Group 1
1. 𝑥2
𝑥+2−1
𝑥−2 =8
𝑥2−4
Solution:
LCD: (𝑥+2)(𝑥−2)
(𝑥+2)(𝑥−2)[ 𝑥2
𝑥+2−1
𝑥−2=8
𝑥2−4](𝑥+2)(𝑥−2)
(𝑥+2)(𝑥−2)( 𝑥2
𝑥+2)−(𝑥+2)(𝑥−2)(1
𝑥−2)=(𝑥+2)(𝑥−2)(8
𝑥2−4)
(𝑥−2)(𝑥2)−(𝑥+2)(1)=8
𝑥3−2𝑥2−(𝑥+2)= 8
𝑥3−2𝑥2−𝑥−2−8 =0
𝒙𝟑−𝟐𝒙𝟐−𝒙−𝟏𝟎= 𝟎
Answer: There is at most one ± real solution.
2. 3
𝑥−2 <1
𝑥
Solution:
3
𝑥−2−1
𝑥<0
3(𝑥)−(𝑥−2)
𝑥(𝑥−2) < 0
3𝑥−𝑥+2
𝑥2−2𝑥 <0
2𝑥+2
𝑥2−2𝑥< 0
Intervals
𝒙< −𝟏
−𝟏< 𝒙< 𝟎
𝟎< 𝒙< 𝟐
𝒙> 𝟐
Test Points
−𝟐
−𝟏
𝟐
𝟏
𝟑
2𝑥+2
𝑥2−2𝑥< 0
2(−2)+2
(−2)2−2(−2)<0
−2
8<0
−𝟏
𝟒<𝟎
2(−1
2)+2
(−1
2)2−2(−1
2)<0
1
5
4<0
𝟓
𝟒<𝟎
2(1)+2
(1)2−2(1)<0
𝟒
−𝟏< 𝟎
−𝟒< 𝟎
2(3)+2
(3)2−2(3)<0
𝟖
𝟑<𝟎
Conclusion
TRUE
FALSE
TRUE
FALSE
Answer: {𝒙 ∈ ℝ|𝒙≤ −𝟏 𝒐𝒓 𝟎 < 𝒙 < 𝟐}
Get the meaningful zeroes:
2𝑥+2=0 𝑥2−2𝑥= 0
2𝑥 = −2 𝑥(𝑥−2)=0
2𝑥
2=−2
2 𝒙= 𝟎 | 𝑥−2= 0
𝒙= −𝟏 𝒙=𝟐
𝒙< −𝟏
−𝟏< 𝒙< 𝟎
𝟎< 𝒙< 𝟐
𝒙>𝟐
