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May 19th

May 19th

# Solving Equations &amp; Inequalities

Lesson 7: Solving Equations & Inequalities

7. Solving Equations & Inequalities

7.1. How to Solve Linear Equations
• Essentially Linear Equations
7.2. Solving Second Degree Equations
• Factoring Methods • Completing the Square • The Quadratic
Formula
7.3. Solving Inequalities
• Tools for Solving Inequalities • Simple Inequalities • Double
Inequalities • The Method of Sign Charts
7.4. Solving Absolute Inequalities
• Solving the Inequality |a| < b • Solving the Inequality
|a| > b

7. Solving Equations & Inequalities

There are two basic tools for solving equations: (1) Adding the same
expression to both sides of an equation and (2) multiplying both sides
of the equation by the same expression. In symbols,

Here the symbol <=> means “if and only if ” which is a fancy way of
saying “is equivalent to .” Equation (2) has an obvious variation

Some care needs to be taken when applying (2) and (3) in the case
where c is an algebraic expression containing unknowns. Usually, students
have no problems when c is a numerical value.

7.1. How to Solve Linear Equations
A linear equation is an equation of the form
ax + b = 0
a ≠0

The solution set would represent the zeros or roots of the linear polynomial
ax + b.

Even though the equation is very simple, student still make errors
solving it. Here are some representative examples. All linear equations
are solved in the same way.

Example 7.1. Solve each linear equation.
(a) 4x+5 = 0
(b) (1/2)x − 4 = 6
(c) 7 − 3x = 2

Strategyfor Solving. Isolate the unknown, x say, on one side of the
equation with other terms of the equation on the other side. Divide
through both sides by the coefficient of the unknown to obtain the
solution.

Exercise 7.1. Solve for x in each of the following using good techniques
(as exhibited in Example 7.1). Passing is 100% correct.

(a) 4x − 8 = 0 (b) 23 − 2x = 3 (c)(4/3)x − 2 = 3 (d)(2/3)x +4/5=8/3

Here is a slight variation on the type of linear equation previously
considered.

Exercise 7.2. Solve each for x.
(a) 3x − 2 = 7x + 3 (b) 9x+3 = 4x − 2 (c)(1/2)x =(4/3)(x − 3)

Sometimes we have equations that have several symbolic quantities .
Examine each of the following problems.

Exercise 7.3. In each of the equations listed below, solve for the
indicated variable. Solve for . . .
(a) x in 5x − 3y = 4 (b) y in 5x − 3y = 4
(c) z in x2z − 12x + y = 1

It is apparent from the above examples and exercises that the operations
of adding (1), multiplying (2), and dividing (3) both sides of an
equation by the same expression are standard and useful tools in your

toolbox of techniques for solving equations. Do not make up your own
methods, use the standard ones and . . . these are they!

When working out mathematical problems it is important to organize
your thoughts on paper properly and clearly. After solving a problem,
recopy it neatly. Copy the style of this tutorial or some other textbook.
Try to improve your handwriting. Use proper notation.

• Essentially Linear Equations
Some equations are ‘disguised’ linear equations. They do not require
any special techniques other than what is needed to solve linear equations.

Example 7.2. Solve for x in each of the following.

Now, consider the following set of exercises.

Exercise 7.4. Solve for x is each of the following.

7.2. Solving Second Degree Equations
We now turn to the problem of solving equations of the form:
ax2+ bx + c = 0.  (4)

There are two standard methods of solving this kind of equation:
(1) by factoring the left-hand side and (2) by applying the Binomial
Formula.

• Factoring Methods
We have already studied techniques of factoring polynomials of degree
two; therefore, it is not necessary to look at a large number of
examples. If necessary, review factoring.
The method of factoring can certainly be applied to any polynomial
to factoring, a major tool used in solving equations is the Zero-
Product Principle.

Zero-Product Principle
ab = 0 => a = 0 or b = 0  (5)

This principle states the obvious property of the real number system:
The only way the product of two numbers can be zero is if one of
them is zero.

The following example illustrates standard reasoning and solution

Example 7.3. Solve each of the following.
(a) x2− 5x+6 = 0 (b) x2+ 4x + 4 = 0 (c) 6x2− x − 2 = 0

Now using the same methods as exhibited in the previous example
solve each of the following. Passing is 100%.

Exercise 7.5. (Skill Level 1) Solve for x in each of the following.
(a) x2− 7x+12 = 0 (b) x2+ 3x = 10

Here are a few more quadratic equations.

Exercise 7.6. Solve for x in each of the following.
(a) 12x2− 17x + 6 = 0 (b) 20x2+ 3x = 2

Factorization techniques are not limited to second degree equations.
Here are a few higher degree equations that can be factored fairly
easily. Some of the fourth degree equations below can be solved using
the factorization techniques for quadratic polynomials.

Exercise 7.7. Solve for x in each of the following.
(a) x3 − 2x2− 3x = 0 (b) x4 − 16 = 0
(c) x4 − 2x2− 3 = 0 (d) x4 − 5x2+6 = 0

• Completing the Square
In addition to factorization methods, the technique of completing
the square can also be used to solve a quadratic equation. Even
though this technique will be used to solve equation, completion of
the square has certain uses in other kinds of mathematical problems.
If you go on to Calculus, for example, you will see it within the
context of integ ration problem .

Completion of the Square Algorithm. Below are the steps for completing
the square with an abstract and a particular equation that
illustrate the steps.

1. Problem: Solve the quadratic equation for x:
ax2+ bx + c = 0
2x2+ 12x − 3 = 0.

2. Associate the x2 term and the x term:
(ax2+ bx) + c = 0
(2x2+ 12x) − 3 = 0

3. Factor out of the coefficient of the x2term.

2(x2+ 6x) − 3 = 0

4. Take one-half of the coefficient of the x and square it:

 take the coefficient of x . . . 6 and compute one-half this . . . 3 and square it: 9

5. Take this number, and add it inside the parentheses; this addition
must be compensated for by subtracting it from outside the
parentheses:

Care must be made here because we are adding the term inside
the parentheses and subtracting an equal quantity outside
the parentheses. Study the abstract version and the particular
example closely to understand what is meant.
6. The trinomial inside the parentheses is a perfect square:

End Complete Square

What does this accomplish? Observe that the equations in (6) can be
rewritten as
AX2= C  (7)

where X is a linear polynomial. This kind of equation can easily be
solved as follows:

This sequence of steps can be carried out in every case.

Let’s illustrate by continuing to solve the equation carried in the completion
of the square algorithm.
Example 7.4. Solve the equation 2x2+ 12x − 3 = 0.
Example 7.5. Solve by completing the square: 3x2+ 2x − 5 = 0.
Exercise 7.8. Solve each of the following by completing the square.
(a) 8x2− 2x − 1 = 0 (b) 3x2+ 5x+2 = 0 (c) x2+ x − 1 = 0

All the above examples and exercises were done exacly the same way.
The method of completion of the square is a useful tool when solving
quadratic equations, but it is not the most efficient method. The next
section on the Quadratic Formula is a more standard tool than is
completing the square.

Despite its inefficiencies, completion of squares is still is useful technique
to know as it is use elsewhere in mathematics.

The solutions of equation (4) can be found in a more direct way than
the method of factorization or completing the square by using the
so-called Quadratic Formula. Let’s state/prove this formula.

ax2+ bx + c = 0  a ≠ 0  (8)

(1) If b2− 4ac < 0, (8) has no solutions;
(2) if b2− 4ac = 0, (8) has only one solution;
(3) if b2− 4ac > 0, (8) has two distinct solutions.
In the latter two cases, the solutions are given by the Quadratic
Formula:
(9)

Proof.
Theorem Notes: The expression b2−4ac is called the discriminant for
the quadratic equation. It can be used, at casual glance, to determine
whether a given equation has one, two, or no solutions.
•The discriminant is a handy way of classifying a polynomial
P(x) = ax2+ bx + c as ir reducible or not . The polynomial P(x) is
irreducible if and only if its discriminant is negative: b2− 4ac < 0.•

Quiz. Using the discriminant, b2−4ac, respond to each of the following
questions.
1. Is the quadratic polynomial x2− 4x + 3 irreducible?
(a) Yes (b) No
2. Is the quadratic polynomial 2x2− 4x + 3 irreducible?
(a) Yes (b) No
3. How many solutions does the equation 2x2− 3x − 2 = 0 have?
(a) none (b) one (c) two

EndQuiz.
Let’s go to the examples.

Example 7.6. Solve each of the following using the Quadratic
Formula

(a) x2− 5x+6 = 0 (b) x2+ 4x+4 = 0
(c) 6x2− x − 2 = 0 (d) 3x2− 3x+1 = 0

Exercise 7.9. Using the quadratic formula, solve each of the following.
(a) 2x2+ 5x − 12 = 0 (b) 3x2− 7x+1 = 0
(c) x2+1 = 0 (d) x2+ x = 3

Recognition. One problem students have is recognizing a quadratic
equation. This is especially true when the equation has several symbolic
expressions in it. Basically, a quadratic equation in x is an
equation in which x2appears as a term and x appears as a term.
The coefficient of the x2term is the value of a; the coefficient of the
x term is the value of b; and all other terms comprise the value of c.
The symbol x may be some other letter like y or z, but it can also be
a compound symbol like x2, y3, or even something like sin(x)!

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