TUTORIALS:

Solving Quadratic Inequalities
To solve inequalities involving polynomials of
higher degree, use the fact that a polynomial can change signs only at its real zeros
(the numbers that make the polynomial zero). Between two consecutive real zeros, a
polynomial must be either entirely positive or entirely negative. This means that when
the real zeros of a polynomial are put in order, they divide the real line into test
intervals in which the polynomial has no sign changes. Thus, if a polynomial has the
factored form
(x  r_{1})(x  r_{2})...(x  r_{n}), r_{1} <
r_{2} < r_{3} < ... < r_{n}
the test intervals are
(∞, r_{1}), (r_{1},
r_{2}), ..., (r_{n1}, r_{n}), and (r_{n},
∞)
To determine the sign of the polynomial in each test interval, you need to test only one
value from the interval.
Example
Solving a Quadratic Inequality
Solve x^{2} < x + 6.
Solution
x^{2} 
< x + 6 
Original inequality 
x^{2}  x  6 
< 0 
Write in standard form. 
(x  3)(x + 2) 
< 0 
Factor. 
The polynomial x^{2}  x  6 has x = 2 and x = 3 as its zeros. Thus, you can solve
the inequality by testing the sign of x^{2}  x  6 in each of the test intervals
(∞, 2), (2, 3),
and (3, ∞). To test an interval, choose any number in the
interval and compute the sign of x^{2}  x  6. After doing this, you will find that the
polynomial is positive for all real numbers in the first and third intervals and negative
for all real numbers in the second interval. The solution of the original inequality is
therefore (2, 3) as shown in the figure below.
