x∣
2
−3∣x∣+2=0
or, (∣x∣−2)(∣x∣−1)=0
or, ∣x∣=2 and ∣x∣=1
or, x=±2 and x=±1.
So we have 4 real solutions
Author:
skyqleo
Rate an answer:
10Answer:
x=2 and x=1
Step-by-step explanation:
x∣
2
−3∣x∣+2=0
or, (∣x∣−2)(∣x∣−1)=0
or, ∣x∣=2 and ∣x∣=1
or, x=±2 and x=±1.
Author:
rogelioasg0
Rate an answer:
5