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

1 Answer
John D.
Feb 13, 2018

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

Explanation:

First, let's find the points where the function EQUALS #0#.

This is because whenever the function is #0#, it will be switching from negative to positive, and then we can plug in test points to figure out which regions are positive, which we are looking for.

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

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

#x = 0, 5, or -5#

This means that the function equals #0# at these points, so between these points, we need to test to find whether the intervals are positive or negative. If the test point is positive, the entire interval must be positive (since it cannot cross #0#), and vice versa.

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Let's use #-6# as a test point for the interval #(-oo, -5)#

#(-6)^2((-6)^2 - 25)#
#36 (36 - 25)#
#396#

This is a positive number, so the interval #(-oo, -5)# is positive.

Let's use #-1# as a test point for the interval #(-5, 0)#

#(-1)^2((-1)^2-25)#
#(1)(1-25)#
#-24#

This is a negative number, so the interval #(-5,0)# is negative.

Let's use #1# as a test point for the interval #(0, 5)#

#(1)^2((1)^2-25)#
#1(1-25)#
#-24#

This is a negative number, so the interval #(0, 5)# is negative.

Let's use #6# as a test point for the interval #(5, oo)#

#(6)^2((6)^2 - 25)#
#36(36 - 25)#
#396#

This is a positive number, so the interval #(5, oo)# is 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 #-5# and #5#:

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

Final Answer

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BONUS: Here's a graph of the function #y = x^2(x^2 - 25)# to illustrate where the function is greater than or equal to #0#:

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

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