SOLUTION: solve the following inequality: x^2|x-5|>6x. (X squared times the absolute value of x-5 is greater than 6x). Show your work.

We get 0 on the right




Let's find all the critical numbers, which are the

solutions to the equality



That breaks into two cases

(1)     

and     

(2)  

-----------------

For case (1)

(1)  

     

     

          

    x=0; x-6=0;  x+1=0
           x=6;    x=-1  

That gives critical numbers 0, 6, -1

----

For case (2)

(2)  

(2)  

     

     

     

          

    x=0; x-3=0;  x-2=0
           x=3;    x=2  

That gives critical numbers 0, 3, 2

So from both cases we have critical numbers
in order from smallest to largest

-1, 0 , 2, 3, 6

We put those all on a number line

----------o--o-----o--o--------o---------
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

We test the original inequality in all intervals
between or beyond all those critical numbers.

Left of -1, we choose test value -2, and substitute it in








That's true so we shade the part of the number line 
left of -1

<=========o--o-----o--o--------o---------
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

---
Between -1 and 0, we choose test value -0.5, and substitute it in








That's also true so we shade the part of the number line 
between -1 and 0

<=========o==o-----o--o--------o---------
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

---

Between 0 and 2, we choose test value 1, and substitute it in







That's FALSE so we do not shade the part of the number line 
between 0 and 2.  So we still have:

<=========o==o-----o--o--------o---------
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

---

Between 2 and 3, we choose test value 2.5, and substitute it in








That's TRUE so we shade the part of the number line 
between 2 and 3

<=========o==o-----o==o--------o---------
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

---

Between 3 and 4, we choose test value 4, and substitute it in








That's FALSE so we do not shade the part of the number line 
between 3 and 6.  So we still have:

<=========o==o-----o==o--------o---------
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

---

Right of 6, we choose test value 7, and substitute it in








That's true so we shade the part of the number line 
right of 6

<=========o==o-----o==o--------o========>
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

Finally we test the critical numbers:

We test -1






So the critical value -1 is a solution so we shade -1

<============o-----o==o--------o========>
-4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9

I'll leave it up to you to test the others critical
numbers.  You'll find that none of the other critical
numbers are solutions by substituting them into the
original inequality.  But you should test them.

In set builder notation the solution is:

{x|x<0,26}

In interval notation, the solution is:

(,0)U(2,3)U(6,) 

Edwin