Necessary Information:
To solve an inequality involving polynomials of degree greater than 1, we shall express each polynomial as a factor (i.e. ax + b and/or ax²+bx+c). If a factor is not zero in an interval then it is either positive or negative throughout the interval. So if we chose any point in the interval, and the factor is positive or negative for x=k , then it is positive or negative throughout the interval. The value of the factor at x=k is called the test value of the factor at test number k. This is shown in the following example
Ex #1 Solving a quadratic inequality
Solve the inequality 2x² - x < 3.
Solution: To use test values you have to have a 0 on one side of the equality sign (exactly like when solving a quadratic equation) do the following
2x² - x < 3 Given
2x² - x - 3 < 0 Make one side 0
(x + 1)(2x - 3) < 0 Factor
The factors x + 1 and 2x - 3 are zero -1 and 3/2, respectively. So the the nonintersecting intervals are
(-inf., -1), (-1, 3/2), and (3/2, inf.)
A way to find the sign of x + 1 and 2x - 3 we can make a sign chart by applying the laws of signs to the product factors the resulting sign is positive or negative according to whether the number of negative signs of factors is even or odd, respectively
Interval "(-inf., -1)" "(-1, 3/2)" "(3/2, inf.)"
sign x + 1 negative positive positive
Sign 2x - 3 negative negative positive
Result positive negative positive
This is a very basic idea on how to solve a quadratic inequality
Alex Florias
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