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

1.4 Polynomial equations 23
We now introduce the method of completing the square. The idea behind completing
the square is to absorb both thex 2 and thexterm into a single squared term. Note that
thisispossible since
and so
x 2 +2kx+k 2 =(x+k) 2
x 2 +2kx=(x+k) 2 −k 2
and finally
x 2 +2kx+A=(x+k) 2 +A−k 2
Thex 2 and thexterms are both contained in the (x +k) 2 term. The coefficient ofxon
the l.h.s.
(
is 2k. The squared
)
term on the right-hand side (r.h.s.) has the form (x +k) 2 ,
coefficient ofx 2
thatis x + . The following example illustratesthe idea.
2
Example1.22 Solve the following quadraticequationsbycompletingthe square:
(a) x 2 +8x+2=0
(b) 2x 2 −4x+1=0
Solution (a) Bycomparingx 2 +8x +2withx 2 +2kx +Aweseek = 4.Thusthesquaredterm
mustbe (x +4) 2 .Now
and so
Therefore
(x+4) 2 =x 2 +8x+16
x 2 +8x=(x+4) 2 −16
x 2 +8x+2=(x+4) 2 −16+2
=(x+4) 2 −14
At this stage we have completed the square. Finally, solvingx 2 + 8x + 2 = 0 we
have
x 2 +8x+2=0
(x+4) 2 −14=0
(x+4) 2 =14
x+4=± √ 14
x=−4± √ 14 = −7.7417,−0.2583
(b) 2x 2 −4x +1 = 0maybeexpressedasx 2 −2x +0.5 = 0.Comparingx 2 −2x +0.5
with x 2 + 2kx +A we see that k = −1. Thus the required squared term must be
(x −1) 2 .Now
and so
(x−1) 2 =x 2 −2x+1
x 2 −2x=(x−1) 2 −1