How do you solve and write the following in interval notation: # x^4+4x^3-21x^2>=0#?

2 Answers
Jun 9, 2016

Answer:

#x le 7 uu x ge 3#

Explanation:

# x^4+4x^3-21x^2>=0->x^2(x^2+4x-21)#

but

#x^2 ge 0 forall x in RR# so
#x^2+4x-21=(x+7)(x-3) ge 0# or finally
#x le 7 uu x ge 3#

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Jun 9, 2016

Answer:

#(-oo,-7]uu[3,+oo)#

Explanation:

Factor this to find when the expression is equal to #0#. From there, we will create a sign chart to see when the expression is greater than #0#.

Factor #x^2# from each term.

#x^2(x^2+4x-21)>=0#

Factor with #7# and #-3#.

#x^2(x+7)(x-3)>=0#

Thus the expression equals #0# at #x=0#, #x=-7#, and #x=3#.

Use one test point in every interval between two pairs of zeros, and another test point between a zero and positive/negative infinities.

Before #x=-7#:

We can plug in #x=-8# into #x^2(x+7)(x-3)#.
#(-8)^2# is positive, #(-8+7)# is negative, and #(-8-3)# is negative, so their product will be positive. Thus the entire interval #(-oo,-7]# will be #>=0#.

Between #x=-7# and #x=0#:

Plugging in #x=-1# yields:

#overbrace((-1)^2)^+overbrace((-1+7))^(+)overbrace((-1-3))^(-) <=0#

This is negative, so the interval is not a part of the solution set.

Between #x=0# and #x=3#:

#overbrace((1)^2)^+overbrace((1+7))^(+)overbrace((1-3))^(-) <=0#

This is also negative, and not a part of the solution set.

From #x=3# to infinity:

Plugging in #x=4# yields:

#overbrace((4)^2)^+overbrace((4+7))^(+)overbrace((4-3))^+ >=0#

Since this is positive, #[3,+oo)# is also a part of the solution set.

Thus the entire solution is the union of the two positive intervals:

#(-oo,-7]uu[3,+oo)#