 + 3(x-2)^2 \) by grouping terms strategically.
- \( p(x) = 9 x^2 - 49 \) by grouping terms strategically.
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Hint
(please, click on the "+" sign to read more)
Apply the tips given in the problem formulation.
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Solution
(please, click on the "+" sign to read more)
- \( p(x) = \left( x - \frac{11 + \sqrt{181}}{24} \right) \left( x - \frac{11 - \sqrt{181}}{24} \right) \)
- \( p(x) = (x-2)^2 \left( 4 x - 5 \right) \).
- \( p(x) = (3x-7) (3x+7) \) .
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Solution Path
(please, click on the "+" sign to read more)
a) We have
\[
x_{1/2} \, \, = \, \, \frac{11 \pm \sqrt{121+60}}{24} \, \, = \, \, \frac{11 \pm \sqrt{181}}{24}
\]
such that
\[
p(x) \, \, = \, \, 12 x^2 - 11 x - 15 \, \, = \, \,
\left( x - \frac{11 + \sqrt{181}}{24} \right) \left( x - \frac{11 - \sqrt{181}}{24} \right)
\]
b) We have
\begin{eqnarray*}
p(x) & = & 4(x-2)^3 + 3(x-2)^2 \, \, = \, \, (x-2)^2 \left( 4(x-2) + 3 \right) \\[1mm]
& = & (x-2)^2 \left( 4 x - 8 + 3 \right) \, \, = \, \, (x-2)^2 \left( 4 x - 5 \right)
\end{eqnarray*}
c) As \( 7^2 = 49 \), we have
\begin{eqnarray*}
p(x) & = & 9 x^2 - 49 \, \, = \, \, (3x)^2 - 7^2 \, \, = \, \, (3x-7) (3x+7)
\end{eqnarray*}
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Exercise 5: Simplifying rational
expressions |
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Give the following as a rational expression in lowest terms:
\[
\left( \frac{x^3 - 7 x^2 + 10 x}{x^2 + 6x + 9} \right) \left( \frac{x+3}{x-5} \right) \, .
\]
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Hint
(please, click on the "+" sign to read more)
Bring everything to a common denominator, factor numerator and denominator, and see which parts cancel.
Remember to keep track of numbers where the expression is not defined.
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Solution
(please, click on the "+" sign to read more)
\( \frac{x (x-2)}{x+3} \) for \( x \notin \{ 5, -3 \} \).
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Solution Path
(please, click on the "+" sign to read more)
We have
\begin{eqnarray*}
\left( \frac{x^3 - 7 x^2 + 10 x}{x^2 + 6x + 9} \right) \left( \frac{x+3}{x-5} \right) & = &
\frac{x (x+3) (x^2 - 7x + 10)}{(x+3)^2 (x-5)} \\[1mm]
& = &
\frac{x (x+3) (x-5) (x-2)}{(x+3)^2 (x-5)} \, \, = \, \,
\frac{x (x-2)}{x+3}\\[1mm]
& &
\hspace{4cm} \text{for $x \notin \{ 5, -3 \} $}
\end{eqnarray*}
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Exercise 6: Simplifying rational
expressions |
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Give the following as a rational expression in lowest terms:
\[
\frac{-2}{x^2 - 1} + \frac{x}{x-1} \, .
\]
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Hint
(please, click on the "+" sign to read more)
Bring everything to a common denominator, factor numerator and denominator, and see which parts cancel.
Remember to keep track of numbers where the expression is not defined.
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Solution
(please, click on the "+" sign to read more)
\( \frac{x+2}{x+1} \) for \( x \notin \{-1, 1\} \)
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Solution Path
(please, click on the "+" sign to read more)
\begin{eqnarray*}
\frac{-2}{x^2 - 1} + \frac{x}{x-1} & = & \frac{-2}{x^2 - 1} + \frac{x}{x-1} \cdot \frac{x+1}{x+1} \, \, = \, \, \frac{-2}{x^2 - 1} + \frac{x^2 + x}{x^2-1} \\[1mm]
& = &
\frac{x^2 + x - 2}{x^2 - 1} \, \, = \, \, \frac{(x+2)(x-1)}{(x+1)(x-1)} \\[1mm]
& = &
\frac{x+2}{x+1} \qquad \text{for $x \notin \{-1, 1\}$} \, .
\end{eqnarray*}
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Copyright Kutaisi International University — All Rights Reserved — Last Modified: 12/ 10/ 2022
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