In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Use algebra to determine the set of values of x for which
∣x+8∣x2−9>6−2x
(6)
解答
Given the inequality:
∣x+8∣x2−9>6−2x
A critical value occurs where the denominator is zero, so x=−8. Since ∣x+8∣>0 for all x=−8, we can multiply both sides by ∣x+8∣ without flipping the inequality sign.
Case 1: x>−8
Here, ∣x+8∣=x+8. The inequality becomes:
x2−9>x2−9>x2−9>3x2+10x−57>(x+8)(6−2x)6x−2x2+48−16x−2x2−10x+480
Factorising the quadratic:
(3x+19)(x−3)>0
The critical values are x=−319 and x=3.
This quadratic is positive for x<−319 or x>3.
Since we are in the case x>−8, the valid regions are −8<x<−319 or x>3.
Case 2: x<−8
Here, ∣x+8∣=−(x+8). The inequality becomes:
x2−9>x2−9>x2−9>x2−9>0>x2+10x−39<−(x+8)(6−2x)(−x−8)(6−2x)−6x+2x2−48+16x2x2+10x−48x2+10x−390
Factorising the quadratic:
(x+13)(x−3)<0
The critical values are x=−13 and x=3.
This quadratic is negative for −13<x<3.
Since we are in the case x<−8, the valid region is −13<x<−8.
Conclusion
Combining the valid regions from both cases, the set of values for x is:
−13<x<−8or−8<x<−319orx>3
