# How do you solve #x^ { 2} ( x ^ { 2} - 25) \geq 0#?

##### 1 Answer

#### Explanation:

First, let's find the points where the function EQUALS

This is because whenever the function is

#x^2(x^2-25) = 0#

#x^2(x-5)(x+5) = 0#

#x = 0, 5, or -5#

This means that the function equals

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Let's use

#(-6)^2((-6)^2 - 25)#

#36 (36 - 25)#

#396#

This is a positive number, so the interval **positive**.

Let's use

#(-1)^2((-1)^2-25)#

#(1)(1-25)#

#-24#

This is a negative number, so the interval **negative**.

Let's use

#(1)^2((1)^2-25)#

#1(1-25)#

#-24#

This is a negative number, so the interval **negative**.

Let's use

#(6)^2((6)^2 - 25)#

#36(36 - 25)#

#396#

This is a positive number, so the interval **positive**.

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Remember, we're looking for when the function is **greater than or equal to** zero. This means that our solution will include all of the ZEROES (-5, 0, and 5) as well as all of the **positive** intervals.

So our solution is:

#x = -5, " "x = 0, " "x = 5#

#x < -5, " " x > 5#

We can combine the equalities and inequalities for

#x le -5, " " x = 0, " " x ge 5#

*Final Answer*

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BONUS: Here's a graph of the function

graph{x^2(x^2 - 25) [-10, 10, -200, 200]}