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  • How to solve simultaneous equations with linear and quadratic equations?

    When solving simultaneous equations with a linear and quadratic equation, there will usually be two pairs of answers. Substitute y = x + 3 into the quadratic equation to create an equation which can be factorised and solved. If the product of two brackets is zero, then one or both brackets must also be equal to zero.
  • What is the difference between linear equation and quadratic equation and polynomial?

    A linear equation produces a straight line when we graph it whereas when we graph a quadratic equation we produce a parabola. The slope of a quadratic polynomial unlike the slope of a linear polynomial, is constantly changing.
  • How do you solve simultaneous polynomial equations?

    To solve a set of simultaneous equations you need to:

    1Eliminate one of the variables.2Find the value of one variable.3Find the value of the remaining variables via substitution.4Clearly state the final answer/s.5Check your answer by substituting both values into either of the original equations.
  • Linear, quadratic and cubic polynomials can be classified on the basis of their degrees.

    1A polynomial of degree one is a linear polynomial. For example, 5x + 3.2A polynomial of degree two is a quadratic polynomial. For example, 2x2 + x + 5.3A polynomial of degree three is a cubic polynomial.

CHAPTER1.FUNCTIONS7

1.3.2Polynomials

Apol ynomialisafunctionPwithageneralfor m

P(x)=a

0 +a 1 x+a 2 x 2 +···+a n x n ,(1.4) wheret hecoecientsa i (i=0,1,...,n)are number sandnisanon- negat ivewholenum- ber.Thehighes tpowerwh osecoecientisnotzeroi scalled thedegreeofth epolynomial P.

Example1.5.

P(x)23x

2 +4x+2 1 1+x p x13x+⇡x 3 2t+4

Polynomial?YesYesNoNoYesYes

Order02N/AN/A31

Importanceofpolynomials:analytic al& computationalpointsofview

Degree0:P(x)=a

0 =a 0 x 0 ,sayP(x)=2. Thi spol ynomialissimp lyaconstant. (a)y=P(x)=2.(b)Entiredomainmappedto onepoint.

Figure1.5:Ast raightlineparalleltoth ex-axis.

Degre1:P(x)=ax+b,a6=0.T hes earecalledlinear,si ncethegraphofy=ax+bisa straightline.Theline arequationax+b=0h assol utionx=b/a.

CHAPTER1.FUNCTIONS8

Figure1.6:Lineargraphgivenbyy=ax+b.

Theslopeorgradientofy=ax+bisa.It canbewor kedoutas follows : slope= changeinheight changeindistan ce changeiny changeinx (av+b)(au+b) vu a(vu) vu =a.(1.5)

Degree2:P(x)=ax

2 +bx+c,a6=0. Thi sisknownasaqu adraticp olynomial.The quadraticequationax 2 +bx+c=0h assol utions x= b± p b 2 4ac 2a .(1.6) Proof.Startbyre-arran gingthe equationanddividingthroughbyasothat x 2 b a x= c a nextweaddb 2 /(2a) 2 toboths ides,hen ce x 2 b a x+ b 2 (2a) 2 c a b 2 (2a) 2 Nowwecan searchf oracommon denominatorontheri ghthands ide (RHS)and completethesquareorfactor iseonthel efthandside(LHS),i .e. x+ b 2a 2 2 2 ac (2a) 2 b 2 (2a) 2

CHAPTER1.FUNCTIONS9

Takingthesquarero otofbothsid eswehave

x+ b 2a s 2 2 ac (2a) 2 b 2 (2a) 2 p 2 2 ac+b 2 2a andfinall yre-arrangingtheab oveequation(orminusb/2afrombothsi desofthe equation)weget 3 x= b± p b 2 4ac 2a

Example1.6.Considerthequadraticpolyn omial

P(x)=x

2

3x+2.(1.7)

Wecanr eprese ntP(x)inadi↵erentwaybyfactori singit,i. e.

P(x)=(x2)(x1).(1.8)

Again,wecanre presen tP(x)inadi↵erentway,thist imebycompleti ngthesquare. When completingthesquare,wedothis basedonthe valueofbasfol lows.Weaddandsubtract b 2 2 toP(x)su chthatP(x)=x 2 3x+ 3 2 2 3 2 2 +2( i. e.wedon'treallych ange theequation ).Nowitiseasytoseex 2 3x+ 3 2 2 isthe sameas x 3 2 2 i.e.itcanbe factorised.Thuswecanfinallywrit e P(x)= x 3 2 2 1 4 .(1.9) Now,(1.7),(1. 8)and(1.9)arealleq uivalentan dtheyar eeachablet ogiveani nsighton whatthegraph ofthequadrat icfuncti onP(x)lo okslike.Thati s (i)Equation(1.7)tellsusth egraphofP(x)is a"cup"r athe rthana"cap"sin cethe coecientofx 2 isposi tive.AlsowecaneasilyseeP(0)=2. (ii)Equation(1.8)tellsusP(x)=0 atx=1an dx=2 (iii)Equation(1.9)tellsusP(x)is minimal atx= 3 2 andP 3 2 1 4 .Note , x 3 2 2 1 4 1 4 sinceanythingsq uaredisalwayspositive! Nowthatw ehavesome suitabl einformationre gardingP(x),weare abletop roduce an informedsketch: 3 strated"inLatin .

CHAPTER1.FUNCTIONS10

Figure1.7:Quadraticgraphgivenbyy=x

2 3x+2.

Degree3:Thingsgetabitmore complic ated !

Ingener al,wehavethealgebraic equation

a 0 +a 1 x+a 2 x 2 +···+a n x n =0,(1.10) whichhasnroots,including realandcomplex(imaginarynumbers, z=↵+i)r oots. n=2w eha veformu laeforroots(qu adratics) n=3w eha veformu laeforroots(cu bic) n=4w eha veformu laeforroots(qu artics) n>4Noge ner alformulaeexist(p rovenby

EvaristeGalois)

Butinanycas e,wema ytryf actorisationtofin dthero ots.Ifyouf actoriseapol ynomial, say

P(x)=(x1)(x+3)( x+4),

thenyoucaneasi lysolve P(x)=0, in thi scasex 1 =1,x 2 =3,x 3 =4. NOTE:thisdepe ndsontheprope rtyof0ontheRHS.Youcan'teas ily solve (x1)(x+3)( x+4)= 1. Conversely,ifyouknowthatP(↵)=0, th eny oumayfactoriseP(x)as

P(x)=(x↵)q(x),

whereq(x)is somepoly nomialofonede greelessthanthatofP(x).

Example1.7.ConsiderP(x)=x

3 8x 2 +19x12.We knowth atx=1 isa solution toP(x)=0, the nit canbeshownth at

P(x)=(x1)q(x)=(x1)(x

2

7x+12).

HereP(x)is acubic andt husq(x)is aquadrat ic.

CHAPTER1.FUNCTIONS11

Example1.8.ConsiderP(x)=x

3 x 2

3x1.By observ ation,weknow

P(1)=( 1)

3 (1) 2

3(1)1=0.

Sox 1 =1is aroot.Le tus write

P(x)=(x+1)( x

2 +ax+b), thenmultipl yingthebracketswehave

P(x)=x

3 +(1+a)x 2 +(a+b)x+b, whichshouldbeequ ivalenttox 3 x 2

3x1.Th us,comparingthec orresponding

coecientswehave

1+a=1,

a+b=3,quotesdbs_dbs14.pdfusesText_20
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