Question 362439
{{{x^4 - 32x^2 - 144 > 0}}}
{{{(x^2-36)(x^2+4)>0}}}
{{{(x-6)(x+6)(x^2+4)>0}}}
Break up the number line into 4 regions using the critical points of the function.
Region 1:({{{-infinity}}},{{{-6}}})
Region 2:({{{-6}}},{{{0}}})
Region 3:({{{0}}},{{{6}}})
Region 4:({{{6}}},{{{infinity}}})
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For each region, choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution region.
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Region 1:{{{x=-7}}}
{{{(x-6)(x+6)(x^2+4)>0}}}
{{{(-7-6)(-7+6)((-7)^2+4)>0}}}
{{{(-13)(-1)(53)>0 }}}
{{{689>0 }}}
True, Region 1 is part of the solution region.
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Region 2:{{{x=-1}}}
{{{(x-6)(x+6)(x^2+4)>0}}}
{{{(-1-6)(-1+6)((-1)^2+4)>0}}}
{{{(-7)(5)(5)>0 }}}
{{{-175>0 }}}
False, Region 2 is part of the solution region.
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Region 3:{{{x=1}}}
{{{(x-6)(x+6)(x^2+4)>0}}}
{{{(1-6)(1+6)((1)^2+4)>0}}}
{{{(-5)(7)(5)>0 }}}
{{{-175>0 }}}
False, Region 3 is not part of the solution region.
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Region 4:{{{x=7}}}
{{{(x-6)(x+6)(x^2+4)>0}}}
{{{(7-6)(7+6)((7)^2+4)>0}}}
{{{(1)(13)(53)>0 }}}
{{{689>0 }}}
True, Region 4 is part of the solution region.
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Solution Region:  ({{{-infinity}}},{{{-6}}}) U ({{{6}}},{{{infinity}}})
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Graphical verification: Look for regions where the function is above the x-axis ({{{y>0}}})
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{{{drawing(300,300,-10,10,-600,600,grid(1),blue(line(-6,600,-6,-600)),blue(line(6,600,6,-600)),blue(line(0,500,0,-500)),graph(300,300,-10,10,-600,600,x^4 - 32x^2 - 144))}}}