TEST BANK FOR Mathematical methods for physics and engineering 3rd Edition By Riley, Kenneth

It can be shown that the polynomial
g(x) = 4x3 + 3x2 − 6x − 1
has turning points at x = −1 and x = 1
2 and three real roots altogether. Continue
an investigation of its properties as follows.
(a) Make a table of values of g(x) for integer values of x between −2 and 2.
Use it and the information given above to draw a graph and so determine
the roots of g(x) = 0 as accurately as possible.
(b) Find one accurate root of g(x) = 0 by inspection and hence determine precise
values for the other two roots.
(c) Show that f(x) = 4x3 + 3x2 − 6x − k = 0 has only one real root unless
−5 ≤ k ≤ 7
4 .
(a) Straightforward evaluation of g(x) at integer values of x gives the following
table:
x −2 −1 0 1 2
g(x) −9 4 −1 0 31
(b) It is apparent from the table alone that x = 1 is an exact root of g(x) = 0 and
so g(x) can be factorised as g(x) = (x−1)h(x) = (x−1)(b2x2+b1x+b0). Equating
the coefficients of x3, x2, x and the constant term gives 4 = b2, b1 − b2 = 3,
b0 − b1 = −6 and −b0 = −1, respectively, which are consistent if b1 = 7. To find
the two remaining roots we set h(x) = 0:
4x2 + 7x + 1 = 0.
1
PRELIMINARY ALGEBRA
The roots of this quadratic equation are given by the standard formula as
α1,2 =
−7 ±
√
49 − 16
8
.
(c) When k = 1 (i.e. the original equation) the values of g(x) at its turning points,
x = −1 and x = 1
2, are 4 and −11
4 , respectively. Thus g(x) can have up to 4
subtracted from it or up to 11
4 added to it and still satisfy the condition for three
(or, at the limit, two) distinct roots of g(x) = 0. It follows that for k outside the
range −5 ≤ k ≤ 7
4 , f(x) [= g(x) + 1 − k] has only one real root.
1.3 Investigate the properties of the polynomial equation
f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0,
by proceeding as follows.
(a) By writing the fifth-degree polynomial appearing in the expression for f(x)
in the form 7x5 + 30x4 + a(x − b)2 + c, show that there is in fact only one
positive root of f(x) = 0.
(b) By evaluating f(1), f(0) and f(−1), and by inspecting the form of f(x) for
negative values of x, determine what you can about the positions of the real
roots of f(x) = 0.
(a) We start by finding the derivative of f(x) and note that, because f contains no
linear term, f can be written as the product of x and a fifth-degree polynomial:
f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0,
f

(x) = x(7x5 + 30x4 + 4x2 − 3x + 2)
= x[ 7x5 + 30x4 + 4(x − 3
8 )2 − 4( 3
8 )2 + 2]
= x[ 7x5 + 30x4 + 4(x − 3
8 )2 + 23
16 ].
Since, for positive x, every term in this last expression is necessarily positive, it
follows that f(x) can have no zeros in the range 0 < x < ∞. Consequently, f(x)
can have no turning points in that range and f(x) = 0 can have at most one root
in the same range. However, f(+∞) = +∞ and f(0) = −2 < 0 and so f(x) = 0
has at least one root in 0 < x < ∞. Consequently it has exactly one root in the
range.
(b) f(1) = 5, f(0) = −2 and f(−1) = 5, and so there is at least one root in each
of the ranges 0 < x < 1 and −1 < x < 0.
There is no simple systematic way to examine the form of a general polynomial
function for the purpose of determining where its zeros lie, but it is sometimes
2
PRELIMINARY ALGEBRA
helpful to group terms in the polynomial and determine how the sign of each
group depends upon the range in which x lies. Here grouping successive pairs of
terms yields some information as follows:
x7 + 5x6 is positive for x > −5,
x4 − x3 is positive for x >1 and x < 0,
x2 − 2 is positive for x >
√
2 and x < −
√
2.
Thus, all three terms are positive in the range(s) common to these, namely
−5 < x < −
√
2 and x > 1. It follows that f(x) is positive definite in these ranges
and there can be no roots of f(x) = 0 within them. However, since f(x) is negative
for large negative x, there must be at least one root α with α < −5.
1.5 Construct the quadratic equations that have the following pairs of roots:
(a) −6,−3; (b) 0, 4; (c) 2, 2; (d) 3 + 2i, 3 − 2i, where i2 = −1.
Starting in each case from the ‘product of factors’ form of the quadratic equation,
(x − α1)(x − α2) = 0, we obtain:
(a) (x + 6)(x + 3) = x2 + 9x + 18 = 0;
(b) (x − 0)(x − 4) = x2 − 4x = 0;
(c) (x − 2)(x − 2) = x2 − 4x + 4 = 0;
(d) (x − 3 − 2i)(x − 3 + 2i) = x2 + x(−3 − 2i − 3 + 2i)
+ (9 − 6i + 6i − 4i2)
= x2 − 6x + 13 = 0.
Trigonometric identities
1.7 Prove that
cos
π
12
=
√
3 + 1
2
√
2
by considering
(a) the sum of the sines of π/3 and π/6,
(b) the sine of the sum of π/3 and π/4.
(a) Using
sinA + sinB = 2sin

A + B
2

cos

A − B
2

,
3
PRELIMINARY ALGEBRA
we have
sin
π
3
+ sin
π
6
= 2sin
π
4
cos
π
12
,
√
3
2
+
1
2
= 2
√1
2
cos
π
12
,
cos
π
12
=
√
3 + 1
2
√
2
.
(b) Using, successively, the identities
sin(A + B) = sinAcosB + cosAsin B,
sin(π − θ) = sinθ
and cos( 1
2π − θ) = sinθ,
we obtain
sin
π
3
+
π
4

= sin
π
3
cos
π
4
+ cos
π
3
sin
π
4
,
sin
7π
12
=
√
3
2
√1
2
+
1
2
√1
2
,
sin
5π
12
=
√
3 + 1
2
√
2
,
cos
π
12