Solving Inequalities: An Overview (page 2 of 3)
The previous inequalities are called "linear" inequalities because we are dealing with linear expressions like "x – 2" ("x > 2" is just "x – 2 > 0", before you finished solving it). 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 where y = 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.
I get x2
+ 2 = (x –
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,
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:
The particular solution format you use will depend on your text, your teacher, and your taste. Each format is equally valid. Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
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.
Then the solution of x2 – 3x + 2 > 0 are the two intervals with the "plus" signs:
(negative infinity, 1) and (2, positive infinity).
First I find the zeroes,
which are the endpoints of the intervals: y
= –2x2 + 5x + 12 =
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
is negative by the Factor Method, I just solve each factor: –2x
– 3 is positive for
– 3 > 0, –3 > 2x, –3/2 > x,
– 4 is positive for
– 4 > 0,
Then the solution to this inequality is all x's in
(negative infinity, –3/2 ] and [4, positive infinity).