In this section, you will learn how so solve inequalities. "Solving'' an inequality means finding all of its solutions. A "solution'' of an inequality is a number which when substituted for the variable makes the inequality a true statement.
Here is an example: Consider the inequality
x – 2 > 5
When we substitute 8 for x, the inequality becomes 8-2 > 5. Thus, x=8 is a solution of the inequality. On the other hand, substituting -2 for x yields the false statement (-2)-2 > 5. Thus x = -2 is NOT a solution of the inequality. Inequalities usually have many solutions.
As in the case of solving equations, there are certain manipulations of the inequality which do not change the solutions. Here is a list of "permissible'' manipulations:
Rule 1. Adding/subtracting the same number on both sides.
Example: The inequality x-2>5 has the same solutions as the inequality x > 7. (The second inequality was obtained from the first one by adding 2 on both sides.)
Rule 2. Switching sides and changing the orientation of the inequality sign.
Example: The inequality 5-x> 4 has the same solutions as the inequality 4 < 5 - x. (We have switched sides and turned the ``>'' into a ``<'').
Last, but not least, the operation which is at the source of all the trouble with inequalities:
Rule 3a. Multiplying/dividing by the same POSITIVE number on both sides.
Rule 3b. Multiplying/dividing by the same NEGATIVE number on both sides AND changing the orientation of the inequality sign.
Examples: This sounds harmless enough. The inequality 2x £ 6 has the same solutions as the inequality x £ 3. (We divided by +2 on both sides).
The inequality -2x > 4 has the same solutions as the inequality x< -2. (We divided by (-2) on both sides and switched ">'' to "<''.)
But Rule 3 prohibits fancier moves: The inequality x2 > xDOES NOT have the same solutions as the inequality x > 1. (We were planning on dividing both sides by x, but we can't, because we do not know at this point whether x will be positive or negative!) In fact, it is easy to check that x = -2 solves the first inequality, but does not solve the second inequality.
Only ``easy'' inequalities are solved using these three rules; most inequalities are solved by using different techniques.
The basic strategy for inequalities and equations is the same: isolate x on one side, and put the "other stuff" on the other side. Following this strategy, let's move +5 to the right side. We accomplish this by subtracting 5 on both sides (Rule 1) to obtain
(2x + 5) – 5 < 7 - 5
after simplification we obtain
2x < 2
Once we divide by +2 on both sides (Rule 3a), we have succeeded in isolating x on the left: (2x/2) < (2/2)
or simplified, x < 1
All real numbers less than 1 solve the inequality. We say that the "set of solutions'' of the inequality consists of all real numbers less than 1. In interval notation, the set of solutions is the interval (-¥, 1)l.
The previous inequalities are called "linear" inequalities because we are dealing with linear expressions like "3x + 2 > 5". When we have an inequality with "x2" as the highest-degree term, it is called a "quadratic inequality". The method of solution is more complicated.
First, I have to find the x-intercepts of the associated quadratic, because the intercepts are wherey = x2 – 3x + 2 is equal to zero. Graphically, an inequality like this is asking me to find where the graph is above or below the x-axis. It is simplest to find where it actually crosses the x-axis, so I'll start there.
Factoring, I get x2 – 3x + 2 = (x– 2)(x– 1) = 0, so x = 1 or x = 2. Then the graph crosses the x-axis at 1 and 2, and the number line is divided into the intervals (negative infinity, 1), (1, 2), and (2, positive infinity). Between the x-intercepts, the graph is either above the axis (and thus positive, or greater than zero), or else below the axis (and thus negative, or less than zero).
There are two different algebraic ways of checking for this positivity or negativity on the intervals. I'll show both.
1) Test-point method. The intervals between the x-intercepts are (negative infinity, 1), (1, 2), and (2, positive infinity). I will pick a point (any point) inside each interval. I will calculate the value of y at that point. Whatever the sign on that value is, that is the sign for that entire interval.
For (negative infinity, 1), let's say I choose x = 0; then y = 0 – 0 + 2 = 2, which is positive. This says that y is positive on the whole interval of (negative infinity, 1), and this interval is thus part of the solution (since I'm looking for a "greater than zero" solution).
For the interval (1, 2), I'll pick, say, x = 1.5; then y = (1.5)2– 3(1.5) + 2 = 2.25 – 4.5 + 2 = 4.25 – 4.5 = –0.25, which is negative. Then y is negative on this entire interval, and this interval is then not part of the solution.
For the interval (2, positive infinity), I'll pick, say, x = 3; then y = (3)2– 3(3) + 2 = 9 – 9 + 2 = 2, which is positive, and this interval is then part of the solution. Then the complete solution for the inequality is x < 1 and x > 2. This solution is stated variously as:
x < 1, x > 2
inequality notation
x Î (-¥, 1)È(2,+¥)
interval, or set, notation
number line with parentheses
(brackets are used
for closed intervals)
number line with open dots
(closed dots are used
for closed intervals)
2) Factor method. Factoring, I get y = x2 – 3x + 2 = (x– 2)(x– 1). Now I will consider each of these factors separately.
The factor x– 1 is positive for x> 1; similarly, x– 2 is positive for x> 2. Thinking back to when I first learned about negative numbers, I know that (plus)×(plus) = (plus), (minus)×(minus) = (plus), and (minus)×(plus) = (minus). So, to compute the sign on y = x2 – 3x + 2, I only really need to know the signs on the factors. Then I can apply what I know about multiplying negatives.
First, I set up a grid, showing the factors and the number line.
Now I mark the intervals where each factor is positive.
Where the factors aren't positive, they must be negative.
Now I multiply up the columns, to compute the sign of y on each interval.
Then the solution of x2 – 3x + 2 > 0 are the two intervals with the "plus" signs:
(negative infinity, 1) and (2, positive infinity).
Solve –2x2 + 5x + 12 < 0.
First I find the zeroes, which are the endpoints of the intervals:y = –2x2 + 5x + 12 =
(–2x – 3)(x – 4) = 0 for x = –3/2 and x = 4. So the endpoints of the intervals will be at –3/2 and 4. The intervals are between the endpoints, so the intervals are (negative infinity,–3/2], [–3/2, 4], and [4, positive infinity). (Note that I use brackets for the endpoints in "or equal to" inequalities, instead of parentheses, because the endpoints will be included in the final solution.)
To find the intervals where y is negative by the Test-Point Method, I just pick a point in each interval. I can use points such as x = –2, x = 0, and x = 5.
To find the intervals where y is negative by the Factor Method, I just solve each factor: –2x – 3 is positive for –2x – 3 > 0, –3 > 2x, –3/2 > x, orx < –3/2; and x – 4 is positive for x – 4 > 0, x > 4. Then I fill out the grid:
Then the solution to this inequality is all x's in
(negative infinity, –3/2 ] and [4, positive infinity).
Today, there have been 143428 visitors (401177 hits) on this page! CONGRATULATION
