Algebra and Geometry of Elementary Functions

Fei Ye

2021-01-27

Introduction

This notebook is intended to give a brief introduction to elementary functions emphasizing on effective thinking in algebra and geometry.

In the first part, we will review mathematical operations including addition, multiplication, $n$ -th root, exponentiation and logarithm.

In the second part, we will study the concepts of functions, algebraic functions and their applications.

In the third part, we will study elementary transcendental functions and applications.

Comments and suggestions are very welcome.

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Part I: Mathematical Operations

Integer Exponents

Don’t Be Tricked

A pizza shop sales 12-inches pizza and 8-inches pizza at the price $12/each and $6/each respectively. With $12, would you like to order one 12-inches and two 8-inches. Why?
A sheet of everyday copy paper is about 0.01 millimeter thick. Repeat folding along a different side 20 times. Now, how thick do you think the folded paper is?

Properties of Exponents

For an integer $n$ , and an expression $x$ , the mathematical operation of the $n$ -times repeated multiplication of $x$ is call exponentiation, written as $x^n$ , that is, $x^n=\underbrace{x\cdot x \cdots x}_{n~\text{factors of}~x}.$

In the notation $x^n$ , $n$ is called the exponent, $x$ is called the base, and $x^n$ is called the power, read as “ $x$ raised to the $n$ -th power”, “ $x$ to the $n$ -th power”, “ $x$ to the $n$ -th”, “ $x$ to the power of $n$ ” or “ $x$ to the $n$ ”.

Property	Example
The product rule $x^m\cdot x^n=x^{m+n}.$	$2x^2\cdot (-3x^3)=-6x^5.$
The quotient rule (for $x\neq 0$ .) $\dfrac{x^m}{x^n}= \begin{cases} x^{m-n} & \text{if} m\ge n.\\[1em] \dfrac{1}{x^{n-m}} & \text{if} m\le n. \end{cases}$	$\frac{15x^5}{5x^2}=3x^3;$ $\frac{-3x^2}{6x^3}=-\frac{1}{2x}.$
The zero exponent rule (for $x\neq 0$ .) $x^0=1.$	$(-2)^0=1;$ $-x^0=-1.$
The negative exponent rule (for $x\neq 0$ .) $x^{-n}=\dfrac{1}{x^n} \quad\text{and}\quad \dfrac{1}{x^{-n}}=x^n.$	$(-2)^{-3}=\frac{1}{(-2)^3}=-\frac18;$ $\frac{x^{-2}}{x^{-3}}=\frac{x^3}{x^2}=x.$
The power to a power rule $\left(x^a\right)^b=x^{ab}.$	$\left(2^{2}\right)^3=2^6=64;$ $\left(x^2\right)^3=x^6.$
The product raised to a power rule $(xy)^n=x^ny^n.$	$\left(-2x\right)^{2}=(-2)^2x^2=4x^2.$
The quotient raised to a power rule (for $y\neq 0$ .) $\left(\dfrac{x}{y}\right)^n=\dfrac{x^n}{y^n}.$	$\left(\dfrac{x}{-2}\right)^{3}=\dfrac{x^3}{(-2)^3}=-\dfrac{x^3}{8}.$

Order of Basic Mathematical Operations

In mathematics, the order of operations reflects conventions about which procedure should be performed first. There are four levels (from the highest to the lowest):

Parenthesis; Exponentiation; Multiplication and Division; Addition and Subtraction.

Within the same level, the convention is to perform from the left to the right.

Example: Simplify. Write with positive exponents. $\left(\dfrac{2y^{-2}z^{-5}}{4x^{-3}y^6}\right)^{-4}.$

Solution. The idea is to simplify the base first and rewrite using positive exponents only.

$\begin{aligned} \left(\dfrac{2y^{-2}z^{-5}}{4x^{-3}y^6}\right)^{-4} =&\left(\dfrac{x^3}{2z^{5}y^8}\right)^{-4}\\ =&\left(\dfrac{2z^{5}y^8}{x^3}\right)^4\\ =&\dfrac{2^4(z^{5})^4(y^8)^4}{(x^3)^4}\\ =&\dfrac{16y^{32}z^{20}}{x^{12}}.\\ \end{aligned}$

Simplify (at least partially) the problem first
To avoid mistakes when working with negative exponents, it’s better to apply the negative exponent rule to change negative exponents to positive exponents and simplify the base first.

Practice

Exercise: Simplify. Write with positive exponents.

$(3a^2b^3c^2)(4abc^2)(2b^2c^3)$
$\dfrac{4y^3z^0}{x^2y^2}$
$(-2)^{-3}$

Exercise: Simplify. Write with positive exponents.

$\dfrac{-u^0v^{15}}{v^{16}}$
$(-2a^3b^2c^0)^3$
$\dfrac{m^5 n^{2}}{(mn)^3}$

Exercise: Simplify. Write with positive exponents.

$(-3a^2x^3)^{-2}$
$\left(\dfrac{-x^0y^3}{2wz^2}\right)^3$
$\dfrac{3^{-2}a^{-3}b^5}{x^{-3}y^{-4}}$

Exercise: Simplify. Write with positive exponents.

$\left(-x^{-1}(-y)^2\right)^3$
$\left(\dfrac{6x^{-2}y^5}{2y^{-3}z^{-11}}\right)^{-3}$
$\dfrac{(3 x^{2} y^{-1})^{-3}(2 x^{-3} y^{2})^{-1}}{(x^{6} y^{-5})^{-2}}$

Exercise: A store has large size and small size watermelons. A large one cost $4 and a small one $1. Putting on the same table, a smaller watermelons has only half the height of the larger one. Given $4, will you buy a large watermelon or 4 smaller ones? Why?

Review of Factoring

Can You Beat a Calculator

Do you know a faster way to find the values?

Find the value of the polynomial $2x^3-98x$ when $x=-7$ .
Find the value of the polynomial $x^2-9x-22$ when $x=11$ .
Find the value of the polynomial $x^3-2x^2-9x+18$ when $x=-3$ .
Find the value of $16^2-14^2$ .

Factor by Removing the GCF

The greatest common factor (GCF) of two terms is a polynomial with the greatest coefficient and of the highest possible degree that divides each term.

To factor a polynomial is to express the polynomial as a product of polynomials of lower degrees. The first and the easiest step is to factor out the GCF of all terms.

Example: Factor $4x^3y-8x^2y^2+12x^3y^3$ .

Solution.

Find the GCF of all terms.
The GCF of $4x^3y$ , $-8x^2y^2$ and $12x^4y^3$ is $4x^2y$ .
Write each term as the product of the GCF and the remaining factor.
$4x^3y=(4x^2y)\cdot x$ , $-8x^2y^2=(4x^2y)\cdot (-2y)$ , and $12x^4y^3=(4x^2y)(3xy^2)$ .
Factor out the GCF from each term.
$4x^3y-8x^2y^2+12x^3y^3=4x^2y\cdot(x-2y+3xy^2)$ .

Factor by Grouping

For a four-term polynomial, in general, we will group them into two groups and factor out the GCF for each group and then factor further.

Example: Factor $2x^2-6xy+xz-3yz$ .

Solution.

For a polynomial with four terms, one can normally try the grouping method.

Group the first two terms and the last two terms. $\begin{aligned} &2x^2-6xy+xz-3yz\\ =&(2x^2-6xy)+(xz-3yz) \end{aligned}$
Factor out the GCF from each group.
$\begin{aligned} =&2x(x-3y)+z(x-3y) \end{aligned}$
Factor out the binomial GCF. $\begin{aligned} =&(x-3y)(2x+z). \end{aligned}$

Example: Factor $ax+4b-2a-2bx$ .

Solution.

Group the first term with the third term and group the second term with the last term. $\begin{aligned} &ax+4b-2a-2bx\\ =&(ax-2a)+(-2bx+4b) \end{aligned}$
Factor out the GCF from each group. $\begin{aligned} =&a(x-2)+(-2b)(x-2) \end{aligned}$
Factor out the binomial GCF. $\begin{aligned} =&(x-2)(a-2b). \end{aligned}$

Guess and check.

Once you factored one group, you may expect that the other group has the same binomial factor so that factoring may be continued.

Factor Difference of Powers

Factoring is closely related to solving polynomial equations. If a polynomial equation $p(x)=0$ has a solution $r$ , then $p(x)$ has a factor $x-r$ . For example, $x^n-r^n=0$ has a solution $x=r$ . So the difference $x^n-r^n$ has a factor $(x-r)$ . Using long division or by induction, we obtain the following equality.

Difference of $n$ -th powers

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +ab^{n-2}+b^{n-1})$

In particular,

$a^2-b^2=(a-b)(a+b).$

Example: Factor $25x^2-16$ .

Solution.

Recognize the binomial as a difference of squares. $\begin{aligned} &25x^2-16\\ =&(5x)^2-4^2 \end{aligned}$
Apply the formula. $\begin{aligned} =&(5x-4)(5x+4). \end{aligned}$

Example: Factor $32x^3y-2xy^5$ completely.

Solution.

$\begin{aligned} 32x^3y-2xy^3 =&2xy(16x^2-y^4)\\ =&2xy((4x)^2-(y^2)^2)\\ =&2xy(4x+y^2)(4x-y^2). \end{aligned}$

Factor Trinomials

If a trinomial $ax^2+bx+c$ , $A\neq 0$ , can be factored, then it can be expressed as a product of two binomials:
$ax^2+bx+c=(mx+n)(px+q).$ By simplify the product using the FOIL method and comparing coefficients, we observe that
$a=\underbrace{mn}_{\mathrm{F}}\quad\quad\quad b=\underbrace{mq}_{\mathrm{O}}~\underset{+}{\underset{}{+}}~\underbrace{np}_{\mathrm{I}} \quad\quad\quad c=\underbrace{nq}_{\mathrm{F}}$

A trinomial $ax^2+bx+c$ is also called a quadratic polynomial. The function defined by $f(x)=ax^2+bx+c$ is called a quadratic function.

Trial and error.

The observation suggests to use trial and error to find the undetermined coefficients $m$ , $n$ , $p$ , and $q$ from factors of $a$ and $c$ such that the sum of cross products $mq+np$ is $b$ . A diagram as shown in the following examples will be helpful to check a trial.

Example: Factor $x^2+6x+8$ .

Solution. One may factor the trinomial in the following way.

Factor $a=1$ : $1=1\cdot 1.$
Factor $c=8$ : $8=1\cdot 8=2\cdot 4.$
Choose a proper combination of pairs of factors and check if the sum of cross product equals $b=6$ : $1\cdot 4+ 1\cdot 2=6.$
This step can be checked easily using the following diagram.
Factor the trinomial $x^2+6x+8=(x+2)(x+4).$

Example: Factor $2x^2+5x-3$ .

Solution. One may factor the trinomial in the following way.

Factor $a=2$ : $1=1\cdot 2.$
Factor $c=-3$ : $-3=1\cdot (-3)=(-1)\cdot 3.$
Choose a proper combination of pairs of factors and if the sum of cross products equals $b=5$ :
$2\cdot 3+1\cdot(-1)=5.$
This step can be checked easily using the following diagram.
Factor the trinomial $2x^2+5x-3=(x+3)(2x-1).$

Use Auxiliary Problem.

Some higher degree polynomials may be rewrite as a trinomial after a substitution. Factoring the trinomial helps factor the polynomial.

Example: Factor the trinomial completely.

$4x^4-x^2-3$

Solution. One idea is to use a substitute.

Let $x^2=y$ . Then $4x^4-x^2-3=4y^2-y-3$ .
Factor the trinomial in $y$ : $4y^2-y-3=(4y+3)(y-1)$ .
Replace $y$ by $x^2$ and factor further. $\begin{aligned} 4x^4-x^2-3&=4y^2-y-3\\ &=(4y+3)(y-1)\\ &=(4x^2+3)(x^2-1)\\ &=(4x^2+3)(x-1)(x+1). \end{aligned}$

Practice

Exercise: Factor out the GCF.

$18x^2y^2-12xy^3-6x^3y^4$
$5x(x-7)+3y(x-7)$
$-2a^2(x+y)+3a(x+y)$

Exercise: Factor by grouping.

$12xy-10y+18x-15$
$12ac-18bc-10ad+15bd$
$5ax-4bx-5ay+4by$

Exercise: Factor completely.

$25x^2-4$
$8x^3-2x$
$25xy^2+x$

Exercise: Factor completely.

$3x^3+6x^2-12x-24$
$x^4+3x^3-4x^2-12x$

Exercise: Factor the trinomial.

$x^2+4x+3$
$x^2+6x-7$
$x^2-3x-10$

Exercise: Factor the trinomial.

$5x^2+7x+2$
$2x^2+5x-12$
$3x^2-10x-8$

Exercise: Factor completely into polynomials with integer coefficients.

$x^3-5x^2+6x$
$4x^4-12x^2+5$
$2x^3y-9x^2y^2-5xy^3$

Exercise: Each of trinomial below has a factor in the table. Match the letter on the left of a factor with a the number on the left a trinomial to decipher the following quotation.

“ $\dfrac{\phantom{A}}{13}$ $\dfrac{\phantom{A}}{10~~2~~9~~15}$ , $\dfrac{\phantom{A}}{9~~5~~14}$ $\dfrac{\phantom{A}}{13}$ $\dfrac{\phantom{A}}{4~~3~~15~~7~~2~~1}$ ; $\dfrac{\phantom{A}}{13}$ $\dfrac{\phantom{A}}{11~~2~~2}$ , $\dfrac{\phantom{A}}{9~~5~~14}$ $\dfrac{\phantom{A}}{13}$ $\dfrac{\phantom{A}}{8~~5~~3~~6}$ ; $\dfrac{\phantom{A}}{13}$ $\dfrac{\phantom{A}}{14~~3}$ , $\dfrac{\phantom{A}}{9~~5~~14}$ $\dfrac{\phantom{A}}{13}$ $\dfrac{\phantom{A}}{12~~5~~14~~2~~15~~11~~1~~9~~5~~14}$ .”

A: $3x-2$	B: $2x+1$	C: $x+6$	D: $x+7$	E: $2x-1$	F: $3x-1$	G: $x+10$
H: $x-8$	I: $2x+9$	J: $x-1$	K: $x+3$	L: $2x-5$	M: $x+5$	N: $x-7$
O: $x-13$	P: $5x-3$	Q: $4x-11$	R: $x-9$	S: $2x+3$	T: $x+4$	U: $7x+1$
V: $3x+5$	W: $3x+4$	X: $8x+3$	Y: $x-14$	Z: $5x-6$

$x^2-2x-24$
$6x^2+x-2$
$x^2-16x+39$
$6x^2+13x-5$
$x^2-5x-14$
$3x^2-5x-12$
$x^2-x-110$
$x^2-9$
$-3x^2+11x-6$
$x^2-10x+16$
$-2x^2+5x+12$
$42x^2-x-1$
$-2x^2-3x+27$
$x^2+14x+49$
$x^2-81$

Algebra of Rational Expressions

Is There Enough Time

Matt is kayaking upstream on a river with his best effort. After 30 minutes, he received an emergency call and has to return in 20 minute. The speed of the current of the river is 1 mph. Under normal condition, a paddler’s average paddling speed is between 2 and 5 mph. Do you think Matt can return on time? Why?
A construction team is building a house. After half of the work was done, to expedite the construction process, the another team joins in the construction. The first team normally takes 7-10 days to build a hours. The second team normally takes 2 extra days to build a house. How many days it takes to build the house?

Rational Expressions

Let $p$ and $q$ be polynomial functions of $x$ and $p$ is not a constant function. We call the function $r(x)=\frac{p(x)}{q(x)}$ a rational function. The domain of $r$ is $\{x\mid Q(x)\neq 0\}$ . The expression $\frac{p(x)}{q(x)}$ is called a rational expression, the polynomial $q(x)$ is called the numerator, and the polynomial $q(x)$ is called the denominator. A rational expression is simplified if the numerator and the denominator have no common factor other than $1$ .

Let $p(x)$ , $q(x)$ be polynomials with $q(x)\neq 0$ and $c(x)$ be a nonzero expression. Then $\dfrac{~p(x)\cdot c(x)~}{~q(x)\cdot c(x)~}=\dfrac{~p(x)~}{~q(x)~}.$

Example: Simplify $\dfrac{x^2+4x+3}{x^2+3x+2}$ .

Solution.

Factor both the top and the bottom. $\dfrac{x^2+4x+3}{x^2+3x+2}=\dfrac{(x+1)(x+3)}{(x+1)(x+2)}.$
Divide out common factors. $\dfrac{(x+1)(x+3)}{(x+1)(x+2)}=\dfrac{x+3}{x+2}.$

Example: Simplify $\dfrac{2x^2-x-3}{2x^2-3x-5}$ .

Solution.

Factor both the top and the bottom. $\dfrac{2x^2-x-3}{2x^2-3x-5}=\dfrac{(x+1)(2x-3)}{(x+1)(2x-5)}.$
Divide out common factors. $\dfrac{(x+1)(2x-3)}{(x+1)(2x-5)}=\dfrac{2x-3}{2x-5}.$

Multiplying Rational Expressions

If $p$ , $q$ , $s$ , $t$ are polynomials with $q\neq 0$ and $t\neq 0$ , then $\dfrac{~p~}{~q~}\cdot\dfrac{~s~}{~t~}=\dfrac{~ps~}{~qt~}.$

Example: Multiply and then simplify. $\dfrac{3x^2}{x^2+x-6}\cdot\dfrac{x^2-4}{6x}.$

Solution.

Factor numerators and denominators. $\dfrac{3x^2}{x^2+x-6}\cdot\dfrac{x^2-4}{6x}=\dfrac{3\cdot x\cdot x}{(x-2)(x+3)}\cdot\dfrac{(x-2)(x+2)}{2\cdot3\cdot x}$
Multiply and simplify. $\dfrac{\cancel{3}\cdot \cancel{x}\cdot x\cdot\cancel{(x-2)}(x+2)}{\cancel{(x-2)}(x+3)\cdot 2\cdot\cancel{3}\cdot \cancel{x}}=\dfrac{x(x+2)}{2(x+3)}$

Example: Multiply and then simplify. $\dfrac{3x^2-8x-3}{x^2+8x+16}\cdot\dfrac{x^2-16}{5x^2-14x-3}.$

Solution.

$\dfrac{3x^2-8x-3}{x^2+8x+16}\cdot\dfrac{x^2-16}{5x^2-14x-3} =\dfrac{(3x+1)\cancel{(x-3)}\cancel{(x+4)}(x-4)}{\cancel{(x+4)}(x+4)(5x+1)\cancel{(x-3)}} =\dfrac{(3x+1)(x-4)}{(x+4)(5x+1)}$

Dividing Rational Expressions

If $p$ , $q$ , $s$ , $t$ are polynomials where $q\neq 0$ , $s\neq 0$ and $t\neq 0$ , then $\dfrac{~p~}{~q~}\div\dfrac{~s~}{~t~}=\dfrac{~p~}{~q~}\cdot\dfrac{~t~}{~s~}=\dfrac{~pt~}{~qs~}.$

Example: Divide and then simplify.
$\dfrac{2x+6}{x^2-6x-7}\div \dfrac{6x-2}{2x^2-x-3}.$

Solution.

Rewrite as a multiplication.
$\dfrac{2x+6}{x^2-6x-7}\div \dfrac{6x-2}{2x^2-x-3}=\dfrac{2x+6}{x^2-6x-7}\cdot \dfrac{2x^2-x-3}{6x-2}$
Factor and simplify.
$\dfrac{2x+6}{x^2-6x-7}\cdot\dfrac{2x^2-x-3}{6x-2} =\dfrac{\cancel{2}(x+3)\cancel{(x+1)}(2x-3)}{\cancel{2}\cancel{(x+1)}(x-7)(3x-1)} =\dfrac{(x+3)(2x-3)}{(x-7)(3x-1)}$

Adding or Subtracting Rational Expressions with the Same Denominator

If $P$ , $Q$ and $R$ are polynomials with $R\neq 0$ , then $\dfrac{~P~}{~R~}+\dfrac{~Q~}{~R~}=\dfrac{~P+Q~}{~R~}\qquad \text{and} \qquad \dfrac{~P~}{~R~}-\dfrac{~Q~}{~R~}=\dfrac{~P-Q~}{~R~}.$

Example: Add and simplify
$\dfrac{x^2+4}{x^2+3x+2}+\dfrac{5x}{x^2+3x+2}.$

Solution.

Determine if the rational expressions have the same denominator. If so, the new numerator will be the sum/difference of the numerators. $\dfrac{x^2+4}{x^2+3x+2}+\dfrac{5x}{x^2+3x+2}=\dfrac{x^2+5x+4}{x^2+3x+2}.$
Simplify the resulting rational expression. $\dfrac{x^2+5x+4}{x^2+3x+2}=\dfrac{(x+1)(x+4)}{(x+1)(x+2)}=\dfrac{x+4}{x+2}.$

Example: Subtract and simplify $\dfrac{2x^2}{2x^2-x-3}-\dfrac{3x+5}{2x^2-x-3}$ .

Solution.

$\begin{aligned} \dfrac{2x^2}{2x^2-x-3}-\dfrac{3x+5}{2x^2-x-3}=&\dfrac{2x^2-3x-5}{2x^2-x-3}\\ =&\dfrac{(2x-5)(x+1)}{(2x-3)(x+1)}\\ =&\dfrac{2x-5}{2x-3}. \end{aligned}$

Adding or Subtracting Rational Expressions with Different Denominators

To add or subtract rational expressions with different denominators, we need to rewrite the rational expressions to equivalent rational expressions with the same denominator, say the LCD.

Equivalent Reduction.
What if all denominators are the same? How to make denominators the same? Reducing the problem to an easier one using equivalent operations helps solve the problem.

Example: Find the LCD of $\dfrac{3}{x^2-x-6}$ and $\dfrac{6}{x^2-4}$ .

Solution.

Factor each denominator. $x^2-x-6=(x+2)(x-3)\qquad x^2-4=(x-2)(x+2)$
List the factors of the first denominator and add unlisted factors of the second factor to get the final list.
First list $(x+2)$ $(x-3)$
Second list $(x+2)$ $(x-2)$
Final list $(x+2)$ $(x-3)$ $(x-2)$
The LCD is the product of factors in the final list.
$(x+2)(x-3)(x-2).$

Example: Subtract and simplify
$\dfrac{x-3}{x^2-2x-8}- \dfrac{1}{x^2-4}$

Solution.

Find the LCD.
1. First factor denominators.
  $x^2-2x-8=(x+2)(x-4)$ $x^2-4=(x-2)(x+2)$
2. Then using the table to find the final list of factors of the LCD.
  First list $(x+2)$ $(x-4)$
  Second list $(x+2)$ $(x-2)$
  Final list $(x+2)$ $(x-2)$ $(x-4)$
The LCD is .
Rewrite each rational expression into an equivalent expression with the LCD as the new denominator.
$\dfrac{x-3}{x^2-2x-8}- \dfrac{1}{x^2-4}=\dfrac{(x-3)(x-2)}{(x+2)(x-2)(x-4)}-\dfrac{(x-4)}{(x+2)(x-2)(x-4)}$
Subtract and simplify.
$\dfrac{(x-3)(x-2)-(x-4)}{(x+2)(x-2)(x-4)}=\dfrac{(x^2-5x+6)-(x-4)}{(x+2)(x-2)(x-4)}=\dfrac{x^2-6x+10}{(x+2)(x-2)(x-4)}$

Simplifying Complex Rational Expressions

A complex rational expression is a rational expression whose denominator or numerator contains a rational expression.

A complex rational expression is equivalent to the quotient of its numerator by its denominator. That suggests the following strategy to simplify a complex rational expression.

Simplify and Change the Viewpoint. A complex rational expression is a quotient of two rational expressions. You may rewrite it as an multiplication by flipping the denominator. However, it’s better to simply the numerator and denominator or you won’t see a good looking new expression.

Example: Simplify $\dfrac{~\dfrac{2x-1}{x^2-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{1}{x^2-1}~}$

Solution.

Simplify the numerator and the denominator. $\begin{aligned} \dfrac{~\dfrac{2x-1}{x^2-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{1}{x^2-1}~} =&\dfrac{ ~\dfrac{2x-1}{(x-1)(x+1)}+\dfrac{(x-1)(x-1)}{(x-1)(x+1)}~ }{~ \dfrac{(x+1)(x+1)}{(x-1)(x+1)}-\dfrac{1}{(x-1)(x+1)}~ }\\[5pt] =&\dfrac{ ~\dfrac{(2x-1)+(x-1)(x-1)}{(x-1)(x+1)}~ }{ ~\dfrac{(x+1)(x+1)-1}{(x-1)(x+1)}~ }\\[5pt] =&\dfrac{ ~\dfrac{(2x-1)+(x^2-2x+1)}{(x-1)(x+1)}~ }{ ~\dfrac{(x^2+2x+1)-1}{(x-1)(x+1)}~ }\\[5pt] =&\dfrac{ ~\dfrac{x^2}{(x-1)(x+1)}~ }{ ~\dfrac{x^2+2x}{(x-1)(x+1)}~ } \end{aligned}$
Rewrite as a product. $\dfrac{ ~\dfrac{x^2}{(x-1)(x+1)}~ }{ ~\dfrac{x^2+2x}{(x-1)(x+1)}~ } =\dfrac{x^2}{(x-1)(x+1)}\cdot \dfrac{(x-1)(x+1)}{x^2+2x}$
Multiply and simplify. $\begin{aligned} \dfrac{x^2}{(x-1)(x+1)}\cdot \dfrac{(x-1)(x+1)}{x^2+2x} =&\dfrac{x\cdot x}{(x-1)(x+1)}\cdot \dfrac{(x-1)(x+1)}{x(x+2)}\\[5pt] =& \dfrac{\bcancel{x}x\cancel{(x-1)(x+1)}}{\bcancel{x}(x+2)\cancel{(x-1)(x+1)}}\\[5pt] =& \dfrac{x}{x+2} \end{aligned}$

Another way to simplify a complex rational expression is to multiply the LCD to both the denominator and numerator and then simplify.

Example: Simplify $\dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~}$

Solution.

Find the LCD of all denominators.
In this case, the LCD is $(x-1)(x+1)$ .
Multiply the complex rational expression by $\dfrac{(x-1)(x+1)}{(x-1)(x+1)}$ and simplify. $\begin{aligned} \dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~} =&\dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~}\cdot \dfrac{(x-1)(x+1)}{(x-1)(x+1)}\\[5pt] =& \dfrac{~(x-1)(x+1)\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right)~}{~(x-1)(x+1)\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)~}\\[5pt] =& \dfrac{~(x+1)^2+(x-1)^2~}{~(x+1)^2-(x-1)^2~}\\[5pt] =& \dfrac{x^2+1}{2x} \end{aligned}$

Practice

Exercise: Simplify.

$\dfrac{3x^2-x-4}{x+1}$
$\dfrac{2x^2-x-3}{2x^2+3x+1}$
$\dfrac{x^2-9}{3x^2-8x-3}$

Exercise: Multiply and simplify.

$\dfrac{x+5}{x+4}\cdot\dfrac{x^2+3x-4}{x^2-25}$
$\dfrac{3x^2-2x}{x+2}\cdot\dfrac{3x^2-4x-4}{9x^2-4}$
$\dfrac{6y-2}{3-y}\cdot\dfrac{y^2-6y+9}{3y^2-y}$

Exercise: Divide and simplify.

$\dfrac{9x^2-49}{6}\div\dfrac{3x^2-x-14}{2x+4}$
$\dfrac{x^2+3x-10}{2x-2}\div\dfrac{x^2-5x+6}{x^2-4x+3}$
$\dfrac{y-x}{xy}\div\dfrac{x^2-y^2}{y^2}$

Exercise: Simplify. $\frac{-x^2+11x-18}{x^2-4x+4}\div \frac{x^2-5x-36}{x^2-7x+12}\cdot \frac{2x^2+7x-4}{x^2+2x-15}$

Exercise: Add/subtract and simplify.

$\dfrac{x^2+2x-2}{x^2+2x-15}+\dfrac{5x+12}{x^2+2x-15}$
$\dfrac{3x-10}{x^2-25}-\dfrac{2x-15}{x^2-25}$
$\dfrac{4}{(x-3)(x+2)}+\dfrac{3x-2}{x^2-x-6}$

Exercise: Find the LCD of rational expressions.

$\dfrac{2x}{2x^2-5x-3}$ and $\dfrac{x-1}{x^2-x-6}$
$\dfrac{9}{7x^2-28x}$ and $\dfrac{2}{x^2-8x+16}$

Exercise: Add and simplify.

$\dfrac{x}{x+1}+\dfrac{x-1}{x+2}$
$\dfrac{x+2}{2x^2-x-3}+\dfrac{1}{x^2+3x+2}$
$\dfrac{4}{x-3}+\dfrac{3x-2}{x^2-x-6}$

Exercise: Subtract and simplify.

$\dfrac{3x+5}{x^2-7x+12}-\dfrac{3}{x-3}$
$\dfrac{y}{y^2-5y-6}-\dfrac{7}{y^2-4y-5}$
$\dfrac{2x-3}{x^2+3x-10}-\dfrac{x+2}{x^2+2x-8}$

Exercise: Simplify. $\frac{x+11}{7x^2-2x-5}+\frac{x-2}{x-1}-\frac{x}{7x+5}$

Exercise: Subtract and simplify. $\frac{x-1}{x^2-3x}+\frac{4}{x^2-2\:x-3}-\frac{1}{x\left(x+1\right)}$

Exercise: Simplify.

$\dfrac{~1+\dfrac{2}{x}~}{~1-\dfrac{2}{x}~}$
$\dfrac{~\dfrac{1}{x^2}-1~}{~\dfrac{1}{x^2}-\dfrac{1}{x}~}$

Exercise: Simplify.

$\dfrac{~\dfrac{x^2-y^2}{y^2}~}{~\dfrac1x-\dfrac{1}{y}~}$
$\dfrac{~\dfrac{2}{(x+1)^2}-\dfrac{1}{x+1}~}{~1-\dfrac{4}{(x+1)^2}~}$

Exercise: Simplify.

$\dfrac{~\dfrac{5x}{x^2-x-6}~}{~\dfrac2{x+2}+\dfrac{3}{x-3}~}$
$\dfrac{~\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}~}{~\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}~}$

Exercise: Tim and Jim refill their cars at the same gas station twice last month. Each time Tim got $20 gas and Jim got 8 gallon. Suppose they refill their cars on same days. The price was $2.5 per gallon the first time. The price on the second time changed. Can you find out who had the better average price?

Radicals and Rational Exponents

Do You Want to Be a Fire Fighter

To reach the 5th floor window of a building that is 25 feet from the location of the turntable aerial ladder truck. How long should the ladder be placed to reach the window? The hight of that window is 50 feet.

Radical Expressions

If $b^2=a$ , then we say that $b$ is a square root of $a$ . We denote the positive square root of $a$ as $\sqrt{a}$ , called the principal square root.

For any real number $a$ , the expression $\sqrt{a^2}$ can be simplified as $\sqrt{a^2}=|a|.$

If $b^3=a$ , then we say that $b$ is a cube root of $a$ . The cube root of a real number $a$ is denoted by $\sqrt[3]{a}$ .

For any real number $a$ , the expression $\sqrt[3]{a^3}$ can be simplified as $\sqrt[3]{a^3}=a.$

In general, if $b^n=a$ , then we say that $b$ is an $n$ -th root of $a$ . If $n$ is even, the positive $n$ -th root of $a$ , called the principal $n$ -th root, is denoted by $\sqrt[n]{a}$ . If $n$ is odd, the $n$ -the root $\sqrt[n]{a}$ of $a$ has the same sign with $a$ .

In $\sqrt[n]{a}$ , the symbol $\sqrt{\phantom{a}}$ is called the radical sign, $a$ is called the radicand, and $n$ is called the index.

If $n$ is even, then the $n$ -th root of a negative number is not a real number.

For any real number $a$ , the expression $\sqrt[n]{a^n}$ can be simplified as

$\sqrt[n]{a^n}=|a|$ if $n$ is even.
$\sqrt[n]{a^n}=a$ if $n$ is odd.

A radical is simplified if the radicand has no perfect power factors against the radical.

Example: Simplify the radical expression using the definition.

$\sqrt{4(y-1)^2}$
$\sqrt[3]{-8x^3y^6}$

Solution.

$\sqrt{4(y-1)^2}=\sqrt{[2(y-1)]^2}=2|y-1|$ .
$\sqrt[3]{-8x^3y^6}=\sqrt[3]{(-2xy^2)^2}=-2xy^2$ .

Rational Exponents

If $\sqrt[n]{a}$ is a real number, then we define $a^{\frac mn}$ as $a^{\frac mn}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m.$

Rational exponents have the same properties as integral exponents:

$a^m\cdot a^n=a^{m+n}$
1. $\dfrac{a^m}{a^n}=a^{m-n}$
$a^{-\frac mn}=\dfrac{~1~}{~a^{\frac mn}~}$
$(a^m)^n=a^{mn}$
$(ab)^m=a^m\cdot b^m$
$\left(\dfrac ab\right)^m=\dfrac{a^m}{b^m}$

Example: Simplify the radical expression or the expression with rational exponents. Write in radical notation.

$\sqrt{x}\sqrt[3]{x^2}$
$\sqrt[3]{\sqrt{x^3}}$
$\left(\frac{x^{\frac12}}{x^{-\frac56}}\right )^{\frac14}$
$\sqrt{\frac{x^{-\frac12}y^2}{x^{\frac32}}}$

Solution.

$\sqrt{x}\sqrt[3]{x^2}=x^{\frac12}x^{\frac{2}{3}}=x^{\frac{1}{2}+\frac{2}{3}}=x^\frac{7}{6}=x\sqrt[6]{x}.$
$\sqrt[3]{\sqrt{x^3}}=(\sqrt{x^3})^\frac{1}{3}=[(x^3)^{\frac{1}{2}}]^{\frac{1}{3}}=x^{3\cdot\frac{1}{2}\cdot\frac{1}{3}}=x^{\frac{1}{2}}=\sqrt{x}.$
$\left(\frac{x^{\frac12}}{x^{-\frac56}}\right)^{\frac14}=(x^{\frac12}x^{\frac56})^{\frac14}=(x^{\frac{1}{2}+\frac{5}{6}})^{\frac{1}{4}}=(x^\frac{4}{3})^{\frac{1}{4}}=x^{\frac{1}{3}}=\sqrt[3]{x}.$
$\sqrt{\frac{x^{-\frac12}y^2}{x^{\frac32}}}=\sqrt{\frac{y^2}{x^2}}=\sqrt{\left(\frac yx\right)^2}=\left|\frac yx\right|.$

In general, rewriting radical in rational exponents helps simplify calculations.

Product and Quotient Rules for Radicals

If $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers, then ${\sqrt[n]a}{\sqrt[n]b}=\sqrt[n]{ab}.$

If $\sqrt[n]a$ and $\sqrt[n]b$ are real numbers and $b\neq 0$ , then $\dfrac{\sqrt[n]a}{\sqrt[n]b}=\sqrt[n]{\dfrac ab}.$

Example: Simplify the expression.

$\sqrt[4]{8xy^4}\sqrt[4]{2x^7y}$ .
$\dfrac{\sqrt[5]{96x^9y^3}}{\sqrt[5]{3x^{-1}y}}$ .

Solution.

$\sqrt[4]{8xy^4}\sqrt[4]{2x^7y}=\sqrt[4]{(8xy^4)\cdot (2x^7y)}=\sqrt[4]{16x^8y^5}=\sqrt[4]{2^4(x^2)^4y^4\cdot y}=2x^2y\sqrt[4]{y}$ .
$\dfrac{ \sqrt[5]{96x^9y^3} }{ \sqrt[5]{3x^{-1}y} }=\sqrt[5]{\dfrac{96x^9y^3}{3x^{-1}y}}=\sqrt[5]{32x^{10}y^2}=\sqrt[5]{(2x^2)^5\cdot y^2}=2x^2\sqrt[5]{y^2}$ .

Combining Like Radicals

Two radicals are called like radicals if they have the same index and the same radicand. We add or subtract like radicals by combining their coefficients.

Example: Simplify the expression.

$\sqrt{8x^3}-\sqrt{(-2)^2 x^4}+\sqrt{2x^5}.$

Solution.

$\sqrt{8x^3}-\sqrt{(-2)^2 x^4}+\sqrt{2x^5}=2x\sqrt{2x}-2x^2+x^2\sqrt{2x}=(x^2+2x)\sqrt{2x}-2x^2.$

Multiplying Radicals

Multiplying radical expressions with many terms is similar to that multiplying polynomials with many terms.

Example: Simplify the expression. $(\sqrt{2x}+2\sqrt{x})(\sqrt{2x}-3\sqrt{x}).$

Solution.

$\begin{aligned} (\sqrt{2x}+2\sqrt{x})(\sqrt{2x}-3\sqrt{x}) =&\sqrt{2x}\cdot\sqrt{2x}-3\sqrt{x}\cdot\sqrt{2x}+2\sqrt{x}\cdot\sqrt{2x}-6\sqrt{x}\cdot\sqrt{x}\\ =&2x-3x\sqrt{2}+2x\sqrt{2}-6x\\ =&-4x-x\sqrt{2}\\ =&-(4+\sqrt{2})x. \end{aligned}$

Rationalizing Denominators

Rationalizing denominator means rewriting a radical expression into an equivalent expression in which the denominator no longer contains radicals.

Example: Rationalize the denominator.

$\dfrac{1}{2\sqrt{x^3}}$
$\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$

Solution.

In this case, to get rid of the radical in the bottom, we multiply the expression by $\dfrac{\sqrt{x}}{\sqrt{x}}$ so that the radicand in the bottom becomes a perfect power. $\dfrac{1}{2\sqrt{x^3}}=\dfrac{1}{2\sqrt{x^3}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}}{2\sqrt{x^2}\sqrt{x}}=\dfrac{\sqrt{x}}{2x^2}.$
In this case, we use the formula $(a-b)(a+b)=a^2+b^2$ . Multiply the expression by $\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$ . $\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\dfrac{x+y+2\sqrt{xy}}{x-y}.$

Complex Numbers

The imaginary unit $\mathbf{i}$ is defined as $\mathbf{i}=\sqrt{-1}$ . Hence $\mathbf{i}^2=-1$ .

If $b$ is a positive number, then $\sqrt{-b}=\mathbf{i}\sqrt{b}$ .

Let $a$ and $b$ are two real numbers. We define a complex number by the expression $a+b \mathbf{i}$ . The number $a $ is called the real part and the number $b$ is called the imaginary part. If $b=0$ , then the complex number $a+b\mathbf{i}=a$ is just the real number. If $b\neq 0$ , then we call the complex number $a+b\mathbf{i}$ an imaginary number. If $a=0$ and $b\neq 0$ , then the complex number $a+b\mathbf{i}=b\mathbf{i}$ is called a purely imaginary number.

Adding, subtracting, multiplying, dividing or simplifying complex numbers are similar to those for radical expressions. In particular, adding and subtracting become similar to combining like terms.

Example: Simplify and write your answer in the form $a+b\mathbf{i}$ , where $a$ and $b$ are real numbers and $\mathbf{i}$ is the imaginary unit.

$\sqrt{-3}\sqrt{-4}$
$(4\mathbf{i}-3)(-2+\mathbf{i})$
$\frac{-2+5\mathbf{i}}{\mathbf{i}}$
$\frac{1}{1-2\mathbf{i}}$
$\mathbf{i}^{2018}$

Solution.

$\sqrt{-3}\sqrt{-4}=\mathbf{i}\sqrt{3}\cdot\mathbf{i}\sqrt{4}=\mathbf{i}^2\cdot \sqrt3\cdot 2=-2\sqrt{3}.$
$\begin{aligned} (4\mathbf{i}-3)(-2+\mathbf{i})&=4\mathbf{i}\cdot(-2)+4\mathbf{i}\cdot \mathbf{i}+(-3)\cdot(-2)+(-3)\cdot\mathbf{i} \\ &=-8\mathbf{i}+(-4)+6+(-3\mathbf{i})=2-11\mathbf{i}. \end{aligned}$
$\begin{aligned} \frac{-2+5\mathbf{i}}{\mathbf{i}}&=\frac{(-2+5\mathbf{i})\mathbf{i}}{\mathbf{i}\cdot \mathbf{i}}=\frac{-2\mathbf{i}+5\mathbf{i}^2}{\mathbf{i}^2}\\ &=\frac{-2\mathbf{i}-5}{-1}=5+2\mathbf{i}. \end{aligned}$
$\begin{aligned} \frac{1}{1-2\mathbf{i}}&=\frac{(1+2\mathbf{i})}{(1-2\mathbf{i})(1+2\mathbf{i})}=\frac{1+2\mathbf{i}}{1-(2\mathbf{i})^2}\\ &=\frac{1+2\mathbf{i}}{5}=\frac{1}{5}+\frac{2}{5}\mathbf{i}. \end{aligned}$
$\mathbf{i}^{2018}=\mathbf{i}^{4\cdot 504+2}=(\mathbf{i}^4)^{504}\cdot \mathbf{i}^2=-1.$

Example: Evaluate the express $z^2+\frac{z-1}{z+1}$ for $z=1+\mathbf{i}$ . Write your answer in the form $a+b\mathbf{i}$ .

Solution.

$\begin{aligned} f(1+\mathbf{i})=&(1+\mathbf{i})^2+\frac{\mathbf{i}}{2+\mathbf{i}}\\ =&1+2\mathbf{i}+\mathbf{i}^2+\frac{\mathbf{i}(2-\mathbf{i})}{4-\mathbf{i}^2}\\ =&2\mathbf{i}+\frac{1+2\mathbf{i}}{5}\\ =&\frac15+\frac{12}{5}\mathbf{i}. \end{aligned}$

Practice

Exercise: Evaluate the square root. If the square root is not a real number, state so.

$-\sqrt{\frac{4}{25}}$
$\sqrt{49}-\sqrt{9}$
$-\sqrt{-1}$

Exercise: Simplify the radical expression.

$\sqrt{(-7x^2)^2}$
$\sqrt{(x+2)^2}$
$\sqrt{25x^2y^6}$

Exercise: Simplify the radical expression.

$\sqrt[3]{-27x^3}$
$\sqrt[4]{16x^8}$
$\sqrt[5]{(2x-1)^5}$

Exercise: Simplify the radical expression. Assume all variables are positive.

$\sqrt{50}$
$\sqrt[3]{-8x^2y^3}$
$\sqrt[5]{32x^{12}y^2z^8}$

Exercise: Write the radical expression with rational exponents.

$\sqrt[3]{(2x)^5}$
$(\sqrt[5]{3xy})^7$
$\sqrt[4]{(x^2+3)^3}$

Exercise: Write in radical notation and simplify.

$4^{\frac32}$
$-81^{\frac 34}$
$\left(\frac{27}{8}\right)^{-\frac{2}{3}}$

Exercise: Simplify the expression. Write with radical notations. Assume all variables represent nonnegative numbers.

$\dfrac{12x^{\frac12}}{4x^{\frac23}}$
$(x^{-\frac35}y^{\frac12})^{\frac13}$
$\left(\frac{x^{\frac12}}{x^{-\frac13}}\right)^4$

Exercise: Simplify the expression. Write in radical notation. Assume $x$ is nonnegative.

$\dfrac{\sqrt{x}}{\sqrt[3]{x}}$
$\sqrt{\sqrt[3]{x}}$
$\sqrt{x}\sqrt[3]{x}$

Exercise: Simplify the expression. Write in radical notation. Assume $x$ is nonnegative.

$\sqrt[5]{32x^{\frac13}}$
$\left(\dfrac{\sqrt[4]{9x}}{3}\right)^{-2}$
$\sqrt{\frac{1}{\sqrt[3]{x^{-2}}}}$

Exercise: Simplify the expression. Write in radical notation. Assume all variables are nonnegative.

$\left(\dfrac{8a^{-\frac{5}{2}}b}{a^{\frac12}b^{-5}}\right)^{-\frac23}$
$\left(\dfrac{y^{-\frac{1}{3}}}{\sqrt[3]{x^{2}}}\right)^{-3}$
$\sqrt[3]{(-x)^{-2}}\sqrt{x^3}$

Exercise: Multiply and simplify.

$\sqrt[3]{4}\sqrt[3]{5}$
$\sqrt{|x+7|}\sqrt{|x-7|}$
$\sqrt[3]{(x-y)^{\frac52}}\sqrt[3]{(x-y)^{\frac72}}$

Exercise: Simplify the radical expression. Assume all variables are positive.

$\sqrt{20xy}\cdot\sqrt{4xy^2}$
$\sqrt[3]{16}\cdot5\sqrt[3]{2}$
$\sqrt[5]{8x^4y^3z^3}\cdot\sqrt[5]{8xy^4z^8}$

Exercise: Divide. Assume all variables are positive. Answers must be simplified.

$\sqrt{\dfrac{9x^3}{y^8}}$
$\sqrt[3]{\dfrac{32x^4}{x}}$
$\dfrac{\sqrt{40x^5}}{\sqrt{2x}}$
$\dfrac{\sqrt[3]{24a^6b^4}}{\sqrt[3]{3b}}$

Exercise: Add or subtract, and simplify. Assume all variables are positive.

$5\sqrt6+3\sqrt6$
$4\sqrt{20}-2\sqrt5$
$3\sqrt{32x^2}+5x\sqrt{8}$

Exercise: Add or subtract, and simplify. Assume all variables are positive

$7\sqrt{4x^2}+2\sqrt{25x}-\sqrt{16x}$
$5\sqrt[3]{x^2y}+\sqrt[3]{27x^5y^4}$
$3\sqrt{9y^3}-3y\sqrt{16y}+\sqrt{25y^3}$

Exercise: Multiply and simplify. Assume all variables are positive.

$\sqrt2(3\sqrt3-2\sqrt2)$
$(\sqrt5+\sqrt7)(3\sqrt5-2\sqrt7)$
$(\sqrt3+\sqrt2)^2$

Exercise: Multiply and simplify. Assume all variables are positive.

$(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)$
$(\sqrt{x+1}-1)(\sqrt{x+1}+1)$
$(2\sqrt[3]x+6)(\sqrt[3]x+1)$

Exercise: Simplify the radical expression and rationalize the denominator. Assume all variables are positive.

$\sqrt[3]{\dfrac2{25}}$
$\sqrt{\dfrac{2x}{7y}}$
$\dfrac{\sqrt[3]{x}}{\sqrt[3]{3y^2}}$
$\dfrac{3x}{\sqrt[4]{x^3y}}$

Exercise: Simplify the radical expression and rationalize the denominator. Assume all variables are positive.

$\dfrac{6\sqrt3}{\sqrt3-1}$
$\dfrac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
$\dfrac{3+\sqrt2}{2+\sqrt3}$
$\dfrac{2\sqrt{x}}{\sqrt x- \sqrt y}$

Exercise: Simplify and rationalize the denominator. Assume all variables are positive.

$\dfrac{\sqrt{x}}{\sqrt x-1}+\dfrac{1}{\sqrt{x}+1}$
$\dfrac{\sqrt{x}+1}{\sqrt x}-\dfrac{1}{\sqrt{x}-1}$

Exercise: Add, subtract, multiply complex numbers and write your answer in the form $a+b\mathbf{i}$ .

$\sqrt{-2}\cdot\sqrt{-3}$
$\sqrt{2}\cdot\sqrt{-8}$
$(5-2\mathbf{i})+(3+3\mathbf{i})$
$(2+6\mathbf{i})-(12-4\mathbf{i})$

Exercise: Add, subtract, multiply complex numbers and write your answer in the form $a+b\mathbf{i}$ .

$(3+\mathbf{i})(4+5\mathbf{i})$
$(7-2\mathbf{i})(-3+6\mathbf{i})$
$(3-x\sqrt{-1})(3+x\sqrt{-1})$
$(2+3\mathbf{i})^2$

Exercise: Divide the complex number and write your answer in the form $a+b\mathbf{i}$ .

$\dfrac{2\mathbf{i}}{1+\mathbf{i}}$
$\dfrac{5-2\mathbf{i}}{3+2\mathbf{i}}$
$\dfrac{2+3\mathbf{i}}{3-\mathbf{i}}$
$\dfrac{4+7\mathbf{i}}{-3\mathbf{i}}$

Exercise: Simplify the expression.

$(-\mathbf{i})^{8}$
$\mathbf{i}^{15}$
$\mathbf{i}^{2017}$
$\dfrac1{\mathbf{i}^{2018}}$

Exercise: Evaluate the function polynomial $2x^2-3x+5$ for $x=1-\mathbf{i}$ . Write your answer in the form $a+b\mathbf{i}$ .

Exercise: Evaluate the polynomial $\mathbf{i} x^2-x+\frac{2}{x-1}$ for $x=\mathbf{i}-1$ . Write your answer in the form $a+b\mathbf{i}$ .

Part 2: Equations and Applications

Solving Polynomial Equations by Factoring

Handshaking Problem

In meeting room, a group of people all shook hands with one another. In total, 15 handshakes occurred. Do you know how many people in the group?

Properties of Equations

An equation is an statement that asserts an equality containing unknown variables. For example, $2x+3=1$ is an equation of the unknown variable $x$ .
Equations often contain variables other than the unknowns. Those variables, which are assumed to be known, are usually called constants, coefficients or parameters. For example, in the linear equation $ax+b=c$ of (the unknown) $x$ , the variables $a$ , $b$ and $c$ are referred as known coefficients or constants.
An identity is an equation that is true for all possible values of the variable(s) it contains. For example, $x^2-y^2=(x+y)(x-y)$ is an identity.
Solving an equation consists of determining values of the variables that make the equality true. Two equations are said to be equivalent if and only if they have the same solution set, that is, a solution of one equation is also a solution of the other equation. For example $2x-6=0$ and $x-1=2$ are equivalent.

When solving an equation, the following operations can be used to transform an equation to an equivalent one:

Adding or subtracting the same quantity to both sides of an equation. For example, $x-1=2$ is equivalent to $x-1+1=2+1$ .
Multiplying or dividing both sides of an equation by a non-zero quantity. For example $2x=4$ is equivalent to $\frac{2x}{2}=\frac{4}{2}$ .
Applying an identity to transform one side of the equation. For example, $x^2-1=0$ is equivalent to $(x-1)(x+1)=0$ , where the identity $x^2-1=(x-1)(x+1)$ was applied.

In general, one may apply any choice of a function to both sides of the equation to make a transformation. The resulting equation still has the solutions of the original equations as it solutions. However, the resulting equation may also have some extra solutions which are called extraneous solutions. For example, taking squares of both sides of the equation $x=1$ produces the equation $x^2=1$ . The new equation $x^2=1$ has two solutions $x=-1$ and $x=1$ , but the original equation $x=1$ only has one solution. The solution $x=-1$ of the equation $x^2=1$ is an extraneous solution of the equation $x=1$ .

Quadratic Equations

A polynomial equation is an equation that can be written in the form $a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{2}x^{2}+a_{1}x+a_{0}=0,$ where $n$ is a positive integer and $a_n\ne 0$ .

A polynomial equation is called a quadratic equation if $n=2$ . For example, $2x^2+5x-3=0$ . We often write a quadratic equation in its standard form $a x^2+bx+c=0,$ where $a$ , $b$ and $c$ are numbers, and $a\neq 0$ .

When solving linear equations, arithmetic operations are enough. In general, one may need to use identity or functional operation. Factoring is one of those frequently used identity operation. Indeed, to solve a problem, a general strategy is to to reduce the original problem to easier problems. Using factoring and the zero product property: $A\cdot B=0 \quad ~~\text{if and only if}~~ \quad A=0 ~~\text{or}~~ B=0,$ one can transform a polynomial equation into smaller degree polynomial equations. In particular, if $ax^2+bx+c=(mx-p)(nx-q)$ , then a solution of the quadratic equation $ax^2+bx+c=0$ is a solution of either $mx-p=0$ or $nx-q=0$ .

Example: Solve the equation $2x^2+5x=3.$

Solution.

Rewrite the equation into “Expression=0” form and factor. $\begin{aligned} 2x^2+5x&=3\\ 2x^2+5x-3&=0\\ (2x-1)(x+3)&=0 \end{aligned}$
Apply the zero product property. $2x-1=0\quad\text{or}\quad x+3=0.$
Solve each equation. $\begin{aligned} 2x-1 =0 & \qquad\text{or} & x+3 =0 \\ 2x =1 & & x =-3 \\ x =\frac12 & \qquad\text{or} & x =-3 \end{aligned}$
The solution set is $\{-3, \frac12\}$ .

Example: Solve the equation $(x-2)(x+3)=-4.$

Solution.

Rewrite the equation into “Expression=0” form and factor. $\begin{aligned} (x-2)(x+3)&=-4\\ x^2+x-6&=-4\\ x^2+x-2&=0\\ (x-1)(x+2)&=0 \end{aligned}$
Apply the zero product property. $x-1=0\quad\text{or}\quad x+2=0.$
Solve each equation. $\begin{aligned} x-1 =0 & \qquad\text{or} & x+2 =0 \\ x =1 & \qquad\text{or} & x =-2 \end{aligned}$
The solution set is $\{-2, 1\}$ .

Example: A rectangular garden is surrounded by a path of uniform width. If the dimension of the garden is $10$ meters by $16$ meters and the total area is 216 square meters, determine the width of the path.

rectangular garden with uniform width pass

Solution.

Suppose that the width of the frame is $x$ meters. Translate given information into expressions in $x$ and build an equation.
Total Width: $2x+10$ Total Length: $2x+16$
Width $\times$ Length=Total Area: $(2x+10)(2x+16)=216.$
Solve the equation. $\begin{aligned} (2x+10)(2x+16)&=216\\ 4x^2+52x+160&=216\\ 4x^2+52x-56&=0\\ x^2+13x-14&=0\\ (x+14)(x-1)&=0 \end{aligned}$ $x = -14\quad \text{or}\quad x = 1$
So the width of the path is $1$ meter.

Understand the Problem
When solving a word problem, you may first outline what’s known and what’s unknown, and restate the problem using algebraic expressions. Once you reformulated the problem algebraically, you may solve it using your mathematical knowledge.

Practice

Exercise: Solve the equation by factoring.

$x^2-3x+2=0$
$2x^2-3x=5$
$(x-1)(x+3)=5$
$\frac13(2-x)(x+5)=4$

Exercise: Find all real solutions of the equation by factoring.

$4(x-2)^2-9=0$
$2x^3-18x=0$
$3x^4-2x^2=1$
$x^3-3x^2-4x+12=0$

Exercise: A paint measuring $3$ inches by $4$ inches is surrounded by a frame of uniform width. If the combined area of the paint and the frame is $30$ square inches, determine the width of the frame.

paint with uniform width path

Exercise: A rectangle whose length is $2$ meters longer than its width has an area $8$ square meters. Find the width and the length of the rectangle.

Exercise: The product of two consecutive negative odd numbers is $35$ . Find the numbers.

Exercise: In a right triangle, the long leg is 2 inches more than double of the short leg. The hypotenuse of the triangle is 1 inch longer than the long leg. Find the length of the shortest side.

Exercise: A ball is thrown upwards from a rooftop. It will reach a maximum vertical height and then fall back to the ground. The height $h(t)$ of the ball from the ground after time $t$ seconds is $h(t)=-16t^2 + 48t + 160$ feet. How long it will take the ball to hit the ground?

Exercise: A toy factory estimates that the demand of a particular toy is $300 -x$ units each week if the price is $ $x$ dollars per unit. Each week there is a fixed cost $40,000 to produce the demanded toys. The weekly revenue is a function of the price given by $R(x)=x(30-x)$

Find the function that models the weekly revenue, $R$ , received when the selling price is $ $x$ per unit.
What the price range so the the revenue is nonnegative.

Quadratic Formula

Estimate a Square Root

Can you estimate the irrational numbers $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ and $\sqrt{7}$ without using a calculator?

Can you estimate the square root $\sqrt{m^2+n}$ , where $m$ and $n$ are positive integers?

Completing the Square

The square root property:
Suppose that $X^2=d$ . Then $X=\sqrt{d}$ or $X=-\sqrt{d}$ , or simply $X=\pm\sqrt{d}$ .

The square root property provides another method to solve a quadratic equation, completing the square. This method is based on the following observations: $x^2+{\mathbf{b}}x+{\mathbf{\left(\frac b2\right)^2}}=\left(x+{\mathbf{\frac b2}}\right)^2,$ and more generally, let $f(x)=ax^2+bx+c$ , and $h=-\frac{b}{2a}$ , then $ax^2+bx+c=a(x-h)^2+f(h)=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}.$

The procedure to rewrite a trinomial as the sum of a perfect square and a constant is called completing the square.

Example: Solve the equation $x^2+2x-1=0$ .

Solution.

Isolate the constant. $x^2+2x=1$
With $b=2$ , add $\left(\frac 22\right)^2$ to both sides to complete a square for the binomial $x^2+bx$ . $\begin{aligned} x^2+2x+\left(\frac22\right)^2&=1+\left(\frac22\right)^2\\\ \left(x+\frac22\right)^2&=1+1\\ (x+1)^2&=2\\ \end{aligned}$
Solve the resulting equation using the square root property. $\begin{aligned} x+1 =\sqrt2 & \qquad\text{or} & x+1 =-\sqrt2 \\ x =-1+\sqrt2 & \qquad\text{or} & x =-1-\sqrt2 \end{aligned}$

Note that the solution can also be written as $x=-1\pm\sqrt2$ .

Example: Solve the equation $-2x^2+8x-9=0$ .

Solution.

Isolate the constant. $-2x^2+8x=9$
Divide by $-2$ to rewrite the equation in $x^2+Bx=C$ form $x^2-4x=-\frac{9}{2}$
With $b=-4$ , add $\left(\frac{-4}{2}\right)^2=4$ to both sides to complete the square for the binomial $x^2-4x$ . $\begin{aligned} x^2-4x+4&=-\frac{9}{2}+4\\\ (x-2)^2&=-\frac{1}{2} \end{aligned}$
Solve the resulting equation and simplify. $\begin{aligned} x-2 =\frac{\mathbf{i}}{\sqrt{2}} & \qquad\text{or} & x-2 =-\frac{\mathbf{i}}{\sqrt{2}} \\ x =2+\frac{\sqrt{2}}{2}\mathbf{i} & \qquad\text{or} & x =2-\frac{\sqrt{2}}{2}\mathbf{i} \end{aligned}$

Another way to complete the square is to use the formula $ax^2+bx+c=a(x-h)^2+f(h)$ , where $f(h)=ah^2+bh+c$ is the value of the polynomial $ax^2+bx+c$ at $x=h$ .

The Quadratic Formula

Using the method of completing the square, we obtain the follow quadratic formula for the quadratic equation $ax^2+bx+c=0$ with $a\neq 0$ : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$

The quantity $b^2-4ac$ is called the discriminant of the quadratic equation.

If $b^2-4ac>0$ , the equation has two real solutions.
If $b^2-4ac=0$ , the equation has one real solution.
If $b^2-4ac<0$ , the equation has two imaginary solutions (no real solutions).

Example: Determine the type and the number of solutions of the equation $(x-1)(x+2)=-3$ .

Solution.

Rewrite the equation in the form $ax^2+bx+c=0$ . $\begin{aligned} (x-1)(x+2)&=-3\\ x^2+x+1&=0 \end{aligned}$
Find the values of $a$ , $b$ and $c$ . $a=1, b=1 ~\text{and}~ c=1.$
Find the discriminant $b^2-4ac$ . $b^2-4ac=1^2-4\cdot 1\cdot 1=-3.$

The equation has two imaginary solutions.

Example: Solve the equation $2x^2-4x+7=0$ .

Solution.

Find the values of $a$ , $b$ and $c$ . $a=2, b=-4 ~\text{and}~ c=7.$
Find the discriminant $b^2-4ac$ . $b^2-4ac=(-4)^2-4\cdot 2\cdot 7=16-56=-40.$
Apply the quadratic formula and simplify. $\begin{aligned} x=&\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ =&\dfrac{-(-4)\pm\sqrt{-40}}{2\cdot 2}\\ =&\frac{4\pm 2\sqrt{10} \mathbf{i}}{4}\\ =&1\pm\frac{\sqrt{10}}{2}\mathbf{i}. \end{aligned}$

Example: Find the base and the height of a triangle whose base is three inches more than twice its height and whose area is $5$ square inches. Round your answer to the nearest tenth of an inch.

Solution.

We may suppose the height is $x$ inches. The base can be expressed as $2x+3$ inches.
By the area formula for a triangle, we have an equation. $\frac12 x(2x+3)=5.$
Rewrite the equation in $ax^2+bx+c=0$ form. $\begin{aligned} x(2x+3)&=10\\ 2x^2+3x-10&=0. \end{aligned}$
By the quadratic formula, we have $x=\frac{-3\pm\sqrt{3^2-4\cdot 2\cdot (-10)}}{2\cdot 2}=\frac{-3\pm \sqrt{89}}{4}.$

Since $x$ can not be negative, $x=\frac{-3+\sqrt{89}}{4}\approx 1.6$ and $2x+3\approx 6.2$ . The height and base of the triangle are approximately $1.6$ inches and 6.2 inches respectively.

Practice

Exercise: Solve the quadratic equation by the square root property.

$2x^2-6=0$
$(x-3)^2=10$
$4(x+1)^2+25=0$

Exercise: Solve the quadratic equation by completing the square.

$x^2+x-1=0$
$x^2+8x+12=0$
$3x^2+6x-1=0$

Exercise: Determine the number and the type of solutions of the given equation.

$x^2+8x+3=0$
$3x^2-2x+4=0$
$2x^2-4x+2=0$

Exercise: Solve using the quadratic formula.

$x^2+3x-7=0$
$2x^2=-4x+5$
$2x^2=x-3$

Exercise: Solve using the quadratic formula.

$(x-1)(x+2)=3$
$2x^2-x=(x+2)(x-2)$
$\frac12 x^2+x= \frac13$

Exercise: A triangle whose area is $7.5$ square meters has a base that is one meter less than triple the height. Find the length of its base and height. Round to the nearest hundredth of a meter.

Exercise: A rectangular garden whose length is $2$ feet longer than its width has an area 66 square feet. Find the dimensions of the garden, rounded to the nearest hundredth of a foot.

Rational Equations

A Problem of the Father of Algebra

The father of algebra, Muḥammad ibn Musa al-Khwarizmi, in his book “Algebra”, answered the following problem:

“… I have divided ten into two parts; and have divided the first by the second, and the second by the first, and the sum of the quotient is two and one-sixth;…”

Do you know what are those two parts?

Solving Rational Equations

A rational equation is an equation that contains a rational expression. One way to solve rational equations is to clear all denominators by multiplying the LCD to both sides. Note that because there are values such that the LCD has the value zero. Clearing denominator in general is not an equivalent transformation, rather a functional transformation. So don’t forget to check possible extraneous solutions.

Example: Solve $\dfrac{5}{x^2-9}=\dfrac{3}{x-3}-\dfrac{2}{x+3}.$

Solution.

Find the LCD. Since $x^2-9=(x+3)(x-3)$ , the LCD is $(x+3)(x-3)$ .
Clear denominators. Multiply each rational expression in both sides by $(x+3)(x-3)$ and simplify: $\begin{aligned} (x+3)(x-3)\cdot\dfrac{5}{x^2-9}&=(x+3)(x-3)\cdot\dfrac{3}{x-3}-(x+3)(x-3)\cdot\dfrac{2}{x+3}\\ 5&=3(x+3)-2(x-3) \end{aligned}$
Solve the resulting equation. $\begin{aligned} 5&=3(x+3)-2(x-3) 5&=x+15\\ -10&=x \end{aligned}$
Check for any extraneous solution by plugging the solution into the LCD to see if it is zero. If it is zero, then the solution is extraneous. $(-10+3)(-10-3)\neq 0$ So $x=-10$ is a valid solution of the original equation.

Reduction With Auxiliary Conditions
Clearing denominator uses the strategy “reduction with auxiliary conditions”. The auxiliary condition used when clearing the denominators is that the LCD is non-zero. Generally, if you don’t know how to solve a problem under the given condition, you may try solve the problem by adding extra conditions first and then try to eliminate the extra conditions and/or their consequences.

Literal Equations

A literal equation is an equation involving two or more variables. When solving a literal equation for one variable, other variables can be viewed as constants.

Example: Solve for $x$ from the equation $\frac{1}{x}+\frac{1}{y}=\dfrac{1}{z}.$

Solution.

The LCD is $xyz$ .
Clear denominators. $\begin{aligned} xyz\cdot\frac{1}{x}+xyz\cdot\frac{1}{y}&=xyz\cdot\dfrac{1}{z}\\ yz+xz&=xy\\ \end{aligned}$
Solve the resulting equation. $\begin{aligned} yz+xz&=xy\\ yz&=xy-xz\\ yz&=x(y-z)\\ \dfrac{yz}{y-z}&=x \quad \text{if} y\neq z \end{aligned}$
The solution is $x=\dfrac{yz}{y-z}$ if $y\neq z$ . If $y=z$ , the equation has no solution.

Another way to solve a rational equation is to rewrite and simplify the equation into the form $\dfrac{\text{A}}{B}=0$ where $\dfrac{A}{B}$ is a reduced fraction. Then the rational equation is equivalent to the equation $A=0$ .

Practice

Exercise: Solve.

$\dfrac1{x+1}+\dfrac1{x-1}=\dfrac4{x^2-1}$
$\dfrac{30}{x^2-25}=\dfrac3{x+5}+\dfrac2{x-5}$

Exercise: Solve.

$\dfrac{2x-1}{x^2+2x-8}=\dfrac1{x-2}-\dfrac{2}{x+4}$
$\dfrac{3x}{x-5}=\dfrac{2x}{x+1}-\dfrac{42}{x^2-4x-5}$

Exercise: Solve a variable from a formula.

Solve for $f$ from $\dfrac1p+\dfrac1q=\dfrac1f$ .
Solve for $x$ from $A=\dfrac{f+cx}{x}$ .

Exercise: Solve for $x$ from the equation.

$2(x+1)^{-1}+x^{-1}=2$ .
$\displaystyle\frac{a^2x +2a}{x^{-1}}=-1$ .

Exercise: David can row 3 miles per hour in still water. It takes him 90 minutes to row 2 miles upstream and then back. How fast is the current?

Exercise: The size of a A0 paper is defined to have an area of 1 square meter with the longer dimension $\sqrt[4]{2}$ meters. Successive paper sizes in the series A1, A2, A3, and so forth, are defined by halving the preceding paper size across the larger dimension. Can you find the dimension of a A4 paper?

Radical Equations

Design a Pendulum clock

A pendulum clock is a clock that uses a pendulum, a swinging weight, as its timekeeping element. Galileo Galilei discovered in early 17th century the relation between the length $L$ of a pendulum and the period $T$ of th pendulum. For a pendulum clock, the relations is approximately determined by the following rule of thumb formula: $T\approx 2\sqrt{L}$ given that $L$ and $T$ are measured in meters and seconds respectively. If the period of a pendulum clock is 2 seconds, how long should be the pendulum?

Solving Radical Equations by Taking a Power

The idea to solve a radical equation $\sqrt[n]{X}=a$ is to first take $n$ -th power of both sides to get rid of the radical sign, that is $X=a^n$ and then solve the resulting equation.

Solve by Reduction
The goal to solve a single variable equation is to isolate the variable. When an equation involves radical expressions, you can not isolate the variable arithmetically without eliminating the radical sign unless the radicand is a perfect power. To remove a radical sign, you make take a power. However, you’d better to isolate it first. Because simply taking powers of both sides may create new radical expressions.

Example: Solve the equation $x-\sqrt{x+1}=1.$

Solution.

Arrange terms so that one radical is isolated on one side of the equation. $x-1=\sqrt{x+1}$
Square both sides to eliminate the square root. $(x-1)^2=x+1$
Solve the resulting equation. $\begin{aligned} x^2-2x+1&=x+1\\ x^2-3x&=0\\ x(x-3)&=0 \end{aligned}$ $\begin{aligned} x =0 & \qquad \text{or} & x-3 =0 \\ x =0 & \qquad \text{or} & x =3 \end{aligned}$
Check all proposed solutions.
Plug $x=0$ into the original equation, we see that the left hand side is $0-\sqrt{0+1}=0-\sqrt{1}=0-1=-1$ which is not equal to the right hand side. So $x=0$ cannot be a solution.
Plug $x=3$ into the original equation, we see that the left hand side is $3-\sqrt{3+1}=3-\sqrt{4}=3-2=1$ . So $x=3$ is a solution.

Example: Solve the equation $\sqrt{x-1}-\sqrt{x-6}=1.$

Solution.

Isolated one radical. $\sqrt{x-1}=\sqrt{x-6}+1\\$
Square both sides to remove radical sign and then isolate the remaining radical. $\begin{aligned} x-1&=(x-6)+2\sqrt{x-6}+1\\ x-1&=x-5+2\sqrt{x-6}\\ 4&=2\sqrt{x-6}\\ 2&=\sqrt{x-6}. \end{aligned}$
Square both sides to remove the radical sign and then solve. $\begin{aligned} \sqrt{x-6}&=2\\ x-6&=4\\ x&=10. \end{aligned}$ Since $10-1>0$ and $10-6>0$ , $x=10$ is a valid solution. Indeed, $\sqrt{10-1}-\sqrt{10-6}=\sqrt{9}-\sqrt{4}=3-2=1.$

Example: Solve the equation $-2\sqrt[3]{x-4}=6.$

Solution.

Isolated the radical. $\sqrt[3]{x-4}=-3$
Cube both sides to eliminate the cube root and then solve the resulting equation. $\begin{aligned} x-4&=(-3)^3\\ x-4&=-27\\ x&=-23 \end{aligned}$ The solution is $x=-23$ .

Equations Involving Rational Exponents

Equation involving rational exponents may be solved similarly. However, one should be careful with meaning of the expression $\left(X^{\frac mn}\right)^{\frac nm}$ . When $m$ is even and $n$ is odd, $\left(X^{\frac mn}\right)^{\frac nm}=|X|$ . Otherwise, $\left(X^{\frac mn}\right)^{\frac nm}=X$ .

Example: Solve the equation $(x+2)^{\frac12}-(x-3)^{\frac12}=1$ .

Solution.

Since there are more than one term involving rational exponents, to solve the equation, we isolate one term and taking power and so on so forth. $\begin{aligned} (x+2)^{\frac12}-(x-3)^{\frac12}=&1\\ (x+2)^{\frac12}=& (x-3)^{\frac12}+1\\ x+2 =& \left((x-3)^{\frac12}+1\right)^2\\ x+2 =& (x-3)+2(x-3)^{\frac12}+1\\ 2(x-3)^{\frac12}=&4\\ (x-3)^{\frac12}=&2\\ x-3=&4\\ x=&7 \end{aligned}$ Check: $(7+2)^{\frac12}-(7-3)^{\frac12}=\sqrt{9}-\sqrt{4}=3-2=1.$

So the equation has one solution $x=7$ .

Example: Solve the equation $(x-1)^{\frac{2}{3}}=4$ .

Solution.

There are different way to solve this equation. One may is to take rational powers of both sides and solve the resulting equation. $\begin{aligned} (x-1)^{\frac{2}{3}}=&4\\ \left((x-1)^{\frac{2}{3}}\right)^{\frac32}=&4^{\frac32}\\ |x-1|=&8\\ x-1=8 \qquad\text{or}&\qquad x-1=-8\\ x=9 \qquad\text{or}&\qquad x=-7 \end{aligned}$ Check: $(9-1)^{\frac23}=8^{\frac23}=(8^{\frac13})^2=2^2=4;$ $(-7-1)^{\frac23}=(-8)^{\frac23}=((-8)^{\frac13})^2=(-2)^2=4.$

So the equation has two solutions $x=9$ and $x=-7$ .

Learn from Mistakes

Example: Can you find the mistakes made in the solution and fix it?

Solve the radical equation. $\sqrt{x-1}+2=x$

Solution (incorrect): $\begin{aligned} \sqrt{x-2}+2&=x\\ (\sqrt{x-2})^2+2^2&=x^2\\ x-2+4&=x^2\\ x+2&=x^2\\ x^2-x-2&=0\\ (x-2)(x+1)&=0\\ x-2=0 \qquad\text{or}& \qquad x+1=0\\ x=2 \qquad\text{or}&\qquad x=-1 \end{aligned}$ Answer: the equation has two solutions $x=2$ and $x=-1$ .

Solution.

When squaring one side of the equation, the other side as a whole should be squared. The mistake occurred at the squaring step. The right way to solve the equation is as follows. $\begin{aligned} \sqrt{x-2}+2&=x\\ \sqrt{x-2}&=x-2\\ x-2&=(x-2)^2\\ (x-2)^2-(x-2)&=0\\ (x-2)(x-2-1)&=0\\ (x-2)(x-3)&=0\\ x=2 \qquad\text{or}&\qquad x=3 \end{aligned}$ Because squaring is not an equivalent transformation in general, the solutions of the resulting equations must be checked. When $x=2$ , the left side of the original equation is $\sqrt{2-2}+2=0+2=2$ . When $x=3$ , the left side is $\sqrt{3-2}+2=1+2=3$ . So both $x=2$ and $x=3$ are solutions of the function $\sqrt{x-1}+2=x$ .