Question 335605
{{{ 2x^2 + 9x + 4 < 0}}}
{{{(2x+1)(x+4)<0}}}
Break up the number line into 3 regions.
Region 1:({{{-infinity}}},{{{-4}}})
Region 2:({{{-4}}},{{{-1/2}}})
Region 3:({{{-1/2}}},{{{-infinity}}})
Choose a point in each region (not an endpoint) and test the inequality.
If the point satisifes the inequality, that region is part of the solution region.
Region 1: {{{x=-5}}}
{{{(2x+1)(x+4)<0}}}
{{{(2(-5)+1)(-5+4)<0}}}
{{{(-9)(-1)<0}}}
{{{9<0}}}
False, this region is not part of the solution.
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Region 2: {{{x=-3}}}
{{{(2x+1)(x+4)<0}}}
{{{(2(-3)+1)(-3+4)<0}}}
{{{(-5)(1)<0}}}
{{{-5<0}}}
True, this region is part of the solution.
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Region 3: {{{x=0}}}
{{{(2x+1)(x+4)<0}}}
{{{(1)(4)<0}}}
{{{4<0}}}
False, this region is not part of the solution.
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The solution region is ({{{-4}}},{{{-1/2}}})
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Graphical verification by graphing {{{2x^2+9x+4}}} and looking for the region where the function is less than zero (below the x-axis)
{{{drawing(300,300,-5,5,-5,5,grid(1),blue(line(-4,-10,-4,10)),blue(line(-1/2,10,-1/2,-10)),graph(300,300,-5,5,-5,5,2x^2+9x+4))}}}