Algebraic Equation

Question

((x−5)(2x+3))

-----------------   ≤0

     (x−1)2 

Answer 1

numerator = 0 when x = -3/2 or x = 5

 

denominator = 0 when x = 1

 

Use the values found above to divide the number line into a collection of intervals:

 

  ____________l_______________l_____________l____________

                     -3/2                        1                      5

 

When x < -3/2,  (2x+3)(x-5)/(x-1)² > 0

 

When -3/2 < x < 1, (2x+3)(x-5)/(x-1)² < 0

 

When 1 < x < 5, (2x+3)(x-5)/(x-1)² < 0

 

When x > 5, (2x+3)(x-5)/(x-1)² > 0

 

(2x+3)(x-5)/(x-1)² = 0 when x=3/2 or 5 and is undefined when x = 1.

 

Solution set:  [-3/2, 1) U (1,5]

 

 

Answer 2

[(x-5)(2x-3)]/(x-1)² ≤ 0

 

Multiply both sides by (x-1)²

 

(x-5)(2x-3) ≤ 0

 

This is equal to zero if

x = 5 or x = 3/2

 

Noting that for a product to be negative the

signs must be opposite .. (+)(-) or (-)(+):

 

This is <0 if:

x-5 <0   and  2x-3 > 0

x < 5  and  2x > 3

x < 5  and x > 3/2

 

Or this is < 0 if:

x-5 > 0   and   2x-3 < 0

x > 5      and   2x < 3

x > 5      and   x < 3/2

 

Note, this last interval does not exist siince

no number is greater than 5 and less than 3/2 so

 

Final answer: interval [3/2, 5] ...

                     3/2 ≤ x ≤ 5