Solving Inequalities: An Overview (page 2 of 3) Sections: Linear inequalities, Quadratic inequalities, Other inequalities 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 "x^{2}" as the highestdegree term, it is called a "quadratic inequality". The method of solution is more complicated.
First, I have to find the xintercepts of the associated quadratic, because the intercepts are where y = x^{2} – 3x + 2 is equal to zero. Graphically, an inequality like this is asking me to find where the graph is above or below the xaxis. It is simplest to find where it actually crosses the xaxis, so I'll start there. Factoring,
I get x^{2}
– 3x
+ 2 = (x –
2)
There are two different algebraic ways of checking for this positivity or negativity on the intervals. I'll show both. 1) Testpoint method. The intervals between the xintercepts 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:
The particular solution format you use will depend on your text, your teacher, and your taste. Each format is equally valid. Copyright © Elizabeth Stapel 19992011 All Rights Reserved 2) Factor method. Factoring, I get y = x^{2} – 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 = x^{2} – 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 x^{2} – 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
= –2x^{2} + 5x + 12 = To find the intervals where y is negative by the TestPoint 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,
or x
< –^{3}/_{2};
and x
– 4 is positive for
x
– 4 > 0,
Then the solution to this inequality is all x's in (negative infinity, ^{–3}/_{2 }] and [4, positive infinity). << Previous Top  1  2  3  Return to Index Next >>



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