8.2 CHAPTER 8. SOLVINGQUADRATIC INEQUALITIES
f(x) sign of f(x)
Region A x < 2 f(1) = 2 +
Region B x = 2 f(2) = 0 +
Region C 2 < x < 3 f(2,5) =− 2 , 5 -
Region D x = 3 f(3) = 0 +
Region E x > 3 f(4) = 2 +
We see that the functionis positive for x≤ 2 and x≥ 3.
Step 5 : Write the final answerand represent on a number line
We see that x^2 − 5 x + 6≥ 0 is true for x≤ 2 and x≥ 3.
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1 2 3 4
Example 3: Solving Quadratic Inequalities
QUESTION
Solve the quadratic inequality−x^2 − 3 x + 5 > 0.
SOLUTION
Step 1 : Determine how to approach the problem
Let f(x) =−x^2 − 3 x + 5. f(x) cannot be factorised so, use the quadratic
formula to determine the roots of f(x). The x-intercepts are solutionsto the
quadratic equation
−x^2 − 3 x + 5 = 0
x^2 + 3x− 5 = 0
∴ x =
− 3 ±
�
(3)^2 − 4(1)(−5)
2(1)
=
− 3 ±
√
29
2
x 1 =
− 3 −
√
29
2
=− 4 , 2
x 2 =
−3 +
√
29
2
= 1, 2
Step 2 : Determine which ranges correspond to the inequality
We need to figure out which values of x satisfy the inequality. From the answers
we have five regions toconsider.
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