082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

24 Chapter 1 Review of algebraic techniques
and
x 2 −2x+0.5=(x−1) 2 −1+0.5
=(x−1) 2 −0.5
Finally, solvingx 2 −2x +0.5 = 0 we have
(x−1) 2 −0.5=0
(x −1) 2 = 0.5
x−1=± √ 0.5
x=1± √ 0.5 = 0.2929,1.7071
1.4.2 Polynomialequationsofhigherdegree
Example1.23 Verifythatx = 1 andx = 2 areroots of
P(x)=x 4 −2x 3 −x+2=0
Solution P(x) =x 4 −2x 3 −x +2
P(1)=1−2−1+2=0
P(2)=2 4 −2(2 3 )−2+2=16−16−2+2=0
SinceP(1) = 0 andP(2) = 0, thenx = 1 andx = 2 are roots of the given polynomial
equation and are sometimes referred to as real roots. Further knowledge is required to
find the two remainingroots,whichareknown as complex roots. This topiciscovered
inChapter9.
Example1.24 Solve the equation
P(x)=x 3 +2x 2 −37x+52=0
Solution AsseeninExample1.21aformulacanbeusedtosolvequadraticequations.Forhigher
degree polynomial equations such simple formulae do not always exist. However, if
one of the roots can be found by inspection we can proceed as follows. By inspection
P(4) = 4 3 +2(4) 2 −37(4) +52 = 0sothatx = 4isaroot.Hencex−4isafactorof
P(x). ThereforeP(x) can bewritten as
P(x)=x 3 +2x 2 −37x+52=(x−4)(x 2 +αx+β)
where α and β mustnow befound.Expanding the r.h.s.gives
Hence
P(x)=x 3 +αx 2 +βx−4x 2 −4αx−4β
x 3 +2x 2 −37x+52=x 3 +(α−4)x 2 +(β−4α)x−4β
Bycomparingconstanttermson the l.h.s. and r.h.s.wesee that
52 = −4β