Theory Of Equations 2

THEORY OF )EQUATIONS )2

CONTENTS  Complete

Division

Remainders and Derivatives Relation between Roots and Coefficients Applications

Complete Division Consider

f(x)

to be a polynomial of order 3 and is divided several times by the factor

(x − r)

f(x) = (x − r) p1 (x) + R 0

= (x - r)((x - r)p2 (x) + R 1 ) + R 0

p2 ( x) = (x - r)R 3 + R2 3

2

f(x) = R 3 (x − r) + R 2 (x − r) + R 1 (x − r) + R 0

Example ًًWrite the polynomial

f(x) = x 4 + 6x 3 − 7x 2 − 36 x + 37 ًًin terms of powers of (x − 2)

2 1 1

6

−7

− 36

2

16

18

8

9

− 18

37 − 36 1 = R0

2 1

6

−7

− 36

1

2 8 2

16 9 20

10 2

29 24

18 − 36 − 18 1 = R0 58 40 = R 1

1 1 1 1

12 53 = R 2 2 14 = R 3 = R4

37

4

3

f(x) = (x − 2) + 14(x − 2) 2

+ 53(x − 2) + 40(x - 2) + 1

Remark ًًThe equation 4

3

2

y + 14 y + 53y + 40 y + 1 = 0 ًًhas roots that are less by “2” than those of the original equation f(x) = 0

x 4 + 6x 3 − 7x 2 − 36 x + 37 = 0

Relation between remainders and derivatives ًًThe division leads to 2

f(x) = R 0 + R 1(x − r) + R 2 (x − 2) 3

+ R 3 (x − 2) + ⋅ ⋅ ⋅

From Taylor’s Expansion

f ′′(r) 2 ′ f(x) = f(r) + f (r)(x − r) + (x − r) 2! f (3) 3 + (x − r) + ⋅ ⋅ ⋅ 3! Rn =

f

(n)

(r) n!

Example 4

3

2

f(x) = x + 6x − 7x − 36 x + 37 Find the derivatives at

x =2 4

3

f(x) = (x − 2) + 14(x − 2) 2

+ 53(x − 2) + 40(x - 2) + 1

f(2) = 1 f ′(2) = 1! R 1 = 40 f ′′(2) = 2! R 2 = 106 f

(3)

(2) = 3! R 3 = 84

f (4)(2) = 4! R 4 = 24

Relation between the roots of an equation and the coefficients Assume that the roots of the equation f(x) = an x

n

+ an−1 x

n−1

+ an−2 x

n−2

+ ⋅ ⋅ ⋅ + a1 x + a0 = 0

are

r1 , r2 , ⋅ ⋅⋅, rn

f(x) = an (x

n

an−1 n−1 an−2 n−2 + x + x +⋅⋅⋅ an an a0 a1 + x+ ) =0 an an

f(x) = an (x − r1 )(x − r2 ) ⋅ ⋅ ⋅ (x − rn )

Comparing the coefficients of the different powers of x

x x

n

n −1

an = an an− 1 = − (r1 + r2 + ⋅ ⋅ ⋅ + rn ) an

an− 2 = r1r2 + r1r3 + ⋅ ⋅ ⋅ + r1rn an

x

n −1

+ r2r3 + ⋅ ⋅ ⋅ + r2rn  + rn−1rn

x

0

a0 n = (− 1) (r1 r2 r3 ⋅ ⋅ ⋅ rn ) an

The quadratic equation 2

ax + bx + c = 0 2

− b ± b − 4ac r1 , r2 = 2a b r1 + r2 = − a c r1r2 = a

Example If the roots of the equation 3

2

x − 7x + cx − 8 = 0 constitute a geometric series, find them and find the value of c

Solution Assume the roots are

a , a, am m

 a − 8 = (− 1)   ( a)( am)  m = -a3 3

a =2

a  − 7 = (− 1) + a + am  m  2  7= + 2 + 2m  m 

2

2m − 5m + 2 = 0 (m − 2)(2m − 1) = 0

m= 2

1,2 ,4

a , a, am m

1 m= 2

4 ,2 ,1

The value of

c

Substitute with one of the roots

r=1 3

2

(1) − 7(1) + c(1) − 8 = 0

c = 14 Or using the relation

c = (−1) ( r1r2 + r1r3 + r2r3 ) 2

Finding an equation with roots that are inverses of those of a given equation 1 1 y= ⇒ x= x y n −1

n

n−2

1  1  an  y   + an−1  y      

a0 y

n

+ a1 y

1  + an−2  y     1  + ⋅ ⋅ ⋅ + a1  y   + a0 = 0  

n−1

+ a2 y

n−2

+ ⋅ ⋅ ⋅ + an-1 y + an = 0

Finding an equation with roots that are multiples of those of a given equation y = mx n

⇒ n−1

y  y  an   + an−1   m  m 

y x = m

n−2

y  + an−2   m  y  + ⋅ ⋅ ⋅ + a1   + a0 = 0 m 

Special case: equation with roots opposite in sign y = −x

Example 4

3

2

x + 6x − 7x − 36 x + 37 = 0

4

3

2

y − 6y − 7y + 36 y + 37 = 0

Finding an equation with roots that are squares of those of a given equation f(x) = (x − r1 )(x − r2 ) ⋅ ⋅ ⋅ (x − rn ) = 0

×

f(x) = (x + r1 )(x + r2 ) ⋅ ⋅ ⋅ (x + rn ) = 0

2

2 1

2

2 2

2

2 n

(x − r )(x − r ) ⋅ ⋅ ⋅ (x − r ) = 0

x

2

⇒

2 1

y

2 2

2 n

(y − r )(y − r ) ⋅ ⋅ ⋅ (y − r ) = 0

Equation with roots 2 1

2 2

2 n

r , r ,⋅ ⋅ ⋅, r

Example Find the equation with roots that are the square of the roots of the equation

x 3 − 4x2 + 2x − 9 = 0

Solution 1)Equation with the opposite sign roots 2)Multiplication of the two equations 3) Set

x2 ⇒

y

3

2

3

2

x − 4x + 2x − 9 = 0

×

− x − 4x − 2x − 9 = 0 − x 6 + 12 x 4 + 68 x 2 + 81 = 0

x

2

⇒

y

− y 3 + 12 y 2 + 68 y + 81 = 0 y 3 − 12 y 2 − 68 y − 81 = 0