By:$Arianne$Diaz$and$Jovanka$Noel$

Quadra&c(Inequali&es(! When%solving%a%quadratic%inequality,%your%goal%is%to%find%all%possible%values%of%the%variable%(x)%which%are%going%to%make%the%inequality%true.%

!  For%example:%x2 + 4x -5 < 0 ! In%this%inequality%you%would%find%all%the%values%of%x%which%make%the%quadratic%less%than%zero.%

Remember:$An%inequality%does%not%say%that%two%values%are%equal,%but%instead%describes%the%relationship%between%them.%

Solving(an(Inequality(Algebraically(!  The%steps%involved%in%solving%a%quadratic%inequality%algebraically%are%as%follows:%

1)  Move%all%values%to%the%left%side%of%the%inequality.%2)  Factor%the%quadratic%into%two%binomials.%

!  From%here%on%out,%the%steps%involved%are%different%depending%on%the%type%of%problem%that%you%have.%

Less(than(problems(•  Let’s%go%back%to%our%sample%problem:%x2 + 4x – 5 < 0 Factored=%(x + 5) (x – 1)%•  There%are%two%ways%in%which%this%quadratic%could%be%less%than%zero%

(negative)%:%M%(x + 5) < 0 and%(x - 1) > 0 or%(x + 5) > 0 and%(x - 1) < 0.%%In%short,%one%factor%has%to%be%negative%while%the%other%is%positive.%%

•  First%case:%(x + 5) < 0 and%(x - 1) > 0 ! This%results%in%%x < -5 and%x > 1 but%this%scenario%is%not$possible$

because%if%x%is%greater%than%one%it%can't%possibly%be%less%than%negative%five.%

•  Second%case:%(x + 5) > 0 and%(x - 1) < 0 !  This%situation%results%in%%x > -5 and%x < 1,%which%is%possible%because%

although%x%is%greater%then%negative%five%that%doesn’t%mean%it%has%to%be%greater%than%one.%

•  Since%the%first%situation%was%not%possible,%the%answer%to%this%inequality%is%-5 < x < 1.%%

Greater(than(problems(•  Now%let’s%use%a%problem%similar%to%our%sample%problem:%x2 + 4x – 5 > 0 Factored=%(x + 5) (x – 1) •  There%are%two%ways%in%which%this%quadratic%could%be%greater%than%zero%

(positive)%–%either%both%of%your%factors%are%positive%or%both%of%them%are%negative.%%Therefore%you%must%test%out%both%scenarios.%%

•  Both%positive:%( x + 5) > 0 and%(x - 1) > 0

! This%results%in%x > -5 and%x > 1,%which%can%be%simplified%to%%%%%%%%%%%%%%%%%x > 1(since%if%x%is%greater%than%one,%it%is%obviously%greater%than%negative%five).%

•  Both%negative:%( x + 5) < 0 and%(x - 1) < 0

! This%results%in%x < -5 and%x < 1,%which%can%be%simplified%to%x < -5%(since%if%x%is%less%than%negative%five%it%is%obviously%less%than%one).%

•  Now%all%you%have%to%do%is%combine%your%two%simplified%answers%and%you%get%the%answer%to%this%question,%which%is%x > 1 or x < -5.%

%%

Sample(problem(#(1(!  So%now%that%you%guys%know%the%basics%about%solving%a%quadratic%inequality%algebraically,%try%a%problem%for%yourselves%"%%

%

x(x+3) > 10

Solu&on(with(work(shown(x(x+3) >10 x2 +3x >10 -10 -10 x2 +3x-10 > 0 Factored:%(x+5)(x-2) > 0 (x+5) > 0 and (x-2) > 0 OR%(x+ 5) < 0 and (x-2) < 0 (x > -5) and (x > 2) OR%(x < -5) and (x < 2) %

(x > -5) and%(x > 2) can%be%further%simplified%because%if%x%is%already%greater%than%2,%then%it%is%surely%greater%than%-5.%This%can%simply%be%stated%as%(x > 2). %(x < -5) and%(x < 2) can%also%be%stated%as%(x < -5) because%is%x%is%less%than%-5 then%it%is%already%less%than%2.%%%Therefore:%(x > 2) or%(x < -5)

Sample(problem(#2(

" Here’s%another%one%to%try%out%^.^%%

-x(2x-14) > 24

Solu&on(with(work(shown(-x(2x-14) > 24 -2x2 +14x > 24 x2 -7x < -12 x2 -7x+12 < 0 Factored:%(x-3)(x-4) < 0 (x-3) < 0 and (x-4) > 0 %OR%(x-3) > 0 and (x-4) < 0 (x < 3) and (x > 4)

OR%(x > 3) and (x < 4)

In%the%first%situation%(x < 3)%and%(x > 4) ,%x%can%not%be%less%than%3%and%greater%than%4%because%there%is%no%such%number.%However,%in%the%second%situation%(x > 3) and%(x < 4),%there%are%numbers%that%are%greater%than%3%and%less%than%4.%These%would%simply%be%all%the%numbers%between%3%and%4.%Therefore,%the%solution%would%be%3 < x < 4.%

Mul&ple(Choice(Ques&ons(1.%%What%is%the%solution%of%the%inequality%x2- x - 6< 0?%1) -3 < x < -2 2) -2 < x < 3 3) 1 < x < 6 4) -3 < x < 2 2.%What%is%the%solution%of%the%inequality%x2 + 2x – 15 < 0?%1) x < -5 or x > 3 2) -5 < x < 3 3) x < -3 or x > 5 4) -3 < x < 5

Mul&ple(Choice(Ques&ons(#2(3.%What%is%the%solution%set%of%x2- 3x -28 > 0?%1) x > 7 or x < -4 2) x < 7 or x > -4 3) -4 < x < 7 4) -4 < x < 7 4.%The%solution%set%of%the%inequality x2- 3x > 10 is….%1) {x| -2 < x < 5} 2) {x| 0 < x < 3} 3) {x| x < -2 or x > 5} 4) {x| x < -5 or x > 2}

Answers(to(mul&ple(choice((1)%%2%2)%2 3)%1 4)%3