text:
Overview
Factoring
Expressions
Equations
Videos
Solved Examples

# Expressions and Equations

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - <NAME>

## Welcome to Our Site

I greet you this day,
First: read the notes. Second: view the videos. Third: solve the questions/solved examples. Fourth: check your solutions with my thoroughly-explained solved examples. Fifth: check your answers with the calculators as applicable.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.
If you are my student, please do not contact me here. Contact me via the school's system.
Thank you for visiting!!!

<NAME> (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

## Objectives

#### Objectives

Students will:

(1.) Define expressions.

(2.) Define equations.

(3.) Translate word problems from English to Math.

(4.) Evaluate algebraic expressions.

(5.) Simplify algebraic expressions.

(6.) Solve linear equations.

(7.) Check the solution of linear equations.

(8.) Factor algebraic expressions.

(9.) Determine the nature of the roots of a quadratic equation.

(11.) Check the solutions of quadratic equations.

(12.) Solve literal equations.

(14.) Check the solutions of quadratic equations.

(15.) Solve absolute value equations.

(16.) Check the solutions of absolute value equations.

(17.) Solve rational equations.

(18.) Check the solutions of rational equations.

(20.) Check the solutions of radical equations.

#### Skills Measured/Acquired

(1.) Use of prior knowledge

(2.) Critical Thinking

(3.) Interdisciplinary connections/applications

(4.) Technology

(5.) Active participation through direct questioning

(6.) Research

## Vocabulary Words

subtracted from
multiplied by
divided by

Bring it to English: vary, constant, express, expression, equate, equal, equation, equality, equanimity, equity, addendum

Bring it to Physics: Prefixes: mono or uni = $1$, di or bi = $2$, tri = $3$, tetra or quad = $4$, penta = $5$, hexa = $6$, hepta = $7$, octa = $8$, nona = $9$, deca = $10$, hendeca = $11$, dodeca = $12$, etc.

Ask students to give examples of real-world scenarios where they have used any of the prefixes.
These include: unity, bilateral, triangle, tetrahedral, quadrilaterals (you can ask students to list examples of quadrilaterals - Geometry!), etc.

Bring it to Math: arithmetic, arithmetic operators, sum, difference, product, quotient, augend, addend, minuend, subtrahend, multiplier, multiplicand, factor, dividend, divisor, positive, negative, nonpositive, nonnegative, constant, number, variable, term, add, subtract, multiply, divide, expression, equation, equal, equality, inequality, linear, quadratic, cubic, quartic, exponent, index, power, degree, order, pentic, hexic, heptic, octic, absolute value, rational, fractional, radical, root, solution, zero, perfect square, prime trinomial etc.

Generally, "linear" implies that the exponent of the variable is $1$
"quadratic" implies that the exponent of the variable is $2$
"cubic" implies that the exponent of the variable is $3$
"quartic" implies that the exponent of the variable is $4$

## Definitions

The basic arithmetic operators are the addition symbol, $+$, the subtraction symbol, $-$, the multiplication symbol, $*$, and the division symbol, $\div$

Augend is the term that is being added to. It is the first term.

Addend is the term that is added. It is the second term.

Sum is the result of the addition.

$$3 + 7 = 10$$ $$3 = augend$$ $$7 = addend$$ $$10 = sum$$

Minuend is the term that is being subtracted from. It is the first term.

Subtrahend is the term that is subtracted. It is the second term.

Difference is the result of the subtraction.

$$3 - 7 = -4$$ $$3 = minuend$$ $$7 = subtrahend$$ $$-4 = difference$$

Multiplier is the term that is multiplied by. It is the first term.

Multiplicand is the term that is multiplied. It is the second term.

Product is the result of the multiplication.

$$3 * 10 = 30$$ $$3 = multiplier$$ $$10 = multiplicand$$ $$30 = product$$

Dividend is the term that is being divided. It is the numerator.

Divisor is the term that is dividing. It is the denominator.

Quotient is the result of the division.

Remainder is the term remaining after the division.

$$12 \div 7 = 1 \:R\: 5$$ $$12 = dividend$$ $$10 = divisor$$ $$1 = quotient$$ $$5 = remainder$$

A constant is something that does not change. In mathematics, numbers are usually the constants.

A variable is something that varies (changes). In Mathematics, alphabets are usually the variables.

A mathematical expression is a combination of variables and/or constants using arithmetic operators.

A mathematical equation is an equality of two terms - the term or expression on the LHS (Left Hand Side) and the term or expression on the RHS (Right Hand Side).
This implies that we should always check the solution of any equation that we solve to make sure the LHS is equal to the RHS.
Whenever we solve for the variable in "any" equation, how do we know we are correct? CHECK!

The solution of an equation is the value of the variable which when substituted in the equation ensures that the LHS is equal to the RHS.

The solution of an equation is also known as the root of the equation or the zero of the function.

The solutions of an equation are the values of the variables which when substituted in the equation ensures that the LHS is equal to the RHS.

The solutions of an equation are also referred to as the roots of the equation or the zeros of the function.

An extraneous root of an equation is a solution that is not a solution of the main equation but a solution of the modified equation.

A linear expression is an expression in which the highest exponent of the independent variable in the expression is $1$

A quadratic expression is an expression in which the highest exponent of the independent variable in the expression is $2$

A cubic expression is an expression in which the highest exponent of the independent variable in the expression is $3$

A quartic expression is an expression in which the highest exponent of the independent variable in the expression is $4$

A linear equation is an equation in which the highest exponent of the independent variable in the equation is $1$

A quadratic equation is an equation in which the highest exponent of the independent variable in the equation is $2$

A cubic equation is an equation in which the highest exponent of the independent variable in the equation is $3$

A quartic equation is an equation in which the highest exponent of the independent variable in the equation is $4$

A rational equation is an equation containing a rational function.
A rational equation is an equation containing a ratio of two polynomials.
A rational equation is also referred to as a fractional equation.

A radical equation is an equation in which at least one of it's terms has a radical.

An absolute value equation is an equation in which at least one of it's terms has an absolute value.

The absolute value of a term is the magnitude or modulus of that term regardless of sign.

A trinomial is a polynomial of three terms.

A prime trinomial is a trinomial that cannot be factored.

A quadratic trinomial is a trinomial of degree $2$.

An equation is said to be in standard form if the equation is written in terms of the descending order of exponents.

The Zero Product Property states that if the product of two or more terms is $0$; then, at least one of those terms is $0$

The Square Root Property states that if the square of a variable or term is equal to a value; to isolate the variable or term; the positive and negative square roots of that value must be considered.

## Identify Variables, Expressions and Equations

##### Identify the variables as applicable. Identify as expressions, equations, polynomials, and/or functions. Specify the type as applicable.
(1.) $3 + 4$
(2.) $3 + 4 = 7$
(3.) $3x - 5$
(4.) $3x = -7 + 10x$
(5.) $f(x) = 7x + 12$
(6.) $p(x) = -7 + 3x^2$
(7.) $f(x) = 7$
(8.) $3p = -12 + 7p^3$
(9.) $2c + 9c^4$
(10.) $f(c) = 2c + 9c^4$
(11.) $|k + 10| = |k|$
(12.) $f(d) = -2\left|3 - \dfrac{d}{3}\right| + 1$
(13.) $f(m) = \dfrac{27m^4 - m}{6m^3 + 10m^2 - 4m}$

Ask students to write examples of several expressions and equations with different variables.

## Translate from English to Math

#### Something About Word Problems ☺

(1.) Word problems are written in standard British English.

(2.) Some word problems are very lengthy and some are unnecessary. Those ones are meant to discourage you from even trying.

(3.) Word problems in Mathematics demonstrate the real-world applications of mathematical concepts.

(4.) Embrace word problems. See it as writing from "English to Math".
Take time to:
(a.) Read to understand. Paraphrase and shorten long sentences as necessary.
(b.) Re-read and note/underline the vocabulary "math" terms written in English.
(c.) Translate/Write "important" sentences one at a time.
(d.) Review what you wrote to ensure correctness.
(e.) Solve the math, and check your solution in the word problem. Does that solution makes sense? If it does, you are probably correct. If it does not, please re-do it.

English Math
$7$ more than a number, $p$ $7 + p$
$7$ increased by a number, $q$ $7 + q$
$7$ added to a number, $r$ $7 + r$
A number, $g$ added to $7$ $g + 7$
The sum of $7$ and a number $t$ $7 + t$
$7$ less than a number, $p$ $p - 7$
$7$ decreased by a number, $q$ $7 - q$
A number, $r$ decreased by $7$ $r - 7$
$7$ subtracted from a number, $e$ $e - 7$
A number, $t$ subtracted from $7$ $7 - t$
The difference of $7$ and a number, $u$ $7 - u$
The difference of a number, $v$ and $7$ $v - 7$
$7$ times a number, $p$ $7 * p$ = $7p$
$7$ multiplied by a number, $q$ $7 * q$ = $7q$
A number, $r$ multiplied by $7$ $r * 7$ = $7r$
The product of a $7$ and a number, $h$ $7 * h$ = $7h$
The product of a number $t$ and $7$ $t * 7$ = $7t$
$7$ divided by a number, $p$ $7 \div p$
A number, $q$ divided by $7$ $q \div 7$
$7$ divides a number, $r$ $r \div 7$
A number, $m$ divides $7$ $7 \div m$
The quotient of $7$ and a number, $t$ $7 \div t$
The quotient of a number, $u$ and $7$ $u \div 7$
A number, $d$ decreased by the sum of the number and five $d - (d + 5)$
Three times the product of negative two and a number, $c$ $3(-2c)$
"is" or "equal to" or "equals" $=$
$3$ consecutive integers beginning with $3$ $3, 4, 5$
$3$ consecutive integers beginning with $-3$ $-3, -2, -1$
$3$ consecutive integers beginning with $p$ $p, (p + 1), (p + 2)$
$3$ consecutive even integers beginning with $8$ $8, 10, 12$
$3$ consecutive even integers beginning with $-8$ $-8, -6, -4$
$3$ consecutive even integers beginning with $q$ $q, (q + 2), (q + 4)$
$3$ consecutive odd integers beginning with $7$ $7, 9, 11$
$3$ consecutive odd integers beginning with $-7$ $-7, -5, -3$
$3$ consecutive odd integers beginning with $r$ $r, (r + 2), (r + 4)$
The smaller of three consecutive integers beginning with $p$ $p$
The middle of three consecutive integers beginning with $p$ $p + 1$
The larger of three consecutive integers beginning with $p$ $p + 2$
The median of five consecutive odd integers beginning with $p$ $p + 4$
The larger of five consecutive even integers beginning with $p$ $p + 8$
The sum of three consecutive integers is $12$ Let the first integer = $p$
$p + (p + 1) + (p + 2) = 12$
The sum of three consecutive even integers is $12$ Let the first even integer = $p$
$p + (p + 2) + (p + 4) = 12$
The sum of three consecutive odd integers is $12$ Let the first odd integer = $k$
$k + (k + 2) + (k + 4) = 12$
Twice a number, $p$ $2 * p$ = $2p$
Double a number, $d$ $d * 2$ = $2d$
Thrice a number, $r$ $3 * r$ = $3r$
Triple a number, $c$ $c * 3$ = $3c$
Quadruple a number, $p$ $4 * p$ = $4p$
Multiplicative inverse or Reciprocal of a number, $p$ $\dfrac{1}{p}$
$5$ less than twice a number, $c$ Twice a number, $c$ = $2c$
$5$ less than $2c$ = $2c -5$
$2c - 5$
$4$ more than twice as many tickets, $t$ Twice as many tickets, $t$ = $2t$
$4$ more than $2t$ = $4 + 2t$
$4 + 2t$
The difference between the product of seven and a number, $k$; and twice the number. $7k - 2k$
The difference between twelve times a number, $d$; and five more than three times the number. $12d - (5 + 3d)$

## Linear Equations

A linear equation is an equation in which the highest exponent of the independent variable in the equation is $1$
An equation is any expression with an equal ($=$) sign.
Ask students the difference(s) between an expression, an equation, and a function.
Are all functions, equations? Why?

The solution of a linear equation is defined as the value of the independent variable which will ensure that the LHS (Left Hand Side) of the equation is the same as the RHS (Right Hand Side) of the equation.
In this regard, it is very important that you always check the solution of any equation.
After you calculate the value(s) of the variable(s), it is necessary that you substitute the value(s) of those variable(s) in the original equation(s) (not the modified equation(s)).
Ask students why they need to check their solutions using the original equation(s) rather than the modified equation(s)

When you substitute the values of the variable(s) into the main equation(s),
If the $LHS = RHS$, then you can drink some water ☺ (not alcohol!)
If the $LHS \ne RHS$, fix it.

Typically, a linear equation has only one solution.

However, a Linear Equation can have:
(1.) One solution.
(2.) Many solutions.
(3.) No solution.

When you get to the final step of solving a linear equation,
If the $variable = some\:\:value$, there is a distinct solution (only one solution). The value can be any real number, including $0$

$\underline{Example} \\[3ex] p - 3 = 4 \\[3ex] p = 4 + 3 \\[3ex] p = 7 \\[3ex] One\:\:solution \\[3ex]$ If the variable cancels out, and $0$ = $0$, there are many solutions. The solutions will be all real numbers.

$\underline{Example} \\[3ex] p - 3 = p - 3 \\[3ex] p - p = -3 + 3 \\[3ex] 0 = 0 \\[3ex] \underline{Review} \\[3ex] p - p = -3 + 3 \\[3ex] p - p = 0 \\[3ex] p\:\:could\:\:be\:\:5 \\[3ex] 5 - 5 = 0 p\:\:could\:\:be\:\:-10 \\[3ex] -10 - (-10) = -10 + 10 = 0 \\[3ex] p\:\:could\:\:be\:\:any\:\:real\:\:number \\[3ex] Many\:\:solutions \\[3ex]$ If the variable cancels out, and $0$ = $any\: value\: besides \: 0$, there is no solution.

$\underline{Example} \\[3ex] p - 3 = p - 2 \\[3ex] p - p = -2 + 3 \\[3ex] 0 = 1 \\[3ex] It\:\:is\:\:not\:\:possible \\[3ex] No\:\:solution \\[3ex]$

When solving a linear equation, you want to isolate the variable. Let the variable stay by itself.
If the variable is being multiplied by any value, divide both sides of the equation by that value to isolate the variable.
If the variable is being divided by any value, multiply both sides of the equation by that value to isolate the variable.
If the variable is being added to a value, subtract that value from both sides of the equation to isolate the variable.
If any value is being subtracted from the variable, add that value to both sides of the equation to isolate the variable.
If the variable is being subtracted from any value, simply move the variable to the other side of the equation.
Alternatively, you may add the variable to both sides of the equation.
Be it as it may, when it comes to equations that deals with additions and subtractions; it will be great if you know how to solve equations simply by moving "stuff" around.
If you move any value or variable from left to right, the sign of that value or variable will change.
If you move any value or variable from right to left, the sign of that value or variable will change.
If you move any value or variable from left to left, or right to right; the sign of that value or variable will not change.
If you swap (make the $LHS$ to be the $RHS$, and the $RHS$ to be the $LHS$), nothing changes.

As always, remember to Check your solution, even if your professor does not require you to check.
Checking the solution lets you know right away if you got it correct or incorrect.
The $LHS$ must be equal to the $RHS$ when the value of the variable is substituted in the original equation. (not the modified equation).

## Applications of Linear Equations

There are many applications of Linear Equations.
Some of them include applications in:

$(1.)\:\:\boldsymbol{Motion} \rightarrow s\:\:\:\:\:\:\:t\:\:\:\:\:\:\:d \\[3ex] s = speed \\[3ex] t = time \\[3ex] d = distance \\[3ex] d = s * t \\[3ex] distance = speed * time$

#### Notable Notes About Motion Problems

(1.) How fast implies Speed

(2.) How long implies Time

(3.) How far implies Distance

(4.) When two people travel in opposite directions, the total distance is the sum of the individual distances.
For example: If Mr. $A$ goes East and $Mr. B$ goes west, then the total distance covered by Mr. $A$ and Mr. $B$ is the sum of the distance covered by Mr. $A$ and the distance covered by Mr. $B$.

(5.) When two people travel in opposite directions, it can be seen as one person traveling that same distance in one direction (beginning from the end point of the first person and reaching the end point of the second person.
For example: If Mr. $A$ goes East and Mr. $B$ goes west, then the total distance covered by Mr. $A$ and Mr. $B$ can be seen as if Mr. $C$ started from the end point of Mr. $A$ and reached the end point of Mr. $B$

(6.) When two people travel in opposite directions, or when one person travels in the same direction but in two intervals; the total distance is the sum of the individual distances.
In other words, the total distance is the arithmetic sum.

(7.) In some cases (depending on the question), the total time is an arithmetic sum.
For example: If Mr. $A$ goes in the same direction at two intervals, then the total time is the sum of the times of the two intervals.
Say Mr. $A$ traveled to the City of Birmingham, Alabama from the City of Raleigh, North Carolina;
Assume that the:
First Interval: is from the City of Raleigh, North Carolina to the City of Atlanta, Georgia for $7$ hours and
Second Interval: is from the City of Atlanta, Georgia to the City of Birmingham, Alabama for $3$ hours
then the total time taken for him to complete the trip from the City of Birmingham, Alabama to the City of Raleigh, North Carolina is $10$ hours. In this case, the time is an arithmetic sum.

(8.) (9.) The total speed is NOT an arithmetic sum regardless of the question.
For example: If Mr. $A$ travels East at $30mph$ and Mr. $B$ travels West at $10mph$, it DOES NOT mean that both Mr. $A$ and Mr. $B$ traveled at $40mph$
It DOES NOT mean that the total speed of Mr. $A$ and Mr. $B$ is $40mph$

(10.) For each question, it is better to:
First: Identify the scenarios. It is usually two scenarios but it could be more depending on the question.
Second: Identify the total distance (and sometimes the total time as applicable).
Third: Identify what you are looking for. That should be the variable.
Fourth: Set up the $s\:\:\:t\:\:d$ table and solve the question.

$(2.)\:\:\boldsymbol{Mixtures} \rightarrow C\:\:\:\:\:\:\:V\:\:\:\:\:\:\:n \\[3ex] C = concentration \\[3ex] V = volume \\[3ex] n = amount \\[3ex] n = C * V \\[3ex] amount = concentration * volume$

#### Notable Notes About Mixture Problems

$(3.)\:\:\boldsymbol{Mathematics\:\:of\:\:Finance} \rightarrow P\:\:\:\:\:\:\:r\:\:\:\:\:\:\:t\:\:\:\:\:\:\:I \\[3ex] P = principal \\[3ex] r = rate \\[3ex] t = time \\[3ex] I = interest(simple\:\:interest) \\[3ex] I = P * r * t \\[3ex] Simple\:\:Interest = principal * rate * time$

#### Notable Notes About Mathematics of Finance Problems

For these applications, it is much better to each one as a table before solving it.
It is important you understand the concept so you can set up the table properly.

## Factor by GCF (Greatest Common Factor)

Factor by GCF (Greatest Common Factor) means that we should find the GCF first.

The Greatest Common Factor (GCF):
Is also known as the as Greatest Common Divisor (GCD), the HCF (Highest Common Factor), or GCM (Greatest Common Measure)
This means that it is a factor of two or more integers.
It is a divisor of two or more integers.
This means that it can divide two or more integers without a remainder.
It is a common factor of two or more integers.
It is the greatest common factor of two or more integers.
There may be other common factors of two or more integers. However, we are interested in the greatest of those factors.
Greatest Common Factor is defined as the greatest positive integer that can divide a set of integers without a remainder.
NOTE: $1$ is a common factor of every integer.
NOTE: If you have a leading negative, factor out the negative first.

There are at least five methods of calculating the $GCF$ of two or more positive integers.
They are:

(1.) Listing Method - can calculate the $GCF$ of two or more integers.
This is the simplest method.
However, it is cumbersome especially for a set of big integers.

(2.) "Nigerian" Method (hmmm... I learnt this method in the elementary school in Nigeria.
I have forgotten the name of the method. It works!) - can calculate the $gcd$ of two or more positive integers.
This method is not recommended for a set of big positive integers.

(3.) Prime Factorization Method - can calculate the $GCF$ of two or more positive integers.
This method is moderate for a set of big positive integers.

(4.) Euclidean Algorithm Method - can calculate the $GCF$ of two positive integers.
This method is recommended for a set of two integers however small or big.
We shall not be discussing this method - it is not included in the curriculum for Algebra
It is for Discrete Mathematics and Computer Science majors.

(5.) Extended Euclidean Algorithm Method - can calculate the $GCF$ of two positive integers.
This method is recommended for a set of two integers however small or big.
We shall not be discussing this method - it is not included in the curriculum for Algebra
It is for Discrete Mathematics and Computer Science majors.

Ask students if they can come up with their own methods after they learn all the five methods. Encourage them to do so.
Emphasize the importance of life-long learning.
Briefly discuss the discoveries made by other humans which are beneficial to them. So, why not discover something that will be beneficial to others?
Emphasize the required application of talents given to each of them by GOD. Use those talents to help one another.

We shall discuss the first three methods in this website.
For the last two methods, please review the Modular Arithmetic and Algorithms website.

Listing Method
In this method; we list all the factors of each integer.
We write the common factors of all of them.
Then, we find the greatest common factor of all of them.

Begin with $1$, and continue listing the factors in ascending order (from least to greatest). $1$ is a factor of every integer.
Then; as you write any small factor, write theother factor as well (in a separate place). That way, you would not forget any factor(s).

(1.) Calculate the $GCF$ of $12$ and $18$
Factors of $12 = 1, 2, 3, 4, 6, 12$
Factors of $18 = 1, 2, 3, 6, 9, 18$
Common Factors of $12 \:and\: 18 = 1, 2, 3, 6$
Highest Common Factor of $12 \:and\: 18 = 6$
$GCF(12, 18) = 6$

(2.) Calculate the $gcd$ of $30$ and $105$
Divisors of $30 = 1, 2, 3, 5, 6, 10, 15, 30$
Divisors of $105 = 1, 3, 5, 7, 15, 21, 35, 105$
Common Divisors of $30 \:and\: 105 = 1, 3, 5, 15$
Greatest Common Divisor of $30 \:and\: 105 = 15$
$GCF(30, 105) = 15$

Nigerian Method
In this method; we set up a table of the positive integers.
We set it up the same way we set up the dividend and divsior in the Long Division method of dividing polynomials.
We begin "trying" each factor in ascending order (from least to greatest).
The factor should be a positive integer that has to divide "all" the positive integers without a remainder.
In other words, it has to be a common factor.
If we cannot find any factor, then the $GCF = 1$.
Remember that $1$ is the smallest factor of every positive integer.
If we find a factor of all the integers (besides $1$), use it. Begin in ascending order.
If we find another factor of all the integers (a number that divides all the integers without a remainder), use it.
Multiply all the factors you listed.
The product gives you the $GCF$.

Let us see how this works!
(1.) Calculate the $GCF$ of $12$ and $18$
$$\begin{array}{c|c c} 2 & 12 & 18 \\ \hline 3 & 6 & 9 \\ \hline & 2 & 3 \end{array}$$ We can not find any common factor of $2$ and $3$
STOP.
The factors listed are $2$ and $3$
Multiply them.
$GCF(12, 18) = 2 * 3 = 6$

(2.) Calculate the $gcd$ of $30$ and $105$
$$\begin{array}{c|c c} 3 & 30 & 105 \\ \hline 5 & 10 & 35 \\ \hline & 2 & 7 \end{array}$$ We can not find any common factor of $2$ and $7$
STOP.
The factors listed are $3$ and $5$
Multiply them.
$GCF(30, 105) = 3 * 5 = 15$

But, we are dealing with Factoring Algebraic Expressions by GCF.
We shall be using the Prime Factorization method.
Prime Factorization Method
Prime Factorization Method requires that each number is written as product of prime numbers.

What are prime numbers?
List all prime numbers between $1$ and $100$
You may refer to this website: Numbers and Notations website

To use this method,
List all the prime factors of each number (factors of the number which are prime numbers). List in ascending order.
Express each number as a product of prime factors
Write each common factor for all the numbers.
If there are no common factors, then do nothing. $1$ is the common factor. In that case, the $GCF = 1$
Multiply the common factors. It gives you the $GCF$.

Calculate the $GCF$ of $12$ and $18$
$12 = 2 * 2 * 3 \\[3ex] 18 = 2 * 3 * 3 \\[3ex] GCF = 2 * 3 = 6$

Calculate the $GCF$ of $30$ and $105$
$30 = 2 * 3 * 5 \\[3ex] 105 = 3 * 5 * 7 \\[3ex] GCF = 3 * 5 = 15$

## Factor Quadratic Trinomials and the Likes

Recall: the Multiplication of Two Binomials
Ask students to multiply $(x + 3)(x + 5)$
They can use any method they like: FOIL Method, Vertical Method, Horizontal Method, or the Box Method

$(x + 3)(x + 5) \\[3ex] x(x) = x^2 \\[3ex] x(5) = 5x \\[3ex] 3(x) = 3x \\[3ex] 3(5) = 15 \\[3ex] = x^2 + 5x + 3x + 15 \\[3ex] = x^2 + 8x + 15 \\[3ex]$ So; Given: to binomials
We multiply the two binomials and we get a trinomial
This is Multiplying

Assume we were given the trinomial
What can we do to the trinomial in order to get the two binomials?
This is Factoring

Discuss comparisons of Multiply and Factor with:

Join and Split
Marriage and Divorce
Assemble and Disassemble
Build up and Tear apart

Given: the general form/standard form of a Quadratic Trinomial: $ax^2 + bx + c$;

$Discriminant = b^2 - 4ac$
The Discriminant:
(1.) tells us whether the quadratic trinomial can be factored or not.
(2.) tells us the nature of the roots of a quadratic equation.

How Do We Know if a Trinomial Can be Factored?
(1.) Calculate the discriminant.

If the discriminant is a perfect square, the trinomial can be factored

If the discriminant is not a perfect square, the trinomial cannot be factored
If the trinomial cannot be factored, it is said to be prime.

What Information Does the Discriminant Provide About the Nature of Roots?
(1.) If the discriminant is positive, the quadratic equation has $2$ real roots.

(1a.)If the discriminant is positive ($b^2 - 4ac \gt 0$) and is a perfect square:
the quadratic trinomial can be factored
the two binomials (got from the factoring) are not repeated

(1b.)If the discriminant is positive ($b^2 - 4ac \gt 0$) and is not a perfect square:
the quadratic trinomial cannot be factored;

(2.) If the discriminant is zero ($b^2 - 4ac = 0$):
the quadratic trinomial can be factored
the two binomials (got from the factoring) are repeated

(3.) If the discriminant is negative ($b^2 - 4ac \lt 0$):
the quadratic trinomial cannot be factored;

Given: the general form/standard form of a Quadratic Trinomial: $ax^2 + bx + c$;

First term = $ax^2$
Second term = $bx$
Third term = $c$
Coefficient of $x^2 = a$
Coefficient of $x = b$

(1.) Make sure it is in the standard form.

(2.) If the coefficient of $x^2$ is negative, factor out the negative.
This means that if $a$ is negative, factor out the negative.
We always want $a$ to be positive.

*(3.) If the coefficient of $x^2$ is one, find two factors of the third term such that the sum of those two factors gives the coefficient of the second term.
This means that if $a = 1$, find two factors of $c$ such that the sum of two factors gives $b$.
Then, write those factors as $(x ... ...)(x ... ...)$

(4.) Multiply the first and third terms.
$ax^2 * c = acx^2$

(5.) Find two factors of the product (in Step 4.) such that the sum of those two factors gives the second term.

(6.) Replace the second term with those two factors.

(7.) Factor by Grouping.
As you factor by Grouping; if you see any leading negative, factor out the negative first.

(8.) Check your work. Multiply the two binomials. It should give you the "original quadratic trinomial".
If it does not give you the "original quadratic trinomial", fix your mistake, re-solve and re-check until you get it correct.

*Step (3.) can be skipped. However, if you are taking a "timed" test, and if $a = 1$, you want to save time by doing it right away.
*See the second method used in Question (15.)*

Are there marriages that cannot be dissolved?
In other words, which entities cannot grant divorce?

Marriage done in the church - The church does not allow divorce.
This means the trinomial is prime - It cannot be factored

Marriage done in the court - The court allows allow divorce
This means the trinomial is not prime - It can be factored

Let us solve some examples

This method is not only used for Quadratic Trinomial: $ax^2 + bx + c$

This method is also used for:

(1.) Quartic Trinomials such as:

$ax^4 + bx^2 + c \\[3ex] where \\[3ex] x \:\:is\:\:the\:\:variable \\[3ex] a, b, c \:\:are\:\:the\:\:constants \\[3ex]$ (2.) Hexic Trinomials and Higher Order Trinomials such as:

$ax^6 + bx^3 + c \\[3ex] where \\[3ex] x \:\:is\:\:the\:\:variable \\[3ex] a, b, c \:\:are\:\:the\:\:constants \\[3ex]$ (3.) Multivariable Quadratic Trinomials such as:

$ax^2 + bxy + cy^2 \\[3ex] where \\[3ex] x, y \:\:are\:\:the\:\:variables \\[3ex] a, b, c \:\:are\:\:the\:\:constants \\[3ex]$ (4.) Multivariable Higher Order Trinomials such as:

$ax^4 + bx^2y^2 + cy^4 \\[3ex] where \\[3ex] x, y \:\:are\:\:the\:\:variables \\[3ex] a, b, c \:\:are\:\:the\:\:constants \\[3ex]$

Student: How do we know if the trinomial is prime or not?
How do we get the discriminant for these other forms?
Teacher: Great question.
We shall use the coefficients respectively...after arranging them well.
Did you notice the degrees in the first, second, and third terms? Please give more examples.

To See if These Other Forms are Prime or Not

(1.) Arrange them in order...as listed

(2.) Use the constants (the coefficients): $a, b, c$ accordingly in the formula

Let us solve some examples

A quadratic equation is defined as an equation in which the highest exponent of the variable in the equation is $2$.

The standard form of a Quadratic Equation: $ax^2 + bx + c = 0$;
where:
First term = $ax^2$
Second term = $bx$
Third term = $c$

If a quadratic equation has all the three terms, it is complete.
If a quadratic equation is missing either the second term or the third term, it is incomplete.
A complete quadratic equation is a quadratic equation that has all the three terms.
An incomplete quadratic equation is a quadratic equation that is missing the second term and/or the third term.

Let us study the cases in which a quadratic may occur.

Case 2: The second term is missing.
Given: incomplete Quadratic Equation where the second term is missing: $ax^2 + c = 0$;

To solve a Quadratic Equation where the second term is missing;
(1.) Make sure only the first term is on the LHS of the equation.
$ax^2 + bx = 0$
$ax^2 = -c$

(2.) Make sure the coefficient of $x^2 = 1$.
If it is not $1$, divide all both sides (all terms) by the coefficient of $x^2$ so as to make it $1$.
In this case, the coefficient of $x^2 = a$
Divide both sides by $a$

$\dfrac{ax^2}{a} = \dfrac{-c}{a} \\[3ex] x^2 = \dfrac{-c}{a}$

(3.) Apply the Square Root Property.

$\sqrt{x^2} = \pm \sqrt {\dfrac{-c}{2a}} \\[5ex] x = \pm \sqrt {\dfrac{-c}{2a}}$

Student: What is the Square Root Property?
Teacher: Good question.
Assume $p^2 = 9$, what is the value of $p$
Student: $3$
Teacher: Correct!
Is that all?
Student: Yes, I guess...
Teacher: Well, what about $p = -3$
$(-3)^2$ still gives $9$
Student: Yes...$p = -3$ also works
Teacher: How many solutions does a quadratic equation have?
Student: A quadratic equation has $2$ solutions
Teacher: Is $p^2 = 9$ a quadratic solution?
Student: Yes, it is
Teacher: So, how many solutions should it have?
Student: It should have two solutions.
Teacher: What are they?
Student: $p = 3$ and $p = -3$
Teacher: That's right.
Here's how we do it.
$p^2 = 9 \\[3ex] p = \pm \sqrt{9} \\[3ex] p = \pm 3 \\[3ex]$ Whenever we have the square of a variable equal to a value;
to isolate the variable;
we have to take the square root of both sides
In taking the square root of both sides;
we have to consider the positive or principal square root
as well as the negative square root.
So, we actually have two square roots of that value:
the positive or principal square root and the negative square root.
This does not apply to only the square of variables.
It only applies to the square of terms.
For example:
That is the Square Root Property.

Square Root Property
This states that if the square of a variable or term is equal to a value; to isolate the variable or term; the positive and negative square roots of that value must be considered.

Case 3: The third term is missing.
Given: incomplete Quadratic Equation where the third term is missing: $ax^2 + bx = 0$;

To solve a Quadratic Equation where the third term is missing;
(1.) Make sure only the first and second terms are on the LHS of the equation.
In other words, set the first and second terms to be equal to zero.
$ax^2 + bx = 0$

(2.) Factor the LHS of the equation by GCF.
$ax^2 + bx = 0$
$x(ax + b) = 0$

(3.) Apply the Zero Product Property.

Student: What is the Zero Product Property?
Teacher: Good question.
What is the number that when multiplied by anything, gives $0$?
Student: $0$
Teacher: Correct!
Say you multiply two numbers and it gives $0$, what does that tell you?
Student: One of the numbers is $0$
Teacher: Okay...but what about if both numbers are $0$?
Student: The result is still $0$
Teacher: That's right. So, what should we say in that case?
At least one (one or more) of the numbers is $0$
This is because either one of the numbers is $0$ or both numbers are $0$'s
Makes sense?
Student: Yes
Teacher: Say you multiplied two things and the product is $0$
What does that tell us?
Student: At least one of those things is $0$
Teacher: Say you multiplied three terms and the product is $0$?
What does it mean?
Student: At least one of those terms is $0$
Teacher: Great!

Zero Product Property
This states that if the product of two or more terms is $0$, at least one of those terms is $0$

$x = 0 \:\:\:OR\:\:\: ax + b = 0 \\[3ex] ax + b = 0 \\[3ex] ax = -b \\[3ex] x = \dfrac{-b}{a} \\[5ex] \Rightarrow x = 0 \:\:\:OR\:\:\: x = \dfrac{-b}{a}$

So, we have discussed $Case\: 2$: where the second term is missing.
We have discussed $Case\: 3$: where the third term is missing.
Are we missing something? Are we missing any other case?
Please note the students' responses. Some might ask about $Case\: 1$.
If they say $Case\: 1$, ask them what they think. Should the first term be missing?
What will happen if the quadratic equation is missing the first term
Will it still be called a quadratic equation? Please review the definition.
If it will not be called a quadratic equation, what is it? What will it be called?
So, may we now classify our $Case\: 1$ as ....?

Case 1: None of the terms are missing.
Everything is complete.
It may not be arranged in the standard form but all the terms are complete.

A complete Quadratic Equation can be solved by:
(1.) Factoring method.
(2.) Completing the Square method.

Let's get back to the three cases we discussed so far.
Is it possible to make both $Case\: 2$ and $Case\: 3$ to look like $Case\: 1$? What do you think?

Yes, we can complete any incomplete quadratic equation

How do we do that? Please note the students' responses.

By writing the missing terms, and putting zeros as the coefficients of those missing terms.
Then, we can still solve those equations using any of the three methods as applicable.

Given: the general form/standard form of a Quadratic Equation: $ax^2 + bx + c = 0$;

$Discriminant = b^2 - 4ac$
The Discriminant:
(1.) tells us whether the quadratic trinomial can be factored or not.
(2.) tells us the nature of the roots of a quadratic equation.

How Do We Know if a Quadratic Equation can be solved by Factoring?
(1.) Calculate the discriminant.
If the discriminant is a perfect square, the quadratic equation can be solved by factoring.
If the discriminant is not a perfect square, the quadratic equation cannot be solved by factoring.

What Information Does the Discriminant Provide About the Nature of Roots?
(1.) If the discriminant is positive, the quadratic equation has $2$ real roots.

(1a.)If the discriminant is positive ($b^2 - 4ac \gt 0$) and is a perfect square:
the quadratic equation can be solved by factoring; and
the quadratic equation has two rational roots.

(1b.)If the discriminant is positive ($b^2 - 4ac \gt 0$) and is not a perfect square:
the quadratic equation cannot be solved by factoring; and
the quadratic equation has two irrational roots.

(2.) If the discriminant is zero ($b^2 - 4ac = 0$):
the quadratic equation can be solved by factoring;
the quadratic equation has one repeated rational root; and
the quadratic trinomial has one repeated factor

(3.) If the discriminant is negative ($b^2 - 4ac \lt 0$):
the quadratic equation cannot be solved by factoring; and
the quadratic equation has two imaginary roots.

Given: the general form/standard form of a Quadratic Equation: $ax^2 + bx + c = 0$;

First term = $ax^2$
Second term = $bx$
Third term = $c$
Coefficient of $x^2 = a$
Coefficient of $x = b$

Factoring method
To solve a Quadratic Equation by Factoring;
(1.) Make sure it is in the standard form.

(2.) If the coefficient of $x^2$ is negative, factor out the negative.
This means that if $a$ is negative, factor out the negative.
We always want $a$ to be positive.

*(3.) If the coefficient of $x^2$ is one, find two factors of the third term such that the sum of those two factors gives the coefficient of the second term.
This means that if $a = 1$, find two factors of $c$ such that the sum of two factors gives $b$.
Then, write those factors as $(x ... ...)(x ... ...) = 0$

(4.) Multiply the first and third terms.
$ax^2 * c = acx^2$

(5.) Find two factors of the product (in Step 4.) such that the sum of those two factors gives the second term.

(6.) Replace the second term with those two factors.

(7.) Factor by Grouping.
As you Factor by Grouping; if you see any leading negative, factor out the negative first.

(8.) Apply the Zero Product property.

(9.) Determine the solutions of the equation.

*Step (3.) can be skipped. However, if you are taking a "timed" test, and if $a = 1$, you want to save time by doing it right away.
See the second method used in Question (11.)

Completing the Square method
Given: the standard form of a Quadratic Equation: $ax^2 + bx + c = 0$
To solve a Quadratic Equation by Completing the Square;
(1.) Set the quadratic equation such that only first and second terms are on the Left Hand Side (LHS) of the equation.
$ax^2 + bx = -c$

(2.) Make sure the coefficient of $x^2 = 1$.
If it is not $1$, divide all both sides (all terms) by the coefficient of $x^2$ so as to make it $1$.

$\dfrac{ax^2 + bx}{a} = \dfrac{-c}{a} \\[3ex] \dfrac{ax^2}{a} + \dfrac{bx}{a} = \dfrac{-c}{a} \\[3ex] x^2 + \dfrac{bx}{a} = \dfrac{-c}{a}$

The LHS is now an incomplete square.
We have to complete the square
(3.) Add to both sides, the square of half of the coeffcient of $x$.

Coefficient of $x = \dfrac{b}{a}$

Half of it = $\dfrac{1}{2} * \dfrac{b}{a} = \dfrac{b}{2a}$

Square it = $\left(\dfrac{b}{2a}\right)^2$

Add to both sides of the equation.

$x^2 + \dfrac{bx}{a} + \left(\dfrac{b}{2a}\right)^2 = \dfrac{-c}{a} + \left(\dfrac{b}{2a}\right)^2 \\[5ex] Condense\: the\: LHS \\[2ex] The\: LHS\: becomes\: \left(x + \dfrac{b}{2a}\right)^2 \\[5ex] \Rightarrow \left(x + \dfrac{b}{2a}\right)^2 = \dfrac{-c}{a} + \dfrac{b^2}{4a^2} \\[5ex] \left(x + \dfrac{b}{2a}\right)^2 = \dfrac{-4ac}{4a^2} + \dfrac{b^2}{4a^2} \\[5ex] \left(x + \dfrac{b}{2a}\right)^2 = \dfrac{-4ac + b^2}{4a^2} \\[5ex] \left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2} \\[3ex]$

(4.) Apply the Square Root Property.

$\sqrt{\left(x + \dfrac{b}{2a}\right)^2} = \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}} \\[5ex] x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} \\[5ex] x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \\[5ex] Isolate\: x \\[2ex] x = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex]$

We just came up with a formula for $x$

What do you think we should call it?
Assess any prior knowledge.

It is known as the Quadratic Formula.
As you can see, Completing the Square method actually led to the Quadratic Formula

Given: the standard form of a Quadratic Equation: $ax^2 + bx + c = 0$
(1.) Make sure it is in the standard form.

(2.) Write the values of $a$, $b$, and $c$.

(3.) Substitute those values in the quadratic formula.

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[2ex]$

(4.) Determine the solutions of the equation.
The two solutions are:

$x = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \\[5ex] and \\[2ex] x = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$

Formative Assessment
Of all these three methods, which one do you prefer? Please note the students' responses.
What are your reasons for the method you choose?

An Interesting Find!
We can express the roots of any quadratic equation in standard form as a Linear System.
Let us see!

Recall: The two solutions of a Quadratic Equation (from the Quadratic Formula) are:

$x = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \\[5ex] and \\[2ex] x = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$

Let us denote the two solutions as: $\alpha$ and $\beta$
Let:
$\alpha = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \\[5ex] \beta = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$

Sum of Roots

$\alpha + \beta \\[3ex] = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} + \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{(-b + \sqrt{b^2 - 4ac}) + (-b - \sqrt{b^2 - 4ac})}{2a} \\[5ex] = \dfrac{-b + \sqrt{b^2 - 4ac} + -b - \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-b + \sqrt{b^2 - 4ac} -b - \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{-2b}{2a} \\[5ex] = \dfrac{-b}{a}$

Product of Roots

$\alpha * \beta \\[3ex] = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} * \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \\[5ex] = \dfrac{(-b + \sqrt{b^2 - 4ac}) * (-b - \sqrt{b^2 - 4ac})}{2a * 2a} \\[5ex] = \dfrac{b^2 + b\sqrt{b^2 - 4ac} - b\sqrt{b^2 - 4ac} - (\sqrt{b^2 - 4ac})^2}{4a^2} \\[5ex] = \dfrac{b^2 - (b^2 - 4ac)}{4a^2} \\[5ex] = \dfrac{b^2 - b^2 + 4ac}{4a^2} \\[5ex] = \dfrac{4ac}{4a^2} \\[5ex] = \dfrac{c}{a}$

Therefore, the:
Sum of roots = $\dfrac{-b}{a}$

Product of roots = $\dfrac{c}{a}$

The Linear System is:

$\alpha + \beta = \dfrac{-b}{a} ...eqn.(1) \\[3ex] \alpha * \beta = \dfrac{c}{a} ...eqn.(2)$

Is this not awesome?
But, of what use is this?

We can call this our Alternative Check
We can check the solution of any quadratic equation by the Sum and Product of Roots
Main Check
(1.) Calculate the solutions of the quadratic equation.
(2.) Substitute those values in the main/original quadratic equation.
(3.) Make sure that the $LHS = RHS$

It is an added knowledge to learn the:
Alternative Check
(1.) Calculate the solutions of the quadratic equation.
(2.) Make sure the original equation is written in the standard form.
(3.) Write the values of $a$, $b$, and $c$
(4.) Add the solutions of your equation. Make sure the sum is equal to $\dfrac{-b}{a}$

(5.) Multiply the solutions of your equation. Make sure the product is equal to $\dfrac{c}{a}$

• Given: a Quadratic Equation in standard form
To Calculate: other details

$x^2$ + $x$ + $= 0$

• Details

## Rational Equations

Recall the definitions:

A rational number is any number that can be written as a fraction where the denominator is not equal to zero.
You can also say that a rational number is a ratio of two integers where the denominator is not equal to zero.
A rational number is a number that can be written as: $$\dfrac{c}{d}$$ where $c, d$ are integers and $d \neq 0$
A rational number can be an integer.
It can be a terminating decimal. Why?
It can be a repeating decimal. Why?
It cannot be a non-repeating decimal. Why?
Ask students to tell you what happens if the denominator is zero.

A rational function is a function of the form: $$\dfrac{c(x)}{d(x)}$$ where $c(x), d(x)$ are functions and $d(x) \neq 0$

Based on the prior definitions, ask students to suggest the definition(s) of a rational inequality.

A rational equation is an equation containing a rational function.
A rational equation is an equation containing a ratio of two polynomials.
You can also say that a rational equation is an equation of the form: $$\dfrac{c(x)}{d(x)} = 0 \\[5ex]$$ where $c(x), d(x)$ are polynomial functions and $d(x) \neq 0$
It is also known as a fractional equation.

Methods of Solving Rational Equations
There are three methods of solving rational equations.
The two main methods are:
First Method (No Fractions Method): This method is recommended for those who do not like dealing with fractions.
The steps are:
(1.) Find the $LCD$ (Least Common Denominator) first.
(2.) Multiply each term by the $LCD$. This removes the fractions.
(3.) Solve the resulting equation.

Second Method (Yes Fractions Method): This method is recommended for those who like dealing with fractions.
(1.) Modify the equation so that there is only one fraction on the $LHS$ (Left Hand Side) and there is only one fraction on the $RHS$ (Right Hand Side).
(2.) Make sure the denominators of the fractions on the $LHS$ and on the $RHS$ are the same.
(3.) Equate the numerators of the fractions.
(4.) Solve the resulting equation.

The third method is used only when there are two fractions (rational functions): one on the $LHS$ of the equation and one on the $RHS$ of the equation. Use this method only when it is that way.
Third Method (Cross Multiply Method or Cross Product Method): This method is used only when there is only one fraction on the $LHS$ of the equation and only one fraction on the $RHS$ of the equation.

(1.) Multiply the numerator of the fraction on the $LHS$ with the denominator of the fraction on the $RHS$. This gives the $LHS$ of the modified equation.
(2.) Multiply the denominator of the fraction on the $LHS$ with the numerator of the fraction on the $RHS$. This gives the $RHS$ of the modified equation.
(3.) Equate both sides of the modified equation.
(4.) Solve the resulting equation.

Regardless of any method used, make sure you Check your solution.

Teacher: This is especially important for rational equations.
Student: Why is that?...speaking like an American lol
Teacher: Because of "extraneous" solutions.
Student: What are "extraneous" solutions?
Teacher: An extraneous solution (also known as extraneous root) is a solution of the modified equation, but not a solution of the original equation. Here is the explanation.

Before we solve any rational equation, we have to modify it.
Whether we solve it by clearing the fractions first (First Method), or whether we solve it as fractions (Second Method); we still have to modify it.
So, we have one equation (original equation); and we modified it in order to solve it (modified equation).
In doing so, we might have a solution which is a solution of the modified equation, but when we substitute that solution in the original equation, it does not the work. The $LHS$ is not equal to the $RHS$; whereas in an equation; the $LHS$ should be equal to the $RHS$.
That kind of solution is known as an extraneous solution or an extraneous root.

Teacher: Hence the need to Check your solution and to check it with the...
Student: Original equation

Note: Rational equations do not always have extraneous roots.
Sometimes, they do. Sometimes, they do not.
Also, it is possible to have an extraneous root that is part of the domain of the main equation.

A radical equation is an equation in which at least one of it's terms has a radical.

(1.) Check to see how many radicals are there.
If there is one radical, place that radical on the LHS (Left Hand Side) of the equation.
If there are two radicals, place one radical on the LHS (Left Hand Side) and another radical on the RHS (Right Hand Side) of the equation.

(2.) If there are square root radicals, square both sides.
If there is only one square root radical, then square both sides one time.
This means that you should multiply the exponents of both sides of the equation by $2$
If there are two square root radicals, you may need to square both sides two times; first before solving, and second before you get the solution(s). It depends on the question. Please review examples.

If there are cube root radicals, cube both sides.
If there is only one cube root radical, then cube both sides one time.
This means that you should multiply the exponents of both sides of the equation by $3$
If there are two cube root radicals, you may need to cube both sides two times; first before solving, and second before you get the solution(s). It depends on the question. Please review examples.

If there are fourth root radicals, raise both sides of the equation to the fourth exponent
and so on...and so forth

Why should you do so?

Remember:
The square of the square root of a term is the term.
Similarly, the square root of the square of a term is the term.
Also, the cube of the cube root of a term is the term.
Similarly, the cube root of the cube of a term is the term.
The square of the square root of a number is the number.
$(\sqrt{3})^2 = 3$
$(\sqrt{3})^2 \\[3ex] = (3^\frac{1}{2})^2 \\[3ex] x^\frac{1}{2} = \sqrt{x} ...\: Fractional\: Exponents\: gives\: radicals \\[3ex] = 3^{\frac{1}{2} * 2} ...\: Law\: of\: Exponents \\[3ex] = 3^1 \\[3ex] = 3$
Similarly, the square root of the square of a number is the number.
$\sqrt{3^2} = 3$
$\sqrt{3^2} \\[3ex] = (3^2)^\frac{1}{2} \\[3ex] = 3^{2 * \frac{1}{2}} ...\: Law\: of\: Exponents \\[3ex] = 3^1 \\[3ex] = 3$

This is very important especially to determine whether there are extraneous roots.
Ask students why it is possible to have extraneous roots for radical equations
Watch out for one particular word: It begins with a "D" and ends with an "n". They should explain.

Before we solve any radical equation, we have to modify it.
We have to remove the radical(s) so we can solve it.
So, we have one equation (original equation); and we modified it in order to solve it (modified equation).
In doing so, we might have a solution which is a solution of the modified equation, but when we substitute that solution in the original equation, it does not the work. The $LHS$ is not equal to the $RHS$; whereas in an equation; the $LHS$ should be equal to the $RHS$.
That kind of solution is known as an extraneous solution or an extraneous root.

Teacher: Hence the need to Check your solution and to check it with the...
Student: Original equation

Note: Radical equations do not always have extraneous roots.
Sometimes, they do. Sometimes, they do not.
Also, it is possible to have an extraneous root that is part of the domain of the main equation.

## Absolute Value Equations

An absolute value equation is an equation in which at least one of it's terms has an absolute value.

The absolute value of a term is the magnitude or modulus of that term regardless of sign.
The absolute value of a term say $x$ is denoted by $|x|$
Some resources/calculators represent it as $abs(x)$

$|-3| = 3 \\[3ex] |3| = 3 \\[3ex]$ Teacher: What do you notice?
Student: The result is positive regardless of what is inside the absolute value.
Teacher: Correct!

Think of it this way:
James walked $3$ miles to school from the left (West Side)
John walked $3$ miles to school from the right (East Side)
The left is the west side of the number line.
The right is the east side of the number line.
The school is in the middle - the center of the number line, $0$
Regardless of whichever direction they took, James and John walked $3$ miles to school.
James approached it from the $LHS$, hence the $-3$
John approached it from the $RHS$, hence the $3$
Let's say we do not care really care about the direction.
We just want to know the "actual" distance - the magnitude of the distance - the displacement (in Physics)
That is the "absolute value".

This implies that the result of the absolute value is always positive.
However, the term inside the absolute value could be positive or negative.

So, whenever we solve absolute value equations, we have to consider two cases.
First Case: The term inside the absolute value is positive.
Second Case: The term inside the absolute value is negative.

And of course, with Mr. C; you have to always Check your solutions

Student: Is it possible to have extraneous roots with absolute value equations also?
Teacher: Absolutely!
Hence another reason to check your work!
Remember: it is possible. Sometimes you have extraneous root(s), sometimes you do not.

#### References

<NAME> (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com

<NAME>., <NAME>., <NAME>., & <NAME>. (2017). Algebra and Trigonometry: Graphs and Models ($6^{th}$ ed.). Boston: Pearson.

<NAME>., & <NAME>. (2014). College Algebra Essentials ($3^{rd}$ ed.). New York: McGraw-Hill

<NAME>., & <NAME>. (2011). Algebra for College Students (Revised/Expanded ed.). Belmont, CA: Brooks/Cole, Cengage Learning.

<NAME>., & <NAME>. (2012). Beginning and Intermediate Algebra (Revised/Expanded ed.). Boston: Pearson Addison-Wesley.

<NAME>., & <NAME>. (2017). Algebra & Trigonometry ($7^{th}$ ed.). Boston: Pearson.


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# take -bytestring

take n, applied to a list xs, returns the prefix of xs of length n, or xs itself if n > length xs: > take 5 "Hello World!" == "Hello" > take 3 [1,2,3,4,5] == [1,2,3] > take 3 [1,2] == [1,2] > take 3 [] == [] > take (-1) [1,2] == [] > take 0 [1,2] == [] It is an instance of the more general Data.List.genericTake, in which n may be of any integral type.
O(log(min(i,n-i))). The first i elements of a sequence. If i is negative, take i s yields the empty sequence. If the sequence contains fewer than i elements, the whole sequence is returned.
O(n) take n, applied to a Text, returns the prefix of the Text of length n, or the Text itself if n is greater than the length of the Text. Subject to fusion.
O(n) take n, applied to a Text, returns the prefix of the Text of length n, or the Text itself if n is greater than the length of the Text. Subject to fusion.
takeWhile, applied to a predicate p and a list xs, returns the longest prefix (possibly empty) of xs of elements that satisfy p: > takeWhile (< 3) [1,2,3,4,1,2,3,4] == [1,2] > takeWhile (< 9) [1,2,3] == [1,2,3] > takeWhile (< 0) [1,2,3] == []
Return the contents of the MVar. If the MVar is currently empty, takeMVar will wait until it is full. After a takeMVar, the MVar is left empty. There are two further important properties of takeMVar: * takeMVar is single-wakeup. That is, if there are multiple threads blocked in takeMVar, and the MVar becomes full, only one thread will be woken up. The runtime guarantees that the woken thread completes its takeMVar operation. * When multiple threads are blocked on an MVar, they are woken up in FIFO order. This is useful for providing fairness properties of abstractions built using MVars.
Get the base name, without an extension or path. > takeBaseName "file/test.txt" == "test" > takeBaseName "dave.ext" == "dave" > takeBaseName "" == "" > takeBaseName "test" == "test" > takeBaseName (addTrailingPathSeparator x) == "" > takeBaseName "file/file.tar.gz" == "file.tar"
Get the directory name, move up one level. > takeDirectory x isPrefixOf x || takeDirectory x == "." > takeDirectory "foo" == "." > takeDirectory "/foo/bar/baz" == "/foo/bar" > takeDirectory "/foo/bar/baz/" == "/foo/bar/baz" > takeDirectory "foo/bar/baz" == "foo/bar" > Windows: takeDirectory "foo\\bar" == "foo" > Windows: takeDirectory "foo\\bar\\\\" == "foo\\bar" > Windows: takeDirectory "C:\\" == "C:\\"
Get the drive from a filepath. > takeDrive x == fst (splitDrive x)
Get the extension of a file, returns "" for no extension, .ext otherwise. > takeExtension x == snd (splitExtension x) > Valid x => takeExtension (addExtension x "ext") == ".ext" > Valid x => takeExtension (replaceExtension x "ext") == ".ext"
Get all extensions > takeExtensions "file.tar.gz" == ".tar.gz"
Get the file name. > takeFileName "test/" == "" > takeFileName x isSuffixOf x > takeFileName x == snd (splitFileName x) > Valid x => takeFileName (replaceFileName x "fred") == "fred" > Valid x => takeFileName (x </> "fred") == "fred" > Valid x => isRelative (takeFileName x)
Takes the right number of bytes from the input.
Return the contents of the TMVar. If the TMVar is currently empty, the transaction will retry. After a takeTMVar, the TMVar is left empty.
O(n) takeWhile, applied to a predicate p and a Text, returns the longest prefix (possibly empty) of elements that satisfy p. Subject to fusion.
O(i) applied to a predicate p and a sequence xs, returns the longest prefix (possibly empty) of xs of elements that satisfy p.
O(i) applied to a predicate p and a sequence xs, returns the longest suffix (possibly empty) of xs of elements that satisfy p. takeWhileR p xs is equivalent to reverse (takeWhileL p (reverse xs)).
O(1) Return the prefix of the Text of n Word16 units in length. If n would cause the Text to end inside a surrogate pair, the end of the prefix will be advanced by one additional Word16 unit to maintain its validity.
O(1) Unchecked take of k Word16s from the front of a Text.
The genericTake function is an overloaded version of take, which accepts any Integral value as the number of elements to take.
A non-blocking version of takeMVar. The tryTakeMVar function returns immediately, with Nothing if the MVar was empty, or Just a if the MVar was full with contents a. After tryTakeMVar, the MVar is left empty.
A version of takeTMVar that does not retry. The tryTakeTMVar function returns Nothing if the TMVar was empty, or Just a if the TMVar was full with contents a. After tryTakeTMVar, the TMVar is left empty.

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text:
Question 204503
# 1

{{{f(x+h)=(x+h)^2+3}}} Replace each "x" with "x+h"

{{{f(x+h)=x^2+2xh+h^2+3}}} FOIL (ie expand)

--------------------------------

{{{(f(x+h)-f(x))/h}}} Move onto the given difference quotient.

{{{(x^2+2xh+h^2+3-(x^2+3))/h}}} Plug in {{{f(x+h)=x^2+2xh+h^2+3}}} and {{{f(x)=x^2+3}}}

{{{(x^2+2xh+h^2+3-x^2-3)/h}}} Distribute

{{{(2xh+h^2)/h}}} Combine like terms.

{{{(h(2x+h))/h}}} Factor out the GCF "h" from the numerator.

{{{(cross(h)(2x+h))/cross(h)}}} Cancel out the common terms.

{{{2x+h}}} Simplify

So {{{(f(x+h)-f(x))/h=2x+h}}} when {{{f(x)=x^2+3}}}

=====================================================================================

# 2

Take note that when {{{x=-3}}} it is less than zero. Because {{{f(x)=-x}}} when {{{x<0}}}, this means that we simply plug in {{{x=-3}}} to get: {{{f(-3)=-(-3)=3}}}.

So {{{f(-3)=3}}}

Also, when {{{x=2}}} it is greater than zero. Since {{{f(x)=x^3}}} when {{{x>=0}}}, we just plug in {{{x=2}}} to get: {{{f(2)=(2)^3=8}}}.

So {{{f(2)=8}}}

================================

# 3

{{{x+2y=3}}} Add 3 to both sides.

{{{2y=3-x}}} Subtract {{{x}}} from both sides.

{{{2y=-x+3}}} Rearrange the terms.

{{{y=(-x+3)/2}}} Divide both sides by {{{2}}} to isolate y.

{{{y=-(1/2)x+3/2}}} Break up the fraction and simplify.

We can see that the equation {{{y=-(1/2)x+3/2}}} has a slope {{{m=-1/2}}} and a y-intercept {{{b=3/2}}}.

Now to find the slope of the perpendicular line, simply flip the slope {{{m=-1/2}}} to get {{{m=-2/1}}}. Now change the sign to get {{{m=2}}}. So the perpendicular slope is {{{m=2}}}.

Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-1/2}}} and the coordinates of the given point *[Tex \LARGE \left$$6,-4\right$$].

{{{y--4=2(x-6)}}} Plug in {{{m=2}}}, {{{x[1]=6}}}, and {{{y[1]=-4}}}

{{{y+4=2(x-6)}}} Rewrite {{{y--4}}} as {{{y+4}}}

{{{y+4=2x+2(-6)}}} Distribute

{{{y+4=2x-12}}} Multiply

{{{y=2x-12-4}}} Subtract 4 from both sides.

{{{y=2x-16}}} Combine like terms.

So the equation of the line perpendicular to {{{x+2y-3=0}}} that goes through the point *[Tex \LARGE \left$$6,-4\right$$] is {{{y=2x-16}}}.

Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(1/2)x+3/2,2x-16)
circle(6,-4,0.08),
circle(6,-4,0.10),
circle(6,-4,0.12))}}}

Graph of the original equation {{{x+2y-3=0}}} (red) and the perpendicular line {{{y=2x-16}}} (green) through the point *[Tex \LARGE \left$$6,-4\right$$].

=================================

# 5

{{{(x^2-8x)+(y^2-6y)=0}}} Group like terms.

{{{(x^2-8x+highlight(16))+(y^2-6y)=0+highlight(16)}}} Take half of the x-coefficient -8 to get -4. Square -4 to get 16. Add this value to both sides.

{{{(x^2-8x+16)+(y^2-6y+highlight(9))=0+16+highlight(9)}}} Take half of the y-coefficient -6 to get -3. Square -3 to get 9. Add this value to both sides.

{{{(x-4)^2+(y-3)^2=25}}} Combine like terms.

{{{(x-4)^2+(y-3)^2=5^2}}} Rewrite 25 as {{{5^2}}}

Now the equation is in the form {{{(x-h)^2+(y-k)^2=r^2}}} (which is a circle) where (h,k) is the center and "r" is the radius

In this case, {{{h=4}}}, {{{k=3}}}, and {{{r=5}}}

So the center is (4,3) and the radius is 5 units.

Here's the graph:

{{{drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10,0),
circle(4,3,5)
)}}}