Solving an inequality

Question

Solve the following:

a) 25 ≤ x²

b) (2x-3)/(x+1) ≥ 1

Answer 1

a. 25≤x²    25-x²≤0

 

               (5-x)(5+x)≤0

 

    (5-x)(5+x)=0 when x = 5 or -5

 

    If x<-5, then (5-x)(5+x)<0

    If -5<x<5, then (5-x)(5+x)>0

    If x>5, then (5-x)(5+x)<0

 

    Solution:  (-∞,-5]∪[5,∞)

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b. (2x-3)/(x+1)≥1

 

   (2x-3)/(x+1) - 1 ≥ 0

 

   [(2x-3)-(x+1)]/(x+1) ≥ 0

 

   (x-4)/(x+1) ≥ 0

 

    Numerator=0 when x=4  Denominator=0 when x=-1

 

   If x<-1, then x-4<0 and x+1<0.  So (x-4)/(x+1) > 0.

 

   If -1<x<4, then x-4<0 and x+1>0.  So, (x-4)/(x+1) < 0.

 

   If x>4, then x-4>0 and x+1>0.  So, (x-4)/(x+1) > 0.

 

   Solution set:  (-∞, -1) ∪ [4,∞)

 

Answer 2

a) 25 ≤ x²

 

This is the same as

 

x² ≥ 25

x ≥ 5

or

- x ≥ 5

x ≤ -5

 

Final answer:  x ≥ 5 or x ≤ -5

 

b.  (2x-3)/(x+1) ≥ 1

 

2x-3 ≥ x+1

 

x ≥ 4