Solving Equations & Inequalities

7.9. Solution to (a): This method begins by factoring the algebraic
expressions:
(x − 1)(x − 2) ≥ 0  (S-3)

The idea is to analyze when each factor is positive and when each
factor is negative. The Sign Chart is a graphical scheme for storing
all the information. Here’s the Sign Chart for (x − 1)(x − 2):
The Sign Chart of (x − 1)(x − 2)

Sign Chart of x − 1. We ask the question, where is x − 1 > 0?
Answer: x > 1. We mark that in blue on the axis. (When using pencil
and paper, I write little ‘+’ signs over the axis.) We now ask the
question, where is x−1 < 0? Answer: x < 1. We mark that in red on
the axis. (Pencil and paper? Write little ‘−’ signs over the axis.) And,
of course, x − 1 = 0 when x = 1. This represents a complete analysis
of the factor x−1: when its positive, when its negative, and when its
zero.

Sign Chart of x − 2. We do the same analysis for the factor x − 2
indicating the results in blue (or a ‘+’) or in red (or a ‘−’ ).

Sign Chart of (x − 1)(x − 2). To obtain the final sign chart for
the expression of interest, we put the other charts together using the
following principles:
(−)(−) = (+), (+)(−) = (−) and (+)(+) = (+)

For example, if we consider a number x < 1, then, according to my
sign charts, the first factor (x − 1) is negative and the second factor
(x − 2) is negative too; the product (x − 1)(x − 2) is positive and we
indicate that on the sign chart for (x − 1)(x − 2). Get the idea?

The last axis tells us exactlywhen (x−1)(x−2) is positive and when
it is negative. We can see that the solution to the inequality
x² − 3x + 2 ≥ 0

is all x in the blue. Members of the solution set are all x ≤ 1 plus all
x ≥ 2. (We include x = 1 and x = 2 because they make the expression
x² − 3x + 2 = 0 and would therefore satisfy the inequality.)

Presentation of Solution:
Set Notation: { x | x ≤ 1 or x ≥ 2 }
Interval Notation: (−∞, 1 ] ∪ [ 2,+∞)
Solution to: (b) Not as much detail will be given in this solution.

Problem: Solve for x in x < x². The first thing we must do is to
take all quantities to one side of the inequality: x − x² < 0. Next, we
must factor, x(1 − x) < 0. Finally, we apply our Sign Chart Method.

The Sign Chart of x(1 − x)

In the (1 − x) sign chart, we asked the question, when is 1 −x > 0.
The answer is when 1 > x, or x < 1. This is depicted in blue.
Presentation of Solution: Solve x(1 − x) < 0.

Set Notation: { x | x < 0 or x > 1 }
Interval Notation: (−∞, 0 ) ∪ ( 1,+∞)

Here, we exclude the endpoints, 0 and 1, because they do not satisfy
the stated inequality. (These points make x(1 − x) equal zero.)

7.10. Solution: Begin by completely factoring the expression.
(S-4)
We now do a Sign Chart on the left-hand side of (S-4).
The Sign Chart of



The solution set is all real numbers that are in red.
Presentation of Solution:

7.11. Solution:
(a) Solve for x: |x − 4| < 3.
 

	given
	from (16)
	add 4 to all sides

Presentation of Solution:
(b) Solve for x: |3x − 1| < 2.

	given
	from (16)
	add 1 to all sides
	multiply all sides by 1/3

Presentation of Solution:

(c) Solve for x: |2 − 4x| ≤ 5.

	given
	from (16)
	add −2 to all sides
	multiply all sides by −1/4

or

In the last step we have multiplied both sides by a negative
number, this will reverse the direction of the inequality!
Presentation of Solution:

7.12. Solution to (a): Solve for x: |x − 3| > 4.
|x − 3| > 4

Use (17) to split the inequality!
 

	upper inequality		lower inequality
	add 3 both sides		add 3 both sides
	solution set		solution set

Now, join the solutions!
Solution Set =(7,+∞) ∪ (−∞,−1)
Presentation of Solution:

Comments That’s wayit goes: split, solve each, and join byunion!

Solution to (b): Solve for x: |5x + 1| ≥ 3.
|5x + 1| ≥ 3

Use (17) to split the inequality!

	upper inequality		lower inequality
	add −1 both sides		add −1 both sides
	divide by 5		divide by 5
	solution set		solution set

Now, join the solutions!
Solution Set
Presentation of Solution:
 

Important Points

Proof: The Quadratic Formula. The quadratic formula is nothing
more than the formula you get when you complete the square and
solve for x just as we did in the section on completing the square. If
you’ve been reading along this tutorial, the following steps would look
familiar.
 

	given
	add −c to both sides
	factor out a from l.h.s.
	add b2/4a to both sides
	perfect square
	divide by a

All the steps above are reversible. This means that the solution set to
the last equation is the same as the solution set to the first equation.

You can see now that if b² − 4ac < 0, the right-hand side of the last
equation is negative and the left -hand side is not; therefore, there can
be no solutions to the equation.

Now let’s continue the development.
 

	take square root of both sides
	extract root in denominator
	subtract b /2a from both sides
	combine fractions

 (I-1)

Now if b² − 4ac = 0, then equation (I-1) becomes that is,
there is onlyone solution as asserted.

Finally, if b² − 4ac > 0, then and formula (I-1) would
clearlylead to two distinct solutions.

The solution to 1 ≤ x ≥ −1? The answer is [ 1,+∞). Sometimes you
see students write inequalities this way, but this is not a good way of
writing a double inequality!

The question is, what does 1 ≤ x ≥ −1 mean? It means
1 ≤ x and x ≥ −1

Turn around the inequalities to get
x ≥ 1 and x ≥ −1

Now we ask ourselves the question: What are all the numbers x that
are greater than (or equal to) 1 and greater than (or equal to) −1?
But any x satisfying x ≥ 1 automatically satisfies x ≥ −1; therefore,
the important condition is x ≥ 1, hence the solution set in interval
notation is [ 1,+∞)

Students who write inequalities like 1 ≤ x ≥ −1 often mean x ≥ 1
or x ≥ −1 and they just meld the two sets of inequalities together.
However, the double inequality is for pairs of inequalities connected
logically by an “and”. This is different from pairs of inequalities connected
by a logical “or”.

Make a distinction in your mind between the two: “and” ≠ “or” in
mathematics or in English.

I often see students write this kind of inequality. The answer is the
empty solution: Recall the meaning of double inequalities:
−1 ≥ x ≥ 1 means − 1 ≥ x and x ≥ 1
or
x ≤ −1 and x ≥ 1

Now what number, x, is there that is less than or equal to −1 and
greater than or equal to 1? The Answer: There is no such number
that satisfies both inequalities.

What the student actually means is x ≤ −1 or x ≥ 1, but theydo
not write it correctly.

Learn to use the notation to express precisely what you mean.
Again, the answer is no solution: If you went through the (routine)
steps of solving, you would have arrived at 2 ≤ x ≤ 1. What
number x satisfies x ≥ 2 and x ≤ 1? Answer: No number.

However, you need not have bothered to solve an inequality having
the empty solution set. It is immediate from the original inequality:
3 ≤ 2x − 1 ≤ 1

Look at the left-most and right-most expressions. Do you see that
3 ≤ 1?

This implies immediately that there are no solutions.