ii sem (csvtu) mathematics unit 5 (theory of equations )solustions

7/28/2019 II sem (csvtu) Mathematics Unit 5 (Theory of Equations )Solustions

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Prepared by Mrityunjoy Dutta

Unit V Theory Of Equations

April -May, 2006

1. I f O, A, B, C are the four point on a straight line such that the distance of A, B and C fromO are the roots of the equation 3 23 3 0ax bx cx d . If B is the middle point of AC show

that3 33 2 0a d abc b

Ans . Let us assume , , are the roots of the equation 3 23 3 0ax bx cx d and

, ,OA OB OC

According to the given condition 2 -------------- (1)According to the property of roots

Now,3b

a

-------------- (2)

3c

a ------------ (3)

d

a

------------- (4)

From 2 and 3 we will have3

3b b

a a

------------- (5)

3 3c c

a a

From 1

2

2 3 32 2c b c

a a a

2

2

32

c b

a a ------------- (6)


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again from 4 we have

d

a

d

a

From 5 and 6 we have

2

2

32

b c b d

a a a a

By solving above we will have

2

2

3 2ca bb d

a

Hence we will have3 33 2 0a d abc b

2. Solve by Cardans method of equation 3 23 3 0x x .Ans: 3 23 3 0x x -------------- (1)

Here we can remove second term of equation (1) by diminishing its roots by

31

3 1

bh

na

We can diminished each root by 1 by synthetic division method

Hence the transformed equation is 3 3 1 0y y ---------(2)where 1y x

Let 1/3 1/3y p q be the solution of equation (2).


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3 1/3 1/3 1/3 1/33y p q p q p q

3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)

By comparing equation (2) and (3) we get

1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q

So, p and q are the roots of the equation2 1 0t t

1 1 4 1 3

2 2

it .

So, let1 3 1 3

,2 2

i ip q

.

So roots of given cubic equation (1) are

(i)1/3 1/3

1/3 1/3 1 3 1 3

2 2

i ip q

1/3 1/32 2 2 2

cos sin cos sin3 3 3 3

i i

2 2 2 2cos sin cos sin

9 9 9 9i i

22cos

9

(ii)1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 32 2 2 2

i i i iwp w q

4/3 4/3

1 3 1 3

2 2

i i


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4/3 4/32 2 2 2


i i

8 8 8 8cos sin cos sin9 9 9 9

i i

82cos9

(iii)1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 3

2 2 2 2

i i i iw p wq

1/3 1/32 2 2 2 2 2 2 2

cos sin cos sin cos sin cos sin3 3 3 3 3 3 3 3

i i i i

2 2 2 2 2 2 2 2cos sin cos sin cos sin cos sin

3 3 9 9 3 3 9 9i i i i

2 2 2 2 2 2 2 2cos sin cos sin

9 3 9 3 9 3 9 3i i


9 9 9 9i i

4 4 142cos 2cos 2 2cos

9 9 9

So, roots of equation (2) are 2 8 142cos , 2cos , 2cos9 9 9

Now, roots of given equation (1) are2 8 14

1 2cos , 1 2cos , 1 2cos9 9 9

(Ans)

3. Solve the equation by Ferraris method: 4 3 212 41 18 72 0x x x x .Ans:

4 3 212 41 18 72 0x x x x

Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ---------- (1)

4 3 2 4 2 2 3 2 2 2 212 41 18 72 36 12 12 2 2x x x x x x x x x m x mnx n

4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n

Equating the coefficients we get


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2 2 2(36 2 ) 41, ( 12 2 ) 18, 72m mn n 2 2 22 5, 9 6 , 72m mn n

222 5 72 9 6

3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0

2( 3)(2 35 147) 0 212

3,7,

By taking 3 1, 9m n equation (1) becomes4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x 2 2( 6 3 9)( 6 3 9) 0x x x x x x 2 2

( 5 6)( 7 12) 0x x x x ( 6)( 1)( 3)( 4) 0x x x x

1,3,4,6x (Ans)

Nov-Dec, 2006



3 13 1

bhna




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3 1/3 1/3 1/3 1/33y p q p q p q

3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q

So, p and q are the roots of the equation 2 1 0t t

1 1 4 1 3

2 2

it

.

So, let1 3 1 3

,2 2

i ip q

.


(i)1/3 1/3

1/3 1/3 1 3 1 32 2

i ip q

1/3 1/32 2 2 2


i i


9 9 9 9i i

22cos

9

(ii)1/3 1/3

1/3 2 1/3 1 3 1 3 1 3 1 3

2 2 2 2

i i i iwp w q

4/3 4/3

1 3 1 3

2 2

i i


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4/3 4/32 2 2 2


i i

8 8 8 8cos sin cos sin9 9 9 9

i i

82cos9

(iii)1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 3

2 2 2 2

i i i iw p wq

1/3 1/32 2 2 2 2 2 2 2


i i i i


3 3 9 9 3 3 9 9i i i i


9 3 9 3 9 3 9 3i i


9 9 9 9i i


9 9 9

So, roots of equation (2) are 2 8 142cos , 2cos , 2cos9 9 9


1 2cos , 1 2cos , 1 2cos9 9 9

(Ans)

5. Solve the equation 5 4 3 26 43 43 6 0x x x x x .Ans: 5 4 3 26 43 43 6 0x x x x x

The given equation is a reciprocal equation of odd degree having coefficient of terms

equidistance from beginning and end.

So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x

Dividing by 2x we get


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2

2

5 66 5 38 0x x

x x

2

2

1 16 5 38 0x x

x x

21 1

6 2 5 38 0x xx x

26 2 5 38 0t t where1

x tx

26 5 50 0t t

10 5(3 10)(2 5) 0 ,

3 2t t t

1 10 1 5

3 2x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 16

6 4x and x

10 8 5 3

6 4x and x

1 13, 2,2 2

x and x

So, roots are1 1

2, 1, , , 32 2

x (Ans)

6. I f , , are the roots of the equation 3 0x qx r . Find the equation whose roots are, ,

.

Ans: Given that , , are the roots of the equation 3 0x qx r .

Now, 0 -------------- (1)

q ------------ (2)

r ------------- (3)

Now 2 2 2 2( ) 2( ) 0 q q ---------- (4)


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Let , ,A B C

A B C

2 2 2 2 2 2

A B C

2 2 2 2 2 2

A B C

2 2 2 2 2 2A B C

( ) ( ) ( )A B C

A B C

as 0

33A B C

------------------ (5)

AB BC CA

2 2 2 2 2 2 2 2 2 2 2 2

AB BC CA

2 2 2 2 2 2 2 2 2 2 2 2AB BC CA

2 2 2 2 2 2q q q q q qAB BC CA

r r r


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2 2 2 2 2 2q q q q q qAB BC CA

r r r

2 2 2 2 2 2 2 2 2 2 2 2 2 2( ) ( ) ( )q q q q q qAB BC CAr r r

2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA

r r r

2 3 2 2 2 2 3 2 2 2 2 3 2 2 2q q q q q q q q qAB BC CA r

2 3 3 3 3 3 3 3( )( ) ( )q q q

AB BC CAr

2 3 2( ) ( ) 0 ( ) ( ) 3 ( )q q q q rAB BC CA

r

3 4 2 3( )

( )

q q q q r qAB BC CA

r r

3 4 2

2

2 3q q q rAB BC CA

r

----------- (6)

ABC

2 2 2 2 2 2

ABC

2 2 2q q qABC

3 2

2 2 2

( ) ( )q q qABC


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3 2 3 2

2 2

0q q r q q rABC

r r

Equation whose roots are A, B, C is

3 2( ) ( ) 0x A B C x AB BC CA x ABC

3 4 2 3 23 2

2 2

2 33 0

q q q r q q rx x x

r r

2 3 2 2 3 4 2 3 23 (2 3 ) ( ) 0r x r x q q q r x q q r (Ans)



31

3 1

bh

na



Let

1/3 1/3

y p q be the solution of equation (2).

3 1/3 1/3 1/3 1/33y p q p q p q

3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


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(iii)1/3 1/3

2 1/3 1/3 1 3 1 3 1 3 1 3

2 2 2 2

i i i iw p wq

1/3 1/32 2 2 2 2 2 2 2


i i i i


3 3 9 9 3 3 9 9i i i i


9 3 9 3 9 3 9 3i i

4 4 4 4

cos sin cos sin9 9 9 9i i


9 9 9

So, roots of equation (2) are2 8 14

2cos , 2cos , 2cos9 9 9


1 2cos , 1 2cos , 1 2cos9 9 9

(Ans)

8. Solve the equation 5 4 3 26 43 43 6 0x x x x x .Ans:

5 4 3 26 43 43 6 0x x x x x



So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x


2

2

5 66 5 38 0x x

x x

2

2

1 16 5 38 0x x

x x


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21 1

6 2 5 38 0x xx x

2

6 2 5 38 0t t where1

x tx

26 5 50 0t t

10 5(3 10)(2 5) 0 ,

3 2t t t

1 10 1 5

3 2x and x

x x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 166 4

x and x

10 8 5 3

6 4x and x

1 13, 2,

2 2x and x

So, roots are1 1

2, 1, , , 32 2

x (Ans)

9. I f , , are the roots of the equation 3 0x qx r . Find the equation whose roots are, ,

.

Ans: Given that , , are the roots of the equation 3 0x qx r .

Now, 0 -------------- (1)

q ------------ (2)

r ------------- (3)

Now 2 2 2 2( ) 2( ) 0 q q ---------- (4)

Let , ,A B C


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Equation whose roots are A, B, C is

3 2( ) ( ) 0x A B C x AB BC CA x ABC

3 4 2 3 23 2

2 2

2 33 0

q q q r q q rx x x

r r

2 3 2 2 3 4 2 3 23 (2 3 ) ( ) 0r x r x q q q r x q q r (Ans)

May-J une, 2007

10. Solve the equation by Ferraris method: 4 3 212 41 18 72 0x x x x .Ans: 4 3 212 41 18 72 0x x x x

Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ---------- (1)


4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n

Equating the coefficients we get2 2 2(36 2 ) 41, ( 12 2 ) 18, 72m mn n

2 2 22 5, 9 6 , 72m mn n

222 5 72 9 6

3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0

2( 3)(2 35 147) 0 212

3,7,

By taking 3 1, 9m n equation (1) becomes4 3 2 2 2 2

12 41 18 72 ( 6 3) ( 9) 0x x x x x x x 2 2( 6 3 9)( 6 3 9) 0x x x x x x 2 2( 5 6)( 7 12) 0x x x x

( 6)( 1)( 3)( 4) 0x x x x

1,3,4,6x (Ans)


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6 36 482, 2, 3 3

2i

So, 2 4, 6

So, roots are -4, 2, 6.

Nov-Dec, 2007

13. Find the number of real roots the equation 3 2x x 4x 4 0 Ans: number of real roots theequation

3 2x x 4x 4 0 are 3 , two positive roots and 1 negative roots.

14. Find the condition that the equation 3 2x x 4x 4 0p had roots , , which satisfy1 0 .

Ans: Let , , are the roots of 3 2 0x px qx r .

Given that 1 0 1 -------------- (1)Now, p -------------- (2)

4 ------------ (3)4 ------------- (4)

From (1) and(4) 4 ---------- (4)

Putting the value of 4 in the equation 3 2 0x px qx r we will have

64 16 16 4 0p

84 21

16 4

p

15. If , , are roots of the equation 3 0x qx r Find the equation whose roots are

2 2 2, ,

Ans: Let , , are the roots of 3 0x qx r (1)

Now, 0 -------------- (2)q ------------ (3)

r ------------- (4)Let

2y

2 2

4

2 2 4


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2 2 4

0r

2 2 4rxx

24r

y xx

3 4 0x yx r ..(5)

Subtracting 4 from 1 we will have

3 0q y x r

3rx

q y

..(6)

Put the value of x in the equation 1 we will have

3

3 30

r rq r

q y q y

2 3327 3 0r rq q y r q y

3 2 2 3 3 2 227 3 2 3 3 0r rq q r rq r q r r q rq 3 3 3 2 2 3 4 3 2 227 3 3 6 3 3 0r rq r q r q rq r r q r q 3 3 3 2 2 427 4 6 9 0r rq r q r q r

16. Solve by Cardans methods: 3 15 126 0x x Ans:

3 15 126 0x x

Here the equation having terms involving 2x term missing.

Let 1/3 1/3x p q be the solution of equation (1).

3 1/3 1/3 1/3 1/33x p q p q p q 3 1/3 1/33x p q p q x 3 1/3 1/33 ( ) 0x p q x p q -------------(2)

By comparing equation(1) and (2) we get1/3 1/3 5, ( ) 126p q p q 125, ( ) 126pq p q

So, p and q are the roots of the equation 2 126 125 0t t

( 1)( 125) 0 1,125t t t . So, let p =1 and q =125.


(i) 1/3 1/3 1 5 6p q


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(ii) 1/3 2 1/3 1 3 1 3 6 4 35 3 2 32 2 2

i i iwp w q i

(iii) 2 1/3 1/3 1 3 1 3 6 4 35 3 2 32 2 2i i i

w p wq i

So, roots are 6, 3 2 3, 3 2 3i i (Ans).

April -May 2008

17. Form the equation whose roots are 1, 2, 3.Ans .The equation will be 1 2 3x x x

3 2

1 2 3 6 11 6x x x x x x

18. Solve the equation 2x2+x2-7x-6 = 0 when the difference of two roots is 3.Ans: Let , , are the roots of 3 22x x 7x 6 0 .

Given that 3 -------------- (1)Now, 1/2 -------------- (2)

7/2 ------------ (3)3 ------------- (4)

From (1) and(2) 2 7/2

-------------

(5)From (4) and(1) . .(3 ) 3 ------------- (6)

From (5) and(6)7

2 .(3 ) 32

From

3 24 24 3 24 24 0

2( 2)( 6 12) 0

6 36 482, 2, 3 32 i

So, 2 4, 6

So, roots are -4, 2, 6.

19. If,, are the roots of the cubic 3 2 0x px qx r find the equation whose roots are


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Now, 1 2 3 3/2r r r -------------- (2)1 2 2 3 1 3 / 2r r r r r r k ------------ (3)1 2 3 1/2r r r ------------- (4)

From (1) and(2) 3 1/2r ------------- (5)From (5) and(4) 1 2 1rr ------------- (6)From (5) and(6)we can construct a quadratic equation that is

2 1 0x x

By solving equation2 1 0x x we will have

2 4 1 3

2 2

b b ac ix

a

1 2 3

1 3 1 3

2, , 1/2

2

i ir r r

23. Find the equation whose roots are the roots of + + = .eachdiminished by 3.

Ans. The above equation can be solved by using synthetic division then we will have

3y x Let , , , are the roots of + + =.

Now, 1 -------------- (1)As roots are diminished by 3 therefore new root will be 3, 3, 3, 3y

3 3 3 3 1 12

3 3 3 3 13

3 1 1 3 1 2

0 3 12 27 78

1 4 9 26 80

0 3 21 90

1 7 30 116

0 3 30

1 10 60

0 3

1 13

Hence the new equation is

4 3 213 60 116 80x x x x


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The transformed equation is 4 3 22 13 2 1 0y y y y ------------(2)

Equation (2) is a reciprocal equation of even degree having coefficients of terms

equidistant from the beginning and end equal.

Dividing equation (2) by 2y we get 22

2 12 13 0y y

y y

2

2

1 12 13 0y y

y y

Now putting2 2

2

1 12y z y z

y y

2 2 2 13 0z z 2 2 15 0 ( 5)( 3) 0z z z z

3,5z

13y

y

15y

y

2 1 3y y 2 1 5y y

2 3 1 0y y 2 5 1 0y y

3 9 4

2y

5 25 4

2y


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The transformed equation is 4 3 22 13 2 1 0y y y y ------------(2)

Equation (2) is a reciprocal equation of even degree having coefficients of terms

equidistant from the beginning and end equal.

Dividing equation (2) by 2y we get 22

2 12 13 0y y

y y

2

2

1 12 13 0y y

y y

Now putting2 2

2

1 12y z y z

y y

2 2 2 13 0z z 2 2 15 0 ( 5)( 3) 0z z z z

3,5z

13y

y

15y

y

2 1 3y y 2 1 5y y

2 3 1 0y y 2 5 1 0y y

3 9 4

2y

5 25 4

2y


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3 2 3 2 3

3

3 3 3 27 9 20 0

r r r r pqr rqp q r

q q q q

3 2 327 9 2 0r pqr rq 2 327 9 2 0r pqr q (Proved).

35. I f ,, are the roots of the equation 03 rqxx , form an equation whose roots are:(i)

,, .

(ii)2 2 2

, ,

Ans: - Here 0, ,q r

i. , ,

Let2

2 2 2 2 2 2 2, , , , , ,

r r r r ry x

x y

Now, 03 rqxx 3 6 4 2 2 22x qx r x qx q x r

3 2

2 22r r r

q q ry y y

3 22 2

3 22

r r rq q r

y y y

2 2 2 32r qry q y ry

3 2 2 22 0ry q y qry r

New eqn is 3 2 2 22 0rx q x qrx r

(ii)222

,,

Ans: - Here 0, ,q r

Let2 2 2 2 2 2

1 1 1, , , ,y

1,

1,

110,

10,

10 y as 0

yx

xy

11


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So, eqn 03 rqxx becomes 01011 32

3

ryqyr

yq

y

0123 qyry

New eqn is 0123 qxrx

36. Solve by Cardans method: 01928 23 xx Ans: - 01928 23 xx ------------------ (1)

Let1 1

y xx y

, then eqn (1) becomes

3 9 28 0y y --------------- (2)Here the equation having terms involving 2y term missing.

Let y u v be the solution of equation (2).3 3 3 3 ( )y u v uv u v 3 3 3 3y u v uvy

3 3 33 0y uvy u v -------------(3)By comparing equation(2) and (3) we get

3 33, 28uv u v 3 3 3 327, 28u v u v

So, 3 3,u v are the roots of the equation 2 28 27 0t t

( 1)( 27) 0 1, 27t t t . So, let 3 31, 27u v .

21, ,u w w , then 23 3, 3 , 3v w wu

Then 2 21 3, 3 , 3y u v w w w w 24,1 2 ,1 2y u v w w

1 3 1 34,1 2 ,1 2

2 2

i iy u v

4,2 3,2 3y i i

1 1 2 3 2 3, ,

4 7 7

i ix

y

So, roots are1 2 3 2 3, ,

4 7 7

i ix

. (Ans).

Nov-Dec 2010


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39. Solve: 5 4 3 26 41 97 97 41 6 0x x x x x .Ans: 5 4 3 26 43 43 6 0x x x x x



So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x


2

2

5 66 5 38 0x x

x x

221 16 5 38 0x xx x

21 1

6 2 5 38 0x xx x

26 2 5 38 0t t where1

x tx

26 5 50 0t t

10 5(3 10)(2 5) 0 ,

3 2t t t

1 10 1 5

3 2x and x

x x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 16

6 4x and x

10 8 5 3

6 4x and x

1 13, 2,

2 2x and x

So, roots are1 1

2, 1, , , 32 2

x (Ans)

40. Solve by Ferraris method: 031052 234 xxxx .


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43. Find the equation of squared difference of the roots of the cubic 3 2x 6x 7x 2 0.

Ans: Let , , are the roots of

3 2

x 6x 7x 2 0.

(1)

Now, 6 -------------- (2)7 ------------ (3)

2 ------------- (4)As sum of the roots is -6 therefore to remove the 2nd term we will make a new equation

whose roots will be increased by 2

Here by removing the term containing2x

We will have

2 1 6 7 2

0 2 8 2

1 4 1 4

0 2 4

1 2 5

0 2

1 0

The transformed equation will be

3-5y 4 0.y and roots of it will be 2, 2, 2 (5)

Let 2, 2, 2a b c

Now, 0a b c -------------- (6)

5ab bc ca ------------ (7)4abc ------------- (8)

2 2

2 2

2 2

b c

c a

a b

And we are here to find 2 2 2, ,

which is equivalent to

2 2 2

, ,a b b c c a if z be the root of the equation squared differences we will have

2 2

4z b c b c bc

2 2

4z b c b c bc from equation 6 and 8 we will have

3

2 2 2 4 164

abc az b c b c bc a

a a


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3c

a ------------ (3)

d

a

------------- (4)

From 2 and 3 we will have3

3b b

a a

------------- (5)

3 3c c

a a

From 1

2

2 3 32 2c b c

a a a

2

2

32

c b

a a

------------- (6)again from 4 we have

d

a

d

a

From 5 and 6 we have

2

2

32

b c b d

a a a a

By solving above we will have

2

23 2ca bb da

Hence we will have3 33 2 0a d abc b

47. Solve by Cardans methods : X3 3X2 + 3 =0.Ans:

3 23 3 0x x -------------- (1)Here we can remove second term of equation (1) by diminishing its roots by

3

13 1

b

h na



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3 1/3 1/3 1/3 1/3

3y p q p q p q

3 1/3 1/33y p q p q y

3 1/3 1/33 ( ) 0y p q y p q -------------(3)


1/3 1/3 1, ( ) 1p q p q

1, ( ) 1pq p q

So, p and q are the roots of the equation 2 1 0t t

1 1 4 1 3

2 2

it

.

So, let1 3 1 3

,2 2

i ip q

.


(i)1/3 1/3

1/3 1/3 1 3 1 3

2 2

i ip q

1/3 1/32 2 2 2


i i


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48. Solve by Ferraris method: 4 3 212 41 18 72 0x x x x .Ans: 4 3 212 41 18 72 0x x x x

Let 4 3 2 2 2 2( ) 12 41 18 72 ( 6 ) ( ) 0f x x x x x x x mx n ---------- (1)


4 3 2 4 3 2 2 2 212 41 18 72 12 (36 2 ) ( 12 2 )x x x x x x m x mn x n


2 2 2

2 5, 9 6 , 72m mn n

222 5 72 9 6

3 2 22 144 5 360 81 36 108 0 3 22 41 252 441 0

2( 3)(2 35 147) 0 212

3,7,

By taking 3 1, 9m n equation (1) becomes4 3 2 2 2 212 41 18 72 ( 6 3) ( 9) 0x x x x x x x

2 2( 6 3 9)( 6 3 9) 0x x x x x x 2 2( 5 6)( 7 12) 0x x x x

( 6)( 1)( 3)( 4) 0x x x x

1,3,4,6x (Ans)

April -May-2012

49. Solve the equation 3 6x 20 0x , one root being 1 3.i Ans . As we know that from the general properties of roots of an equation that In an equation

with real coefficients, if i is a root then i is also one of the root, therefore two

roots of the equation3( ) 6x 20f x x are 1 3.i and 1 3.i

( ) 1-3i 1+3if x x x x a

( ) 1 3 1 3f x x i x i x a

( ) 1 3 1 3f x x i x i x a


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2

( ) 1 9f x x x a

2( ) 2 1 9f x x x x a

2( ) 2 8f x x x x a

2

( )

2 8

f xx a

x x

3

2

6x 20

2 8

xx a

x x

2 2x x a a

Roots of the equation

3

6x 20x are 1 3.,1 3 2i i and and50. Solve the equation Solve the equation 3 2x 4x 20x 48 0 , given that the roots

are connected by the relation 0

Ans . Let us assume , , are the roots of the equation 3 2x 4x 20x 48 0 and

According to the given condition 0 -------------- (1)According to the property of roots

Now, 4 -------------- (2)From 1 we will have 0 4 4

20 ------------ (3)48 ------------- (4)

From 2 and 4 we will have 4 48 12 ------------- (5)

By using the equation 1 and 5 we 0 and 12

We can have quadratic equation from above i.e 2-12 0y and 2 3y

2 3 2 3and and three roots of the equation 3 2x 4x 20x 48 0 are

2 3 , 2 3 4and

51. Solve the equation 3 26x 11x 3x 2 0 , given that its roots are in H.P.Ans. We know that H.P is reciprocal of A.P then by taking 1/y x the new equation will be

3 2

3 21 1 16 11 3 2 0 2 3y 11 6 0y yy y y

now roots of the equation

3 22 3y 11 6 0y y are in A.P As , ,d d are the roots of the cubic equation3 22 3y 11 6 0y y .


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By properties of relation between roots and coefficient of the equation we will have

3/ 2d d

3 3/2 1/ 2

Again we will have 3d d

2 2 3d 2

21 1 32 2

d

21 64

d

2 1 2564 4

d

5

2d by putting the value of 5/ 2d in , ,d d we will have roots of the

equation3 22 3y 11 6 0y y are 2,1/2,3 and roots of the equation

3 26x 11x 3x 2 0 are 1/2, 2 1/3and

52. Solve by Cardans methods: 3x 18x 35 0. Ans: 3x 18x 35 0.

Here the equation having terms involving2x term missing.

Let x u v be the solution of equation (1). 3 3 3 3x u v uv u v 3 3 3 3x u v uvx

3 3 33 0x uvx u v -------------(2)By comparing equation(1) and (2) we get

3 36, ( ) 35uv u v ------------(3)3 3 3 3 3 36, ( ) 35 216, ( ) 35uv u v u v u v

So,3u and 3v are the roots of the equation 2 35 216 0t t

2

35 216 ( 27)( 8) 27, 8t t t t t .

So, let 3u =-27 and 23, 3 , 3u w w therefore from 3 we will have 22, 2 , 2v w w


(iv) 3 2 5u v


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(v) 2 1 3 1 3 5 33 2 3 22 2 2

i i iu v w w

(vi) 2 1 3 1 3 5 33 2 3 22 2 2i i i

u v w w

So, roots are5 3 5 3

5, ,2 2

i i

(Ans).

Dec J an -2012

53. From the equation of the fourth degree whose roots are 3 + i and 7 .Ans . As we know that from the general properties of roots of an equation that In an equationwith real coefficients, if i is a root then i is also one of the root, and for the

irrational roots if an equation has ba is one of the root, then ba is also another root.

Therefore two roots of the equation we will have 3 i and 3-i , 7 7and and the

equation will be ( ) 3+i 3-i 7 7f x x x x x

( ) 3 3 7 7f x x i x i x x

22 2 2( ) 3 7f x x i x

22 2 2( ) 3 7f x x i x

2 2( ) 9 6 1 7f x x x x 4 2 2 3 2( ) 7 9 63 6 42 7f x x x x x x x 4 3 2( ) 6 3 42 70f x x x x x

54. Solve the equations: 5 4 3 26x x 43x 43x 6 0 .x Ans:

5 4 3 26 43 43 6 0x x x x x


equidistance from beginning and end.So, 1x is its root.

4 3 2( 1)(6 5 38 5 6) 0x x x x x 4 3 2(6 5 38 5 6) 0x x x x

Dividing by2x we get


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2

2

5 66 5 38 0x x

x x

2

2

1 16 5 38 0x x

x x

21 1

6 2 5 38 0x xx x

26 2 5 38 0t t where1

x tx

26 5 50 0t t

10 5(3 10)(2 5) 0 ,

3 2t t t

1 10 1 5

3 2x and xx x

2 23 3 10 2 2 5x x and x x 2 23 10 3 0 2 5 2 0x x and x x

10 100 36 5 25 16

6 4x and x

10 8 5 3

6 4x and x

1 13, 2,2 2

x and x

So, roots are1 1

2, 1, , , 32 2

x (Ans)

55. Solve by Cardans methods : 3 27x 54=0x Ans: 3 27x 54=0x

Here the equation having terms involving 2x term missing.

Let x u v be the solution of equation (1).

3 3 3 3x u v uv u v 3 3 3 3x u v uvx

3 3 33 0x uvx u v -------------(2)By comparing equation(1) and (2) we get


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3 39, ( ) 54uv u v ------------(3)3 3 3 3 3 39, ( ) 54 729, ( ) 54uv u v u v u v

So, 3u and 3v are the roots of the equation 2 54 729 0t t 2 2 254 27 ( 27) 27t t t t .

So, let3u =-27 and 23, 3 , 3u w w therefore from 3 we will have 23, 3 , 3v w w


(i) 3 3 6u v (ii) 2 1 3 1 3 63 3 3 3 3

2 2 2

i iu v w w

(iii) 2 1 3 1 3 63 3 3 3 32 2 2

i iu v w w

So, roots are 6, 3, 3 (Ans).

56. Solve by Ferrari;s method : X4 4X3 X2 + 16X 12=0.Ans: 4 3 2 4x 16x 12 0.x x

Let 4 3 2 2 2 2( ) 4x 16x 12 ( 2 ) ( ) 0f x x x x x mx n ---------- (1)

4 3 2 4 2 2 3 2 2 2 2

4x 16x 12 4 4 4 2 2x x x x x x x m x mnx n 4 3 2 4 3 2 2 2 2 4x 16x 12 4 (4 2 ) ( 4 2 )x x x x m x mn x n


2 2 22 5, 8 2 , 12m mn n ..(2)

222 5 12 8 2

3 2 22 24 5 60 64 4 32 3 22 8 4 0

By solving the above equation we will have

2, 2, 1/ 2

By taking 2, in the equation 2 we will have2 29, 12, 16m mn n

23, 12, 4m mn n

3, 4 3, 4m n or m n


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By putting the value of above in the equation 1 we will have

4 3 2 2 2 2( ) 4x 16x 12 ( 2 2) (3 4) 0f x x x x x x

4 3 2 2 2( ) 4x 16x 12 ( 2 2) (3 4) ( 2 2) (3 4)f x x x x x x x x x 4 3 2 2 2( ) 4x 16x 12 2 2 3 4 2 2 3 4f x x x x x x x x x

4 3 2 2 2( ) 4x 16x 12 5 6 2f x x x x x x x

2 25 6 3 2 , 2 2 1x x x x x x x x

4 3 2( ) 4x 16x 12 3 2 2 1f x x x x x x x

Roots of the equation4 3 2 4x 16x 12 0.x x

3,2,-2and 1

ii sem (csvtu) mathematics unit 5 (theory of equations )solustions

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