Question

Solution set of x(x^2+x+1)≤0 is​

Answer 1

Answer:

Step-by-step explanation:

Answered Mar 1, 2015

The question is  

x(x-1)(x+2)>0

Firt ignore the '>' sign:

so:   x(x-1)(x+2)=0

Since 3 numbers are multiplied to give 0, then each may potentially be zero, so:  

  x = 0

  x - 1 = 0,  which means x = 1

  x + 2 = 0, which means x = -2

So, you have 3 numbers  -2, 0, 1  

Now, draw a straight line and mark these 3 points on the line. You can see now that it these numbers have divided the line into 4 different regions. Region 1 is (-inf, -2) -- that is all numbers to the left of -2.

Region 2 is (-2,0) ---  that is all numbers between -2 and 0

Region 3 is (0,1) --- that is all numbers between 0,1

Region 4 is (1,inf)--- that is all numbers to the right of 1

 

Your strategy now, is to test a number inside of each of these 4 regions. Basically you pick a number inside, but not the end values. You substitute the number into the original inequality for x, and see which ones are true.

Hence:

Region 1: (-inf, -2),  

I can choose -5 since its in the region to the left of -2.

I get:        

(-5)(-5-1)(-5+2) >? 0       I used '?' because I am not sure if it is true

so:  (-5)(-6)(-3) >? 0  

Finally   -90 >? 0  is false. SO Mark this region with an X on your line

Region 2: (-2,0),  

I can choose -1 since its between -2 and 0.

I get:        

(-1)(-1-1)(-1+2) >? 0       I used '?' because I am not sure if it is true

so:  (-1)(-2)(1) >? 0  

Finally   2 >? 0  is True. SO Mark this region with an check mark on your line

Region 3: (0,1),  

I can choose 0.5 since its between 0 and 1.

I get:        

(0.5)(0.5-1)(0.5+2) >? 0       I used '?' because I am not sure if it is true

so:  (0.5)(-0.5)(2.5) >? 0  

Finally   -0.625>? 0  is false. SO Mark this region with an X on your line

Finally:  

Region 4: (1,inf),  

I can choose 5 since its in the region to the right of 1.

I get:        

(5)(5-1)(5+2) >? 0       I used '?' because I am not sure if it is true

so:  (5)(4)(7) >? 0  

Finally   140 >? 0  is true. SO Mark this region with a check on your line

Finally: We have 2 regions that were true (check marks)

Regions 2 and 4:  (-2,0) and (1,inf)

We write them together using the U (for union)

Final answer:   (-2,0) U (1,inf)