Algebra Practice Problems for Precalculus and Calculus

1 AlgebraPracticeProblemsforPrecalculusand CalculusSolve 3=5 (x 3)+x=17+3(4 x) (4x 1)( 3x+2)6.(x 1)(x2+x+1)7.(x+1)(x2 x+1)8.(x 2)(x+2)9.(x 2)(x 2)10.(x3+2x 1)(x3 5x2+4) (x)= 1+ (x)=11+ (x)=1 (x)=1 1+ (x)=11+ (x)=x2 3x+4, findandsimplifyf(3),f(a),f( t), andf(x2+1). x 10x+ +10x+ +8x +11x 322. 2x2+7x+ 21 Solve thefollowingquadraticequationsinthreeway s:1)factor, 2)quadraticformula,3) +6x 16=025. x2 3x 2= +2x 4=0 Solve thefollowingsmorgasbordofequationsandine qualities27. x= 2x 128. x2 3= 2x29.|x 5| = +4 331. 2x+4 +4x 3= x 2>0 Add/ +2+3x +1(x 1)(x 2) x3x 3 Simplifythefollowingrationalexpressions( ifpossible) +x 2x2 +5x+6x2 3x+ +2+3x+1x 16,add16tobothsidestoget7x= dividebothsidesby7 togetx= ,weseethat7(3) 16=21 16= 3=5 x, addxtobothsidesandthenadd3 tobothsidestoget3x= dividebothsidesby3 togetx=8/3=2. 6. Checking,weseethat2(8/3) 3=163 3=163 93=73and5 (8/3)=153 83= (x 3)+x=17+3(4 x), wefirstsimplifytheleftandrighthandsidesu singthedistributive propertytoget12x 32+x=17+12 3x.

2 Combininglike termsonbothsidesgives32x 32=29 3x. Now weadd3xand32tobothsides,obtaining92x=612 . Dividingbothsidesby92(ormultiplyingboths idesby29) givesx=612 29=619. CheckingweseethatL H S=12(619 279)+619=12349+619=179+619=789andR H S=17+3(369 619)=17+3 ( 259)=17 759=1539 759= 3, we cross-multiply toobtain5(x 3)=2x. Distributingthe5 gives5x 15=2x. Subtracting2xandadding15tobothsidesgives 3x= givesx= ,weseethat55=1 and25 3=22= s inthedenominatoroffractionsinequations,i tis possiblethatyourfinal solution doesn t satisfytheoriginalequation(becauseyouwou lddividebyzero) soit is reallynota multiplyandsimplify Problems ,wemustmultiplyeachtermintheleft -handfactorwitheachtermintheright-handfa ctor, andthensimplifybycombininglike nomi al bi nomi al, (4x 1)( 3x+2)= 12x2+8x+3x 2= 12x2+11x (x 1)(x2+x+1)=x3+x2+x x2 x 1=x3 (x+1)(x2 x+1)=x3 x2+x+x2 x+1=x3+ (x 2)(x+2)=x2+2x 2x 4=x2 (x 2)(x 2)=x2 2x 2x+4=x2 4x+ (x3+2x 1)(x3 5x2+4)=x6 5x5+4x3+2x4 10x3+8x x3+5x2 4=x6 5x5+2x4 7x3+5x2+8x numberxtobeinthedomainofthefunctionf(x)= 1+x, werequire1+x 0 (sowedon t take thesquarerootofanegative number).

3 Subtracting1 frombothsidesofthisinequalityyieldsx ,ininterval notation,thedomainis theset[ 1, ). numberxtobeinthedomainofthefunctionf(x)= 11+x, werequire1+x6=0 (sowedon t dividebyzero).Thus,werequirex6= notation,thedomainis theset( , 1) ( 1, ). numberxto bein thedomainofthefunctionf(x)=1 x, werequirex>0 (sowedon t dividebyzeroortake thesquarerootofa negative number).Ininterval notation,thedomainis theset(0, ). numberxtobeinthedomainofthefunctionf(x)= 1 1+x, werequire1+x>0 (sowedon t dividebyzeroortake thesquarerootofa negative number).Subtracting1 frombothsidesofthisinequalityyieldsx> ,ininterval notation,thedomainis theset( 1, ). numberxtobeinthedomainofthefunctionf(x)= 11+x2, werequirethat1+x26=0 (sowedon t dividebyzero).Butnomatterwhatxis,1+x2>0. Therefore,thedomainisR, (3)=(3)2 3(3)+4=9 9+4=4,f(a)=a2 3a+4,f( t)=( t)2 3( t)+4=t2+3t+4, andf(x2+1)=(x2+1)2 3(x2+1)+4=x4+2x2+1 3x2 3+4=x4 x2+ , x 20=(x 5)(x+4) 10x+21=(x 3)(x 7) +10x+16=(x+8)(x+2) +8x 105=(x+15)(x 7) +11x 3=(4x 1)(x+3)22.]

4 2x2+7x+15= (2x2 7x 15)= (2x+3)(x 5) 2=(x 2)(x+ 2)tricky, tricky, tricky!!!:) +6x 16=0(a)Byfactoring:x2+6x 16=0 impliesthat(x+8)(x 2)=0. Thus,x= 8 orx=2.(b)Bythequadraticformula:x= b b2 4ac2a= 6 36 4(1)( 16)2(1)= 6 1002= 6 102=2 or 8(c)Bycompletingthesquare:byadding16tobo thsidesofx2+6x 16=0, wegetx2+6x= , adding9=32=(6/2)2to bothsidesmakestheleft-handsidea perfectsquare:x2+6x+9= canfactorthelefthandsidetoget(x+3)2= take thesquarerootofbothsides,allowingforthet wo squarerootsof25ontherighthandsidetoobtai nx+3= frombothsidesgivesx= 3 ,x= 8 orx= :( 8)2+6( 8) 16=64 48 16=0 and(2)2+6(2) 16=4+12 16= x2 3x 2=0(a)Byfactoring: x2 3x 2=0 impliesthatx2+3x+2=0 (multiplybothsidesby-1).Factoringgives(x +1)(x+2)=0. Thus,x= 1 orx= 2.(b)Bythequadraticformula:x= b b2 4ac2a=3 9 4( 1)( 2)2( 1)=3 1 2=3 1 2= 2 or 1(c)Bycompletingthesquare:byadding2 tobothsidesof x2 3x 2=0, weget x2 3x=2.

5 Now, multiplybothsidesby 1 togetx2+3x= 2. Now, adding9/4=(3/2)2tobothsidesmakestheleft- handsidea perfectsquare:x2+3x+94= 2+94=14. We canfactorthelefthandsidetoget(x+32)2=14. Now take thesquarerootofbothsides,allowingforthet wo squarerootsof1/4 ontherighthandsideto obtainx+32= 12. Subtracting3/2 frombothsidesgivesx= 32 12. Inotherwords,x= 1 orx= : ( 1)2 3( 1) 2= 1+3 2=0 and ( 2)2 3( 2) 2= 4+6 2= +2x 4=0(a)Byfactoring:2x2+2x 4=0 impliesthatx2+x 2=0 (dividebothsidesby2). Factoringgives(x+2)(x 1)= ,x= 2 orx=1.(b)Bythequadraticformula:x= b b2 4ac2a= 2 4 4(2)( 4)2(2)= 2 364= 2 64=1 or 2(c)Bycompletingthesquare:byadding4 tobothsidesof2x2+2x 4=0, weget2x2+2x=4. Nowdividebothsidesby2 togive usx2+x=2. Now, adding14=(1/2)2tobothsidesmakestheleft-h andsidea perfectsquare:x2+x+14=2+14=94. We canfactorthelefthandsidetoget(x+12)2=94. Nowtake thesquarerootofbothsides,allowingforthet wo squarerootsof94ontherighthandsidetoobtai nx+12= 32.

6 Subtracting1/2frombothsidesgivesx= 12 32. Inotherwords,x=1 orx= :2(1)2+2(1) 4=2+2 4=0 and2( 2)2+2( 2) 4=8 4 40= x= 2x 1 givesx=2x :L H S= 1=1,R H S= 2(1) 1= 1= x2 3= 2xgivesx2 3=2x. Subtract2xfrombothsidestogetx2 2x 3=0. Theleft-handsidecannowbefactoredtogive(x 3)(x+1)=0, sox=3 orx= :L H S= (3)2 3= 9 3= 6= 2(3)=R H S, however, if youplugx= 1 intoeithersideofthisequation,yougetthesq uarerootofa negative number. Therefore,forus,x= 1 is nota solution(eventhoughtheLHSandRHSareequalt heimaginarynumber 2=i 2).429.|x 5| =4 impliesthateitherx 5=4 orx 5= 4. Thus,eitherx=9 orx= :|9 4| = |4| =4 and|1 5| = | 4| = frombothsidesof2x+4 3 gives2x dividebothsidesby2 togetx 12. Thelogicalsoworksintheotherdirection:ifx 12, thiswillimplythat2x+4 ,thesolutionsetis theinterval[ 12, ). frombothsidesof 2x+4 3 gives 2x dividebothsidesby-2andswitchthedirection oftheinequalitytogetx 12.]

7 Thelogicalsoworksintheotherdirection:ifx 12, thiswillimplythat 2x+4 ,thesolutionsetis theinterval( ,12]. theequationx+4x 3=2 andmultiplybothsidesbyx 3 togetx+4=2(x 3). Nowsolve thisequation:x+4=2x 6= x=10. Checking:10+410 3=14/7= x 2>0 impliesthat(x 2)(x+1) >0. Thus,eitherx 2>0 andx+1>0 ORx 2<0 andx+1<0. Thus,eitherx>2 andx> 1 ORx<2 andx< ,eitherx>2 ORx< :ifx>2 ORx< 1, then(x 2)(x+1) >0 sox2 x 2>0. Thus,thesolutionsetis( , 1) (2, ). needtogeta commondenominator. Thesimplestonetochooseis(x+2)(x 4). Multiplyingthetopandbottomofthefirstfrac tionbyx 4 andmultiplyingthetopandbottomofthesecond fractionbyx+2 andcombiningthefractionsproduces:xx+2+3x 4=x(x 4)(x+2)(x 4)+3(x+2)(x+2)(x 4)=(x2 4x)+(3x+6)(x+2)(x 4)Simplifyingthetopandbottomgivesus:x2 x+6x2 2x 8If therewerea commonfactoronthetopandbottom,wewouldcan celit , ,thisis needtogeta commondenominator. Thesimplestonetochooseis(x 1)(x 2)(x 3).)

8 Multiplyingthetopandbottomofthefirstfrac tionbyx 3 andmultiplyingthetopandbottomofthesecond fractionby(x 1)(x 2)andcombiningthefractionsproducesthefol lowingexpressions(whichareequaltotheorig inal):(x2+1)(x 3)(x 1)(x 2)(x 3) x3(x 1)(x 2)(x 1)(x 2)(x 3)=(x3 3x2+x 3) (x5 3x4+2x3)(x 1)(x 2)(x 3)Simplifyingthetopandbottomgivesus: x5+3x4 x3 3x2+x 3(x 1)(x2 5x+6)= x5+3x4 x3 3x2+x 3x3 6x2+11x 6 Again,wewouldtechnicallyneedtocheckforco mmonfactorstosimplifythis completely .To dothis,it isenoughtodeterminewhether1,2,or3 arezerosofthepolynomialinthenumerator(si ncethey arethezerosofthepolynomialinthedenominat or).Let s callthenumeratorP, soP(x)= x5+3x4 x3 3x2+x (1)= 1+3 1 3+1 3= 46=0,P(2)= 32+48 8 12+2 3= 56=0, andP(3)= 243+243 27 27+3 3= 546=0. Therefore,thereanocommonfactors,sotheans werabove is canfactorthetopandbottomtoobtainx2+x 2x2 1=(x+2)(x 1)(x+1)(x 1)Thereis a commonfactorofx 1, sowecancancelthistoobtainourfinalanswer: x+2x+1It shouldbepointedoutthatthisexpressionis equaltothefirstaslongasx6= factorthenumeratoranddenominatortoobtain :x2+5x+6x2 3x+2=(x+3)(x+2)(x 1)(x 2) , toimmediatelymultiplythetopandbottomofth is double-deckerfraction by(x+2)(x 1).

9 Doingthis,andcancelingthings,weobtain:(x x+2+3)x+1x 1(x+2)(x 1)(x+2)(x 1)=x(x 1)+3(x+2)(x 1)(x+1)(x+2)Simplifying:x2 x+3x2+3x 6x2+3x+2=4x2+2x 6x2+3x+2 Thetopofthiscanbefactoredas2(2x+3)(x 1).