Question 327286
Solved earlier today
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{{{x^3-16x>0}}}
{{{x(x^2-16)>0}}}
{{{x(x-4)(x+4)>0}}}
Break up the number line into 4 regions:
Region 1 : ({{{-infinity}}},{{{-4}}})
Region 2 : ({{{-4}}},{{{0}}})
Region 3 : ({{{0}}},{{{4}}})
Region 4 : ({{{4}}},{{{infinity}}})
Choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution.
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Region 1:{{{x=-5}}}
{{{x(x-4)(x+4)>0}}}
{{{-5(-5-4)(-5+4)>0}}}
{{{-45>0}}}
False, this region is not part of the solution.
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Region 2:{{{x=-2}}}
{{{x(x-4)(x+4)>0}}}
{{{-2(-6)(2)>0}}}
{{{24>0}}}
True, this region is part of the solution.
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Region 3:{{{x=2}}}
{{{x(x-4)(x+4)>0}}}
{{{2(-2)(6)>0}}}
{{{-24>0}}}
False, this region is not part of the solution.

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Region 4:{{{x=5}}}
{{{x(x-4)(x+4)>0}}}
{{{5(1)(9)>0}}}
{{{45>0}}}
True, this region is part of the solution.
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The solution region is ({{{-4}}},{{{0}}}) U ({{{4}}},{{{infinity}}}).
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Verified graphically
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{{{ graph( 300, 300, -10, 10, -50, 50, x*(x-4)*(x+4))) }}} 
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Similarly,
{{{x^2-5x<=-6}}}
{{{x^2-5x+6<=0}}}
{{{(x-2)(x-3)<=0}}}
Break up the number line into 3 regions:
Region 1 : ({{{-infinity}}},{{{2}}})
Region 2 : ({{{2}}},{{{3}}})
Region 3 : ({{{3}}},{{{infinity}}})
Choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution.
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Region 1:{{{x=0}}}
{{{(x-2)(x-3)<=0}}}
{{{(-2)(-3)<=0}}}
{{{6>0}}}
True, this region is part of the solution.
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Region 1:{{{x=0}}}
{{{(x-2)(x-3)<=0}}}
{{{(-2)(-3)<=0}}}
{{{6<=0}}}
False, this region is not part of the solution.
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Region 2:{{{x=2.5}}}
{{{(x-2)(x-3)<=0}}}
{{{(0.5)(-0.5)<=0}}}
{{{-0.25<=0}}}
True, this region is part of the solution.
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Region 3:{{{x=4}}}
{{{(x-2)(x-3)<=0}}}
{{{(2)(1)<=0}}}
{{{2<=0}}}
False, this region is not part of the solution.
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The solution region is [{{{2}}},{{{3}}}].
Since the inequality includes the "=" the brackets denote that 2 and 3 are included as part of the solution.
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{{{graph(300,300,-2,6,-1,1,(x-2)*(x-3))}}}