Unit l: Algebra, Vectors and Geometry
PTER
Solution of Equations
1. Introduction. 2. General
properties. 3. Transformation of 4. Reciprocal equations. 5. Solution ot
equations-Cardan's method. 6. Solution of biquadratic equations.
equations-Ferrari's method; Descarte's meth0d. cubic
7. Graphical solution of equations. 8.
Objective Type of Questions.
1.1 INTRODUCTION
The expression flr) =
at" +a,x- +..
+-1*+ ,
where a's are constants (a, + 0) and
n is a positive
flx) = 0 is called
integer, is called a polynomial in x of degree
an algebraic equation of degree n. If flx) contains some other functions
n. The polynomial
such as trigonometric,
logarithmic, exponential etc. ; then flx) = 0 is called a transcendental equation.
The value of x which satisfies flx) 0, =
..(1)
is called its root. Geometrically, a root of (1) is that value of x where the graph of y = flx) crosses the x-axis. The
process of finding the roots of an equation is known as solution of that equation. This is a problem of basic
importancein applied mathematies. We often come across problems in deflection of beams, electrical circuits and
mechanical vibrations which depend upon the solution of equations. As such, a brief account of solution ofequa
tions is given in this chapter.
1.2 GENERALPROPERTIES
I. Ifa is a root ofthe equation fla) = 0, then the polynomial fla) is exactly divisible by x- a and conversely.
For instance, 3 is a root of the equation ax- 6x-8x -3 = 0, becausex = 3 satisfies this equation.
X - 3 divides - 6x2-8x-3 completely, i.e., x -3 is its factor.
II. Every equation of the nth degree has n roots (real or imaginary).
Conversely if a,, Gay. o, be the roots of the nth degree equation flw) = 0, then
fx)=A (t-a,) (t-a,)..(x-a,) whereA is a constant.
Obs. If a polynomial of degree n vanishes for more than n values ofx, it must be identically zero.
Example 1.1. Solve the equation 2x" +x*- 13x + 6 = 0.
Solution. By inspection, we find x = 2 satisfies the given equation.
2 is its root, i.e. x - 2 is a factor of 2x" + x*- 13x + 6. Dividing this polynomial by x - 2, we get the
quotient 2 + 5x-3 and remainder 0.
Equating the quotient to zero, we get 2: +5x-3 =0.
getx
-5tVI5-4.(2).(-3
-3)-5--3,5
Solving this quadratic,
=
we
2x2 4
2, -3, 1/2.
Hence, the roots of the given equation are
1
, HIGHER ENGINEERING MATHEMATICS
6) Let the roots be alr, a, ar, so that the product of the roots = a" = .
Putting oa = (n), in (), we get n-In3+mn-n = 0 or m = In/8
Cubing both sides, we get m" = /n, which is the required condition.
Example 1.6. Soloe the equation x-2a"-21x + 22x + 40 =0 whose roots are in A.P.
Solution. Let the roots be a-3d, a -d, a + d, a + 3d, so that the
a 1/2
sum of the roots =4a =
2,
Also produet of the roots =
(a- 9d) (a2- d) = 40
=40 or 144d-40d2-639 =0
d= 9/4 or -7/36
Thus, d = + 3/2, the other value is not admissible.
Hence the required roots are -
4, - 1, 2, 5.
Example 1.7. Solve the equation 2x- 15x" +
35x -30x +8 =
0, whose roots are in G.P.
Solution. Let the roots be alr", alr,
ar, ar", that product of the roots a'
so
Also the product of
alr3, ar and alr, = =
4.
a r are each =a2 2. =
T h e factors
corresponding alrs,
to ars and alr, a r are x2 +
px +2, a2 + 2.
Thus, 2x 15x +35x2- 30x +8 qx +
Equating the coefficients of a and a2
2(x2+ px +2) (x2 =
+ qx + 2)
-
15 2p +2q and 35 8+2pq
p=-9/2, q = - 3.
Thus the given equation is 2
2 (x-3x +2)=0
Hence the
required roots are 1/2, 4 and 1, 2 i.e.,,1,2, 4.
2
Example 1.8. 1f oa, B, y be the roots of the equation x +
(a) Ecp px + q =
0, find the value of
6)Eo (c) EoPB.
Solution. We have a+B+y = 0
oß +By + Yo = p ..)
oBy =-q ...(iii)
(a)
Multiplying (i) and (ii), we get
aB +oy+ By + Boa + ya + y2B
or
+3cßy =0
6)
XoB=-3aßy =3q By ii)
Multiplying the given equation by x, we get a + px2 + qx =
0
Putting =a, B, 7 successively and adding, we get
x
or
Eo + pEa +qEa = 0
Now
Ea=-pEo-q(0) ..iu)
squaring (i), we get o + B2 + y2 +2(aß + By + yo) =0
or
Ea=-2p By (i))
Hence, substituting the value of Eo in (iv), we obtain
2o=-p-2p) =2p.
(c) ExB= Ea Eaß-ßyEa=-2p(p)--g) (0) = -
2p2.
PTER
Solution of Equations
1. Introduction. 2. General
properties. 3. Transformation of 4. Reciprocal equations. 5. Solution ot
equations-Cardan's method. 6. Solution of biquadratic equations.
equations-Ferrari's method; Descarte's meth0d. cubic
7. Graphical solution of equations. 8.
Objective Type of Questions.
1.1 INTRODUCTION
The expression flr) =
at" +a,x- +..
+-1*+ ,
where a's are constants (a, + 0) and
n is a positive
flx) = 0 is called
integer, is called a polynomial in x of degree
an algebraic equation of degree n. If flx) contains some other functions
n. The polynomial
such as trigonometric,
logarithmic, exponential etc. ; then flx) = 0 is called a transcendental equation.
The value of x which satisfies flx) 0, =
..(1)
is called its root. Geometrically, a root of (1) is that value of x where the graph of y = flx) crosses the x-axis. The
process of finding the roots of an equation is known as solution of that equation. This is a problem of basic
importancein applied mathematies. We often come across problems in deflection of beams, electrical circuits and
mechanical vibrations which depend upon the solution of equations. As such, a brief account of solution ofequa
tions is given in this chapter.
1.2 GENERALPROPERTIES
I. Ifa is a root ofthe equation fla) = 0, then the polynomial fla) is exactly divisible by x- a and conversely.
For instance, 3 is a root of the equation ax- 6x-8x -3 = 0, becausex = 3 satisfies this equation.
X - 3 divides - 6x2-8x-3 completely, i.e., x -3 is its factor.
II. Every equation of the nth degree has n roots (real or imaginary).
Conversely if a,, Gay. o, be the roots of the nth degree equation flw) = 0, then
fx)=A (t-a,) (t-a,)..(x-a,) whereA is a constant.
Obs. If a polynomial of degree n vanishes for more than n values ofx, it must be identically zero.
Example 1.1. Solve the equation 2x" +x*- 13x + 6 = 0.
Solution. By inspection, we find x = 2 satisfies the given equation.
2 is its root, i.e. x - 2 is a factor of 2x" + x*- 13x + 6. Dividing this polynomial by x - 2, we get the
quotient 2 + 5x-3 and remainder 0.
Equating the quotient to zero, we get 2: +5x-3 =0.
getx
-5tVI5-4.(2).(-3
-3)-5--3,5
Solving this quadratic,
=
we
2x2 4
2, -3, 1/2.
Hence, the roots of the given equation are
1
, HIGHER ENGINEERING MATHEMATICS
6) Let the roots be alr, a, ar, so that the product of the roots = a" = .
Putting oa = (n), in (), we get n-In3+mn-n = 0 or m = In/8
Cubing both sides, we get m" = /n, which is the required condition.
Example 1.6. Soloe the equation x-2a"-21x + 22x + 40 =0 whose roots are in A.P.
Solution. Let the roots be a-3d, a -d, a + d, a + 3d, so that the
a 1/2
sum of the roots =4a =
2,
Also produet of the roots =
(a- 9d) (a2- d) = 40
=40 or 144d-40d2-639 =0
d= 9/4 or -7/36
Thus, d = + 3/2, the other value is not admissible.
Hence the required roots are -
4, - 1, 2, 5.
Example 1.7. Solve the equation 2x- 15x" +
35x -30x +8 =
0, whose roots are in G.P.
Solution. Let the roots be alr", alr,
ar, ar", that product of the roots a'
so
Also the product of
alr3, ar and alr, = =
4.
a r are each =a2 2. =
T h e factors
corresponding alrs,
to ars and alr, a r are x2 +
px +2, a2 + 2.
Thus, 2x 15x +35x2- 30x +8 qx +
Equating the coefficients of a and a2
2(x2+ px +2) (x2 =
+ qx + 2)
-
15 2p +2q and 35 8+2pq
p=-9/2, q = - 3.
Thus the given equation is 2
2 (x-3x +2)=0
Hence the
required roots are 1/2, 4 and 1, 2 i.e.,,1,2, 4.
2
Example 1.8. 1f oa, B, y be the roots of the equation x +
(a) Ecp px + q =
0, find the value of
6)Eo (c) EoPB.
Solution. We have a+B+y = 0
oß +By + Yo = p ..)
oBy =-q ...(iii)
(a)
Multiplying (i) and (ii), we get
aB +oy+ By + Boa + ya + y2B
or
+3cßy =0
6)
XoB=-3aßy =3q By ii)
Multiplying the given equation by x, we get a + px2 + qx =
0
Putting =a, B, 7 successively and adding, we get
x
or
Eo + pEa +qEa = 0
Now
Ea=-pEo-q(0) ..iu)
squaring (i), we get o + B2 + y2 +2(aß + By + yo) =0
or
Ea=-2p By (i))
Hence, substituting the value of Eo in (iv), we obtain
2o=-p-2p) =2p.
(c) ExB= Ea Eaß-ßyEa=-2p(p)--g) (0) = -
2p2.
