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# 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 - r1)(x - r2)...(x - rn), r1 < r2 < r3 < ... < rn

the test intervals are

(-∞, r1), (r1, r2), ..., (rn-1, rn), and (rn, )

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 x2 < x + 6.

Solution

 x2 < x + 6 Original inequality x2 - x - 6 < 0 Write in standard form. (x - 3)(x + 2) < 0 Factor.

The polynomial x2 - x - 6 has x = -2 and x = 3 as its zeros. Thus, you can solve the inequality by testing the sign of x2 - 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 x2 - 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.