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Question-1

If z1, z2 C, show that (z1 +z2)2 = z12 + 2z1z2 + z22

Solution:
Let z1 = x1 + iy1

z2 = x2 + iy2

(z1+ z2)2 =

               =

               = (x1+ x2)2 + 2i (x1+ x2) (y1+y2) - (y1+y2)2

               = x12 + 2x1x2+ x22 + 2ix1y1 + 2ix1y2 + 2ix2y1 + 2ix2y2 - y12-2y1y2 - y22

               = x12 + 2ix1y1 - y12 + x22 + 2ix2y2 - y22 + 2(x1+iy1) (x2 +iy2)

               = x12 +2ix1y1+(iy1)2 + x22 +2ix2y2 + (iy2)2 + 2(x1+iy1) (x2 +iy2)

               = (x1+iy1)2 + (x2+iy2)2 + 2 z1z2

               = z12 + 2z1z2 + z22

Question-2

Write the following as complex numbers

        i.  

        ii.  1 +

        iii. – 1 -

        iv

        v. , (x>0)

        vi. – b + , (a, c > 0)


Solution:
i. = = = i

ii. 1 + = 1 + i

iii. – 1 - = -1 - = -1 - i

iv. ==

v. = + i0

vi. – b + = – b + = -b + 2i

Question-3

Obtain a quadratic equation whose root are 2 and 3.

Solution:
Let α, β be the roots of the equation.

Sum of the roots α +β = 2+3 = 5

Product of the roots αx β= 2x3 = 6

The equation is given by

x2 - (sum of roots)x + product of roots = 0

Equation is x2 - 5x + 6 = 0.

Question-4

Solve the following equation: 25x2–30x+9 = 0.

Solution:

25x2–30x+9 = 0

D = b2-4ac = 900 - 4× 25× 9 =900 – 900=0

Hence the two real equal roots of the equation are :

i.e

Question-5

Write the real and imaginary parts of the following complex numbers below:

         i.

         ii.

         iii.

         iv.

         v. 7

         vi. 3i


Solution:
i. Let z =
   Re z = , Im z =

ii. Let z =
Re z = , Im z =

iii. Let z ==
Re z = , Im z =

iv.
Re z = , Im z =

v. 7
Re z = 7, Im z = 0

vi. 3i
Re z = 0, Im z = 3

Question-6

Without computing the roots of 3x2 + 2x+6 = 0, find (i) (ii) α2 +β2   (iii) α3+β3

Solution:
If α and β an the root of the equation 3x2 +2x+6 = 0
Sum of the roots α +β = 
Product of roots αβ = = 2

(i) = 

(ii)  α2 + β2 = (α + β )2 - 2 αβ =  

(iii .

Question-7

Show that (1-i)2 = -2i.

Solution:
(1-i)2 = 12 - 2(i) (1) +(i)2 = 1-2(-i) +(i)2  = 1-2i-1 = -2i

Question-8

Solve the following equation: 2x2-2 3x + 1 =0.

Solution:
2x2-2 3x + 1 =0

 

D = b2-4ac = 12 - 4× 2× 1 =12 – 8= 4 > 0

= 2

Hence the two real and unequal roots are :

i.e

Question-9

Solve the equation = (x - 2) in C.

Solution:
Squaring both sides
=(x-2)2 
x = x2 - 2(x)(2)+4
0 = x2 - 4x+ 4 - x
   = x2 - 5x + 4

x2 - 5x + 4 = 0
x2 - 4x - x + 4 = 0
x(x - 4) - (x - 4) = 0
x = 4 or x = 1

x=1 doesn't satisfy the equation
x = 4 .

Question-10

Find the conjugate of the following complex numbers  

        i. 3 + i

        ii. 3 – i

        iii.

        iv. -i

        v. 4/5

        vi. 49 – i/7


Solution:
i. Conjugate of 3 + i is 3 – i.

ii. Conjugate of 3 – i is 3 + i.

iii. Conjugate of is

iv. Conjugate of -i is i.

v. Conjugate of 4/5 is 4/5.

vi. Conjugate of 49 – i/7 is 49 + i/7.

Question-11

Solve the following equation: .

Solution:

Squaring,

(3x+1)+(x-1) - 2= 4

4x - 2= 4

= 2x-2

Squaring,

3x2 –2x-1 = (2x-2)2

3x2-2x-1= 4x2 –8x+4

x2-6x+5=0

D = b2-4ac = 36 - 4× 1× 5 =16 > 0

= 4

Hence the two real and unequal roots are :

i.e 5,1

Question-12

Find the conjugate of .

Solution:
=.
Conjugate of = i

Question-13

Show that if a,b,c,d, R, = (a-ib) (c-id).

Solution:
(a+ib) (c+id) = ac + iad + ibc +i2bd = ac + i(ad+bc) - bd = (ac-bd) + i(ad+bc)
 

ac - bd + i(ad + bc)

= (ac-bd) -i (ad+bc) ---- (1)

(a-ib) (c-id) = (ac-bd) -i (bc+ad) ------(2)

From (1) and (2) 

Question-14

Find the value of x and y, if 4x + i(3x - y) = 3 – i6.

Solution:
4x + i(3x - y) = 3 – i6

Equating the real and imaginary, we have

4x = 3

  x = ¾

3x – y = -6

3(3/4) – y = -6

    9/4 – y = -6

– y = -6 – 9/4

   y = 33/4

Question-15

Solve the following equation: 2x2+1 =0.

Solution:

2x2+1 =0 x2 =-1/2

Hence the complex roots of the equation are +i,-i.

Question-16

Does the equation 2x2 – 4x + 3 = 0 have equal roots? Find the roots.

Solution:
The given equation is 2x2 – 4x + 3 = 0.
Comparing with ax2 + bx + c = 0

a = 2, b = -4, c = 3

b2 – 4ac = (-4)2 – 4×2× 3 = 16 – 24 = -8 < 0
The roots are not equal.
Hence the roots of the given equation is =

Question-17

Find the value of x and y, if (3y - 2) + i(7 – 2x) = 0.

Solution:
(3y - 2) + i(7 – 2x) = 0

Equating the real and imaginary, we have

3y – 2 = 0
       y = 2/3

7 – 2x = 0
     2x = 7
       x = 7/2

The value of x = 7/2 and y = 2/3.

Question-18

If two complex numbers z1z2 are such that |z1| = |z2|, is it then necessary that z1 = z2 ?

Solution:


x12 + y12 = x22 + y22 

x12 = x22 and y12 = y22

x1= ±x2 y1 = ±y

z1 need not be z2

Question-19

Solve the following equation: x2–4x+7 = 0 .

Solution:

x2–4x+7 = 0

D = b2-4ac = 16 - 4× 1× 7 =16 -28= -12 <0

= 2i

Hence the two complex roots are :

i.e 2+i , 2+i

Question-20

For what values of a is one of the roots of the equation x2 + (2a +1)x + a2 + 2 = 0 twice the value of the other.

Solution:
Let the roots be α, 2α

α + 2α
3α = - 2α - 1

α

α .2α

2α 2

= α2 +2

2(4α2 + 4α + 1) = 9(α2 +2)
8α2 +8α + 2 = 9α2 + 18
-α2 + 8α - 16 = 0
α2 - 8α + 16 = 0
α2 -4α - 4α + 16 = 0
α(α - 4) - 4(α - 4) = 0
∴α = 4 or α = 4

Question-21

If the difference of the root of x2 – bx + c = 0 is the same as that of the roots of x2 – cx + b = 0 then b+c+4 = 0 unless b - c = 0.

Solution:
Let α,β be the root of the equation x2-bx+c = 0; γ, δand be the roots of the equation x2– cx + b = 0.

Then a + b = b, αβ=c , γ+ δ=c and γ δ = b

Given that a - b =  g - δ

(

b2 - 4c = c2 - 4b

b2-c2+4b-4c = 0

(b-c)(b+c)+4(b-c) = 0

(b-c) (b+c+4) = 0

Hence b-c = 0 or b+c+4 = 0

(ie) b+c+4 = 0 or b = c

Question-22

Find the value of x and y, if + i2y =.

Solution:
+ i2y =

Equating the real and imaginary, we have
=

x = + 5
x =(+ 5)/3

2y = 0
y = 0

The value of x =(+ 5)/3 and y = 0.

Question-23

If z1, z2, z3 are 3 complex numbers such that there exists a z with |z1- z|=|z2- z| =|z3- z| show that z1 , z2 , z3 lie on a circle in the plane diagram.

Solution:
Let z1, z2, z3 be x1 + iy1, x2 + iy2 and x3 + iy3 respectively.

Representing points P, Q, R

Let the z be point O given by x + iy.

= = OP

Similarly = OQ

and = OR

= =

OP = OQ = OR = r

This means P, Q, R are points on a circle with centre O and radius r. 

Or z1, z2, z3 lie on a circle.

Question-24

Solve the following equation: x2+x+1=0.

Solution:

x2+x+1=0

D = b2-4ac = 1 - 4 =-3 <0

= i

Hence the two complex roots are :

Question-25

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: 4 – i3.

Solution:
Conjugate of 4 – 3i is 4 + 3i.


The absolute value of 4 – 3i
                              =

                              =

                             =

                             = 5

 

 

Question-26

A group of students decided to buy a tape-records from 170 to 195 rupee. But at the last moment two student backed out of the decision so that the remaining student had to pay 1rupee more than they had planned. What was the price of the tape recorder if the student paid equal shares ?

Solution:
Let the price of the tape recorder be Rs. x
Let no. of student be n.

At the last moment
No. of students = (n-2) 
Increased contribution = 
Original contribution = 

According to the question
= +1


nx = (n-2)(x+n) = nx + n2 - 2x - 2n

n2 -2n = 2x
x = , Also 170 < x < 195

170 <  < 195

340 < n2 - 2n < 390

Either 340 < n2 -2n
n2 - 2n -340 0

Roots are given by

n =  

n 1+ or n < 1 -


n = 20 or n2 -2n -390 < 0 ----- (1)

n=  n =   
1- < n < 1 + 

Since n is a natural no. n = 1, 2, 3 ……..20 ---- (2) 
From (1) and (2),
n = 20
Cost of tape - recorder x = = = Rs. 180.

Question-27

Solve the following equation: x2 +2x +2 =0.

Solution:

x2 +2x +2 =0
D = b2-4ac = 4 - 4× 2 =-4 <0
= 2i
Hence the two complex roots are :
i.e -1-i , -1+i

Question-28

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: –3 + i5

Solution:
Conjugate of –3 + i5 is –3 - i5.

The absolute value of –3 + i5

                        =

                      =

                      =









 

 

Question-29

Solve (x2-5x+7)2 - (x -2)(x-3) =1.

Solution:
(x2 - 5x + 7)2 - (x - 2) (x - 3) = 1

(x2 - 5x + 7)2 - [x2 - (2 + 3)x +2 × 3] = 1

(x2-5x+7)2 - [x2 -5x+6] -1 = 0

Let x2 - 5x = y -------------(1)

(y+7)2 - (y+6) -1 = 0

y2 + 14y + 49 - y - 6 - 1 = 0

y2 + 13y + 42 =0
y =

  =
  =
  =  
  =  
  = -6 or -7

y = -6 or -7

Substituting y = -6 in (1)
x2 - 5x = -6
x2 - 5x + 6=0
x = 3 or x = 2

Substituting y = -7 in (1)
x2 - 5x = -7
x2 - 5x + 7=0
x = .

Question-30

Solve the following equation: 25x2–30x +11=0.

Solution:

25x2–30x +11=0

                 D = b2-4ac = 900 - 4× 25× 11 =-200 <0

               = 10i

                Hence the two complex roots are :

i.e

Question-31

Prove that x4+4 = (x+1+i)(x+1-i)(x-1+i)(x-1-i).

Solution:
(x+1+i) (x+1-i) (x-1+i) (x-1-i) = [(x+1)2-i2] [(x-1)2 –i2]

                                             = (x2+2x+1+1) (x2-2x+1+1)

                                             = [(x2+2)+2x] [(x+2)-2x]

                                             = (x2+2)2- 4x2

                                             = x4+4x2+4-4x2

                                             = x4+4

Question-32

Solve the following equation: 5x2 - 6x +2 = 0.

Solution:

5x2 - 6x + 2 = 0

D = b2-4ac = 36- 4× 5× 2 =- 4 <0

= 2i

Hence the two complex roots are :

i.e

Question-33

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: 5

Solution:
Conjugate of 5 is 5.

The absolute value of 5 =

                                   =

                                   = 5

 

Question-34

If (1+x)n = p0+p1x+p2 x2+ ………….pnxn, prove that :p0+p3+p6+……….. = .

Solution:
(1+x)n = po+p1 x + p2x2+ ............. pnxn ………………….(1)

Put x = 1,w,w2 in (1) and add

[1+w = -w2 and 1+w2 = -w]

3(po+p3+p6 ............) = 2n+(-w2)n + (-w)n ............... (2)

Now w =
-w = -i= cos -isin, (... r = 1, θ = )
-w2 = cos+isin
(-w)n +(-w2)n = 2cos (Demoivre’ s Theorem)
Substituting in (2),3(po+p3+p6..............) = 2n + 2cos
or po+p3+p6..............=.

Question-35

From an equation whose roots are the squares of the sum and difference of the roots of
2x2 + 2(m + n)x + m2 + n2 = 0.

Solution:
Let α , β be the roots of the equation 2x2 + 2(m + n)x + m2 + n2 = 0.

Then α + β = -2(m + n)/2 = -(m + n)
αβ = (m2 + n2 )/2

The roots of the required equation are (α +β )2 and (α - β )2
Sum of the roots = (α + β )2 + (α - β )2 = (α + β )2 + [(α + β )2 - 4αβ ]

                 = (m + n)2 + [(m + n)2 - ]
                = 4mn

Product of the roots = (α + β )2 (α - β )2 = (α + β )2 [(α + β )2 - 4αβ ]
                                                          = (m + n) 2 [(m + n)2 - ]
                                                          = (m + n) 2 [2mn – m2 – n2]

The required equation is x2 - 4mnx + (m + n)2[2mn – m2 - n2] = 0

or x2 - 4mnx + (m + n)2[– (m – n)2] = 0

or x2 - 4mnx - (m2 - n2) 2 = 0

Question-36

Find the values of the root .

Solution:
1-i = -isin

     =

     = -sin

= 21/4[cos(8n+1)

     = 21/4 (cos-isin for n = 0

     = 21/4(cos(-isin for n = 1

     = -21/4(cos-isin where

cos, sin

Question-37

Solve the following equation: 3x2 - 7x + 5 = 0.

Solution:

3x2 - 7x + 5 = 0

D = b2-4ac = 49- 4× 3× 5 =- 11 <0

= i

Hence the two complex roots are :

Question-38

Solve the equation 25x2 - 30x + 9 = 0.

Solution:
x = = =
x = ,

Question-39

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: 2i

Solution:
Conjugate of 2i is -2i.

The absolute value of 2i
=

                                   =

                                   = 2




 

Question-40

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: -1/2 - 3i

Solution:
Conjugate of is

The absolute value of

                              
=

                            =

                            =

 

Question-41

If the roots of x2 – lx + m = 0 differ by 1, then prove that l2 = 4m + 1.

Solution:
Let a , b be the roots of the equation x2 – lx + m = 0.
α + β = l

αβ = m

α - β = 1

(α +β )2 = (α - β )2 + 4αβ
l2 = 1+ 4m

Question-42

Solve the following equation: 13 x2–7x + 1=0.

Solution:

13 x2–7x + 1=0

D = b2-4ac = 49- 4× 13× 1 =- 3 <0

= i

Hence the two complex roots are :

Question-43

If z = x+iy and z1/3= a-ib then show that .

Solution:
z = x+iy and 21/3 = a-ib

(x+iy)1/3 = a-ib

Cubing both sides,

x+iy = (a-ib)3

      = a3+b3i-3abi(a-ib)

      = a3+b3i-3a2bi-3ab2

Equating the real and imaginary,

x = a3-3ab2

y = b3-3a2b

         = 4(a2-b2)

Question-44

(1-w+w2) (1-w2+w4) (1-w4+w8)………. to 2n factors = 22n.

Solution:
(1-w+w2) (1-w2+w4) (1-w4+w8) ................. 2n factors.

= (1-w+w2) (1-w2+w) (1-w+w2) ................... 2n factors.

( since w4=w, w8 = w2 ........)

= (-2w)(-2w2)(-2w)(-2w2) ....................... 2n factors.

= (22w3) (22w3) ................. n factors.

= (22)n = 22n

Question-45

Plot the following number and their complex conjugates on a complex number plane and find their absolute values:  (-3)

Solution:
   = 3i
Conjugate of 3i is –3i.

The absolute value of 3i = = = 3







 

 

Question-46

Solve the following equation: 9x2+10x+3 =0.

Solution:

9x2+10x+3 =0

D = b2-4ac = 100- 4× 9× 3 =- 8 <0

= 2i

Hence the two complex roots are :
i.e

Question-47

A number of points are marked on a plane and are connected pair wise by a line segment. If the total number of line segments is 10, how many points are marked on the plane ?

Solution:
Let the number of points marked on a plane be n. Since each point is joined with each other point except itself.

The number of line segments =
Each point will be joined to each of the remaining (n-1) point leaving itself. But these line segments include those also which are in the reverse order.

Hence actual number of line segments = = 10
= 10
n2 - n = 20

n2 - n - 20 = 0

n2 - 5n + 4n -20 =0

n(n - 5) + 4(n - 5) = 0

n = 5 or n = -4
Since line segment cannot be negative.
Number of points on the plane = 5 .

Question-48

Solve the following equation: 8x2+9x+3 =0.

Solution:

8x2+9x+3 =0

D = b2-4ac = 81- 4× 8× 3 =- 15<0

= i

Hence the two complex roots are :

Question-49

Solve the equation x2 + px + 45 = 0, given that the square of the difference of its roots is equal to 144.

Solution:
Let α, β be the two root of the equation,
x2 +px+45 = 0
Sum of roots α+β = -p
Product of roots α β = 45
(α- β)2 = 144 (given)

We know
(α+β)2 - (α- β )2 = 4αβ 
(-p)2 - (α- β)2 = 4 x 45
p2 - 144 = 180
p2 = 144 + 180 = 324
p = =

Case (i) p = 18
Then the given equation is
x2 + 18x+45 = 0
x2 +15x+3x+45 = 0
x(x+15) +3 (x+15) = 0
(x+15)(x+3) = 0
x = -15 0r -3

Case (ii) p = -18
Then the given equation is
x2 - 18x+45 =0
x2 -15x-3x+45 = 0
x(x-15)- 3 (x-15) = 0
(x - 15)(x - 3)= 0
x = 15 or 3

Hence the roots are -3,-15, or 3,15.

Question-50

If 2cos θ= x + and 2cos φ = y+, then prove that = 2 cos(θ+ φ).

Solution:
2cosθ = x+ and 2cosφ = y +

x = cosθ + isinθ

y = cosφ+ i sinφ

= (cosθ +isinθ ) (cosφ - isinφ )  = cos(θ -φ ) + i sin(θ -φ )

Similarly =cos-sin

Question-51

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: –4i/3.

Solution:
Conjugate of – 4i/3 is + 4i/3.

The absolute value of – 4i/3
=

                                       =

                                       = 4/3

 

Question-52

Plot the following number and their complex conjugates on a complex number plane and find their absolute values:

Solution:
Conjugate of is .

The absolute value of
                     
=

                      =

                     =

                     = 1

Question-53

Find the value of a for which the roots α,β of the equation x2 - 6x +a = 0 satisfy the relation 3α +2β = 20.

Solution:
x2 - 6x+a = 0
Sum of the root α+β = 6 ----- 1
Product of the root αβ = a ----- 2

Also 3α +2β = 20 ------ 3
(1) × 2   2α +2β = 12 ------ 4

(3) - (4) 
α = 8
Substituting in (1)
Now 8 + β = 6
β = -2
8(-2) = a
a = -16.

Question-54

Solve the following equation: 17x2–28x +12=0.

Solution:

17x2–28x +12=0

D = b2-4ac = 784 - 4× 17× 12 =- 32<0

= 4i

Hence the two complex roots are :

i.e

Question-55

If cos α+ cos β + cosγ = sin α + sin β + sin γ = 0, prove that cos3α + cos3β+ cos3γ = 3sin(α + β+ γ).

Solution:
cosα+ cosβ +cosγ = 0 -------------- (i)

sinα + sin β +sinγ= 0 -------------- (ii)

Multiply (ii) by i and add to (i),

(cosα + isinα ) + (cosβ +isinβ ) + (cosγ + sinγ ) = 0 -------- (iii)

Let z1 = cosα + isinα

z2 = cosβ + isinβ

z3 = cosγ + i sinγ

From( iii),

z1+z2+z3 = 0 --------------------- (iv)

z13 +z23+z33 = 3z1z2z3
(cosα + isinα )3 + (cosβ +isinβ)3 + (cosγ + sinγ )3= 3(cosα + isinα )( cosβ +isinβ )( cosγ +isinγ )

or (cos3α + isin3α ) +(cos3β +isin3β )+(cos3γ + isin3γ )= 3[cos(α +β +γ )+i sin(α +β +γ )] ............(v)

Equation of the real parts,

cos3α + cos3β + cos3γ = 3cos(α +β +γ )

Question-56

The number of straight lines y that can connect x points in a plane is given by: y = (x/2)(x - 1). How many points does a figure have if only 15 lines can be drawn connecting its vertices ?

Solution:
Number of lines given is 15.

(x/2)(x - 1) = 15

x(x - 1) = 30

x2 - x – 30 = 0

x2 - 6x + 5x – 30 = 0

x(x – 6) + 5(x – 6) = 0

(x + 5)(x – 6) = 0

x = -5 or x = 6

A figure has 6 points if only 15 lines can be drawn connecting its vertices.

Question-57

Prove that

Solution:

Question-58

Solve the following equation: 21x2+9x+1=0.

Solution:

21x2+9x+1=0

D = b2-4ac = 81 - 4× 21× 1 = - 3 <0

= i

Hence the two complex roots are :

Question-59

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: 1

Solution:

Conjugate of 1 is 1.

The absolute value of 1 =

                                  =

                                  =  1

 

Question-60

Solve the equation 5x+1 + 5 2-x = 53 + 1.

Solution:
Put 5x = y

y ×51+ = 125 + 1

5y + = 126

5y2 + 25 - 126y = 0

5y2 - 126y + 25 = 0

5y2 - 125y - 1y + 25 =0

5y(y-25) - 1 (y-25) = 0

y = 25 or 5y -1 = 0

y = 25 or 1/5

Put y = 25
Then 5x = 52
x = 2

Put y = 5-1
Then 5x = 5-1
x = -1

the solution is 2 and -1.

Question-61

Plot the following number and their complex conjugates on a complex number plane and find their absolute values: i

Solution:
Conjugate of i is -i.

The absolute value of i
=

                                 =

                                 = 1

Question-62

Solve the following equation: 17x2-8x+1=0.

Solution:

17x2-8x+1=0

D = b2-4ac = 64 - 4× 17× 1 = - 4 <0

= 2i

Hence the two complex roots are :

 

i.e

Question-63

Solve the following equation: 21x2-29x+11=0.  

Solution:

21x2-29x+11=0

D = b2-4ac = 841- 4× 21× 11 = - 83 <0

= i

Hence the two complex roots are :

Question-64

Write in the form A + iB :

Solution:

=

                         =

                         =

                         =

                         = A+iB

When A = 0 and B =

Question-65

Write all the complex numbers whose absolute value is 2.

Solution:
The complex number whose absolute value is 2 are 2, -2, 2i, -2i,+i, -i, -+i
and--i

Question-66

For what values of k,(4 - k)x2 + (2k +4) x + (8k+1) = 0, is a perfect square.

Solution:
For a perfect square b2 - 4ac = 0
(2k +4)2 - 4(8k+1)(4 - k) = 0
4k2 + 2 x 2k x 4 + 16 - 4(32k - 8k2 + 4 - k) = 0
4k2 + 16k + 16 - 4(31k - 8k2 + 4) = 0
4(k2 + 4k + 4) - 4(31k - 8k2 + 4) = 0
k2 + 4k + 4 - 31k +8k2 - 4 = 0
9k2 - 27 k = 0
k2 = 3k

k(k - 3) = 0
k = 0 or k =3 .

Question-67

Show that the root of x2 - 2x(m + 1/m) +3 = 0 are real for real values of m.

Solution:
The roots of the equation are real if D > 0.
b2 - 4ac > 0.
22 (m+ )2 - 4(3) (1) > 0
4 (m + )2 > 12
(m +)2 > 3 ----- 1

Square of any real number is always +ve

(1) is true for real m.

Question-68

Solve the following equation: 21x2+28x+10=0.

Solution:

21x2+28x+10=0

D = b2-4ac = 784- 4× 21× 10 = - 56 <0

= i

Hence the two complex roots are :

i.e .

Question-69

Write in the form of x + iy.

           i. (2i)3

           ii. (8i)(-i/8)
           iii. i6 + i8

           iv. i +i2+i3+i4

           v. i9 + i10 + i11 + i12

Solution:

i. (2i)3= 0-8i
ii. (8i)= -i2 = 1+0i
iii. i6 + i8 = i6 (1 + i2) = i6 (1 – 1) = 0+0i

iv. i +i2+i3+i4 = i(1 + i + i2 + i3) = i(1 + i - 1 - i) = 0+0i

v. i9(1 + i + i2 + i3) = i9 (1 + i - 1 - i) = 0+0i

Question-70

Write the square root of the following: x + i.

Solution:
x+i = x+i

                       =

                       =

                       =

Hence .

Question-71

Solve the following equation: 27x2+10x+1=0.

Solution:

27x2+10x+1=0

D = b2-4ac = 100- 4× 27× 1 = - 8 <0

= i

Hence the two complex roots are :

i.e .

Question-72

Write in the form of x + iy.

          i. i4 + i8 + i12 + i16

          ii. i + i5 + i9 + i13

          iii. i-38


Solution:
i. (i2) 2 + (i2) 4 + (i2) 6 + (i2) 8 = (-1) 2 + (-1) 4 + (-1) 6 + (-1) 8 = 1 + 1 + 1 + 1 = 4+0i

ii.i(1 + i4 + i8 + i12) = i[1 + (i2) 2 + (i2)4 + (i2)6]

                                = i[1 + (-1) 2 + (-1)4 + (-1)6]

                                = i[1 + 1 + 1 + 1]

                                = 0+4i

iii. i-38 = (i2)-19= (-1)-19 =(-1)-1 = -1+0i

Question-73

The sum of the square of 2 positive numbers is 100 and one number is 2 more than the other. Find the numbers.

Solution:
Let one of the numbers be x.
Then the other number = x + 2

As given x2 + (x + 2)2 = 100

x2 + x2 + 4x + 4 = 100

2x2 + 4x - 96 = 0

x2 + 2x - 48 = 0
x = ===6, -8
Since as given x is + ve.

x = 6, x + 2 = 8

Hence the numbers are 6, 8.

Question-74

Find real θ such that is (i) real (ii) imaginary.

Solution:

             = ------------------------- (i)

Hence (i) is real if sinθ = 0

This gives θ = nπ , where n is an integer.

If (i) is purely imaginary,

3-4sin2θ = 0 sin2θ =

or sinθ = ±
This gives θ = n where n is an integer.

Question-75

Solve the following equation: x2-(3i)x - 6=0.

Solution:

x2-(3i)x - 6=0

D = b2-4ac = (3i)2- 4× 1× (-6) =18-4-12+24 = 14+12

Let = a+bi

14+12= a2-b2+2abi

a2-b2=14 -----------(i)

ab = 6

(a2+b2)2 = (14)2+4(72)= 196 + 288 = 484

a2+b2 = 22 ------------(ii)

From (i) and (ii),

a2 = 18 and b2 = 4 a = ± 3 and b = ± 2

a+bi = 3+2i or -3-2i

Hence the two complex roots are :

i.e 3,-2i

Question-76

If x = a+b , y = a α + b β and z = a Z β +b a where a and b are complex cube roots of unity. Show that xyz = a3 + b3.

Solution:
x = a+b, y = aα + bβand z = aβ +bα

xyz = (a+b)(aα +bβ ) (aβ +bα ) where α = w and β= w2

     = (a+b)(aw+bw2)(aw2+bw)

     = (a+b) (a2w3+abw2+abw4+b2w3)

     = (a+b) [a2+b2+ab(w2+w4)]

     = (a+b) [a2+b2+ab(-1)] = (a+b)(a2-ab+b2)

     = a3+b3

Question-77

Solve 8  

Solution:
Put = y ………………….(i)
8y - = 2
8y2 – 1 = 2y

8y2 – 2y – 1 = 0

8y2 + 2y - 4y – 1 = 0

2y(4y + 1) - (4y + 1) = 0

(2y - 1)(4y + 1) = 0

y = 1/2 or y = -¼

Substitute y = 1/2 in (i)

Squaring both sides we have

=

4x = x + 3

3x = 3

x = 1
= -¼ which is impossible since the expression on LHS is not negative x ¹ -¼ and y = ¼ gives
x = 1 which is the required solution of the given solution.

Question-78

Write in the form of x + iy

          i. (5+i4)+(5-i4)

          ii. –(-1 + i) + i7 –5

          iii. 3(7 + i7) + i(7 + i7)

          iv. (1 - i) – (-1 + i6)


Solution:
i. (5+i4)+(5-i4) =10+0i

ii. –(-1 + i) + i7 –5 = 1 - i + i7 –5 = –4 + 6i

iii. 3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i2 = 21 + 21i + 7i – 7 = 14 + 28 i

iv.  (1 - i) – (-1 + i6) = 1 – i + 1 – 6i = 2 – 7i

Question-79

Without solving , find the sum and product of the roots of each of the following equations:

(i) 2x2-3x+5=0

(ii) (3k-1)x2-mx+(a-b) = 0, k .

(iii) x+ =7

(iv) 0.


Solution:

(i) 2x2-3x+5=0

The sum of the roots : =

The product of the roots : c/a = 5/2

(ii) (3k-1)x2-mx+(a-b) = 0, k .

The sum of the roots : =

The product of the roots : c/a = (a - b)/(3k - 1)

(iii) x+ =7

x2-7x+1=0

The sum of the roots : = 7

The product of the roots : c/a = 1

(iv) 0.

x2+kx- k2 = 0

The sum of the roots : = -k

The product of the roots : c/a = -k2

Question-80

Write the following in the form of x + iy

           i. 

           ii.  (7 – i2) – (4 + i) + (-3 + i5)


Solution:
i.= =

ii. (7 – i2) – (4 + i) + (-3 + i5) =(7 -4 – 3) + (– i2 – i + i5) = 0+2i

Question-81

Write the following in the form of x + iy  

Solution:
=

                                       =

                                       =

Question-82

Form the equations whose roots are

             (i)

            (ii) 7i,2i

           (iii)

           (iv) 3-4i,2+3i

            (v)

           (vi) 3-i,-1+2i


Solution:

(i)

The sum of the roots : = 3

The product of the roots :

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2 - (3)x + 7/4 = 0

4x2 –12x +7 = 0

(ii) 7i,2i

The sum of the roots : 7i+ 2i = 9i

The product of the roots : 7i × 2i = -14

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2 - (9i)x –14 =0

(iii)

The sum of the roots : = 0

The product of the roots : =

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2 - (0)x + 1/16 = 0

16x2 +1 = 0

(iv) 3-4i,2+3i

The sum of the roots : 3-4i+2+3i =5-i

The product of the roots : (3-4i)× (2+3i) = 18+i

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2 - (5+i)x +18+i =0

(v)

The sum of the roots :

The product of the roots : =

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2 - ()x + =0

2x2 -(3+7i) +(-3+9i)= 0

(vi) 3-i,-1+2i

The sum of the roots : (3-i)+(-1+2i) = 2 + i

The product of the roots : (3-i)× (-1+2i) = -3+2 +i(6+1)= -1+7i

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2 – (2+i)x +(-1+7i) =0

Question-83

Find k so that one root of the equation 2x2-16x+k=0 is twice the other.

Solution:

2x2-16x+k=0

Let the roots be α and 2α .

The sum of the roots : α + 2α = 16/2
                                       3α
=8 ------------(i)

The product of the roots : α × 2α = k/2
                                          2α
2 = k/2 ---------------(ii)

From (i) and (ii) ,

2(8/3)2 = k/2

k = 256/9

Question-84

If x = (a + b), y = (aω + bω 2) and z = (aω 2 + bω ), where ω is a cube root of unity, prove that:
   x2 + y2 + z2 = 6ab.

Solution:

x2 + y2 + z2= (a + b)2 + (aω + bω 2)2 + (aω 2 + bω )2

= a2 + b2 + 2ab + a2ω 2 + b2ω 4 + 2abω 3 + a2ω 4 + b2ω 2 + 2abω 3

= a2 + b2 + 2ab + a2ω 2 + b2ω
+ 2ab + a2ω + b2ω2 + 2ab          [ω 3 = 1]

= a2 + a2ω + a2ω 2 + b2 + b2ω+ b2ω 2+ 2ab + 2ab + 2ab

= a2(1 + ω + ω 2) + b2(1 + ω + ω 2) + 6ab

= a2(0) + b2(0) + 6ab                                                               [1 + ω + ω 2 = 0]

= 6ab

x2 + y2 + z2 = 6ab

Question-85

Write the following in the form of x + iy

           i3 + (6 + i3) – (20 + i5) + (14 + i3)

Solution:

i3 + (6 + i3) – (20 + i5) + (14 + i3) = -i + (6 + i3) – (20 + i5) + (14 + i3)

                                                   = (6 - 20 + 14) + i(-1 + 3 - 5 +3)

                                                   = 0 + 0i

Question-86

Find m so that one root of the equation x2+(2m+1)x +m2+2=0 is twice the other.

Solution:

x2+(2m+1)x+m2+2=0

Let the roots be α and 2α .

The sum of the roots : α + 2α = -(2m+1)
                                       3α = -(2m + 1)------------(i)

The product of the roots : α × 2α = m2+2
                                           2α2 = m2+2 -------------(ii)

From (i) and (ii) ,

2( )2 = m2+2

8m2+8m+2 = 9m2 +18

m2 –8m + 16 = 0

m = 4.

Question-87

If the roots of the equation x2 – lx + m = 0 differ by 1, then prove that l2 = 4m + 1.

Solution:

Let α ,β be the roots of the equation x2 – lx + m = 0.

Then α + β = l and α β = m.

α - β = 1

(α - β )2 = 1

(α + β )2 - 4α β = 1

l2 - 4m = 1

l2 = 1 + 4m

Question-88

Write the following in the form of x + iy

         


Solution:
= - 3

Question-89

Find k, if the roots of the quadratic equation 2x2+3x+k = 0 are equal.

Solution:

2x2+3x+k = 0

Let the roots be α and α .

The sum of the roots : α + α = -(3/2) ------------(i)

The product of the roots : α × α = k/2 ------------(ii)

From (i) and (ii) ,

k =

Question-90

Write the following in the form of x + iy

          (1 + i)4

Solution:
(1 + i)4= (1 + i)2 (1 + i)2

           = (12 + i2 + 2i) (12 + i2 + 2i)

          = (12 - 1 + 2i) (12 - 1 + 2i)

          = (2i) (2i)

          = 4i2

          = -4

Question-91

Prove that the roots of the quadratic equation ax2+bx+a= 0 are reciprocals of each other.

Solution:

ax2+bx+a= 0

Let the roots be α and β .

The sum of the roots : α + β = -(b/a) ------------(i)

The product of the roots : α × β = 1 ------------(ii)

From (ii),

α =

or the roots of the quadratic equation ax2+bx+a= 0 are reciprocals of each other.

Question-92

If α and β are the complex α 4 + β 4 + α -1β -1 = 0.

Solution:

Since α , β are the complex roots of unity,
α = ω and β = ω 2
Hence α 4 + β 4 + α -1β -1 = ω 4 + ω 8 + ω -1ω -2

                  = ω 3ω + ω 6ω 2 + (ω 3)-2             [ω 3 = 1]

                  = ω + ω 2 + 1

                  = 0

Question-93

Write the following in the form of x + iy

          (7 + i5)(7 – i5)


Solution:
(7 + i5)(7 – i5) = 7(7 – i5) + i5(7 – i5)
                      = 49 – i35 + i35 + 25
                      = 74

Question-94

Find k, so that the difference between the roots of the quadratic equation x2-4x+k = 0 is 2.

Solution:

x2- 4x+k = 0

Let the roots be α and β .

The sum of the roots : α + β = 4 ------------(i)

The product of the roots : α × β = k ------------(ii)

α - β = 2 (given) ---------------(iii)

From (i) and (iii) ,

α = 3 and β = 1

k = α × β = 3.

Question-95

If α and β are roots of x2 + px + 1 = 0 and γ and δ are the roots of x2 + qx + 1 = 0, show that

(α - γ ) (β - γ ) (α + δ ) (β + δ ) = q2 – p2


Solution:

α +β = p ; α β = 1

γ + δ = q ; γ δ = 1

L.H.S. = (α - γ ) (β - γ ) (α + δ ) (β + δ )

 

= (α β - γ α - γ β + γ 2) (α β + α δ + β δ + δ 2)

= (1 - γ p + γ 2) (1 + pδ + δ 2)

= 1 + pδ + δ 2 - γ p - γ δ p2 - γ δ 2p + γ 2 + pγ 2δ + γ 2δ 2

= 1 + pδ + δ 2 - γ p - p2 - δ p + γ 2 + pγ +1

= 1 + δ 2 + γ 2 + 1 – p2

= 2 + δ 2 + γ 2 - P2

= 2δ γ + δ 2 + γ 2 - P2

= (δ + γ )2 – P2

= q2 – p2

Question-96

Write the following in the form of x + iy

         3i3 (15i6)

Solution:
3i3 (15i6) = 3(i2)i(15(i2) 3)

              = 3(-1)i(15(-1) 3)

              = 3(-1)i15(-1)

              = 45i

Question-97

Find k, so that one root of the quadratic equation 2kx2-20x+21 = 0 exceeds the other by 2.

Solution:

2kx2-20x+21 = 0

Let the roots be α and α -2.

The sum of the roots : α + α -2 = 10/k ------------(i)

The product of the roots : α × (α -2) = 21/2k ------------(ii)

From (i) ,

α = 5/k + 1

( + 1)( – 1) = 21/2k

() =

2(25 – k2) = 21k

2k2 +21k –50 = 0

2k2 +25k –4k –50 = 0

(k-2)(2k+25) = 0

k =2 or k =

Question-98

If 1, ω , ω 2 are the three cube roots of unity, show that (a + b + c) (a + bω + cω 2)(a + bω 2 + cω ) = a3 + b3 + c3 - 3abc.

Solution:

(a + b + c)(a + bω + cω 2)(a + bω 2 + cω )

= (a + b + c)[a2 + abω 2 + acω + abω + b2ω 3 + bcω 2 + acω 2 + bcω 4 + c2ω 3]
= (a + b + c)[(a2 + b2 + c2) + ab(ω + ω 2) + ac(ω + ω 2) + bc(ω 2 + ω )]
= (a + b + c)[(a2 + b2 + c2) + ab(-1) + ac(-1) + bc(-1)]
= (a + b + c)(a2 + b2 + c2 ab – ac – bc)

= a3 + b3 + c3 - 3abc

(a + b + c)(a + bω + cω 2)(a + bω 2 + cω ) = a3 + b3 + c3 - 3abc

Question-99

Prove that [ 2 (cos 56o 15’ + i sin56o 15’)]8 = 16i.

Solution:

[ 2 (cos 56o 15’ + i sin56o 15’)]8

= 16[cos (8× 56o 15’) + i sin (8 × 56o 15’)]
= 16[cos 450o + i sin 450o]
= 16[cos 90o + i sin 90o]

= 16[0 + i]
= 16i

[ 2 (cos 56o 15’ + i sin56o 15’)]8 = 16i.

Question-100

Find the equation whose roots are larger by 2 than the roots of the equation x2-3x+2 =0.

Solution:

x2 - 3x+2 =0

If the roots be α and β .

Then the sum of the roots : α + β = 3 ------------(i)

The product of the roots : α × β = 2

The equation whose roots are larger by 2 are α +2 and β +2.

The sum of the roots : α +2 + β +2 = α + β + 4 =7 [From (i)]

The product of the roots :(α +2) (β +2 ) = α β +2(α + β ) +4

                                                          = 2+2(3)+4

                                                          = 12[From (i) and (ii)]

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2-(7)x + 12 = 0

Question-101

Write the following in the form of x + iy

         


Solution:


=

=

=

=

Question-102

Find the equation whose roots are n times the roots of the equation x2 +px + q =0

Solution:

x2 +px+q =0

If the roots be α and β .

Then the sum of the roots : α + β = -p ------------(i)

The product of the roots : α × β = q ------------(ii)

The equation whose roots are n times are nα and nβ .

The sum of the roots : nα + nβ = n(α + β ) =-np [From (i)]

The product of the roots :(nα ) ( nβ ) = n2 α β = n2 q [From (ii)]

The equation is: x2-(sum of the roots)x + product of the roots = 0

x2+npx + n2 q = 0

Question-103

If α , β are the roots of the quadratic equation 3x2-5x-8 =0, then find the values of

           (i) α 2 + β 2      (ii) α 3 + β 3       (iii)        (iv)        (v) α β 2 + α 2 β

Solution:

α , β are the roots of the quadratic equation 3x2-5x-8 =0;

Then the sum of the roots : α + β = 5/3 ------------(i)

The product of the roots : α × β = -8/3 ------------(ii)

(i) α 2 + β 2 = (α + β )2 - 2α β

                  = (5/3)2 –2(-8/3) [From (i) and (ii)]

                  ==

(ii) α 3 + β 3 = (α + β )3 - 3α β (α +β )

                   = (5/3)3- 3(-8/3)(5/3) [From (i) and (ii)]

                   = =

(iii) = =[From (i) and (ii)]

               =

(iv) = = [From (i) and (ii)]

                  =

(v) α β 2 + α 2 β = α β (α + β ) = (5/3)(-8/3) [From (i) and (ii)]

                         = -40/9

Question-104

Write the following in the form of x + iy ,  (5 + i9) ÷ (-3 + i4)

Solution:
(5 + i9) ÷ (-3 + i4)   =

                             =
                          
                            =
                            =

Question-105

If α , β are the roots of the quadratic equation x2+3x+6 =0, then find the values of
          (i) (ii) α 2 + β 2.

Solution:

α , β are the roots of the quadratic equation x2+3x+6 = 0;

Then the sum of the roots : α + β = -3 ------------(i)

The product of the roots : α × β = 6 ------------(ii)

(i) = =[From (i) and (ii)]

              ==

 

(ii) α 2 + β 2 = (α + β )2- 2α β = (-3)2-2(6) [From (i) and (ii)]

                   = -3

Question-106

Write the following in the form of x + iy , (-2 – i5) ÷ (3 – i6)

Solution:
(-2 – i5) ÷ (3 – i6) =

                          =

                          =  

                          =

                          =

Question-107

If α + β =1 and α 2 + β 2=2, then find the values of

           (i) α 3 + β 3 (ii) α 4 + β 4


Solution:

α + β = 1 and α 2 + β 2 = 2 (given) ------------(i)

                   Þ 2α β = (α + β )2 –(α 2 + β 2) = 1-2 = -1;

                  Þ a β = -1/2 ------------(ii)

(i) α 3 + β 3 = (α + β )3 - 3α β (α +β ) [From(i) and (ii)]

                  = 1- 3(-1/2)(1) = 1+3/2 =5/2

(ii) α 4 + β 4 = (α 2 + β 2)2-2(α β )2

= (2)2-2(-1/2)2 [From(i) and (ii)]

= 4 – (1/2) = 7/2

Question-108

Write the following in the form of x + iy ,

Solution:

=

                                  =

                                  =

                                 =

                                =

                                =

                               =

                               =

                               

Question-109

If α , β are the roots of the quadratic equation 3x2-4x+1 =0, then find the equation whose roots are
          (i) 3α ,3β (ii) α 2, β 2.

Solution:

α , β are the roots of the quadratic equation 3x2-4x+1 =0

Then the sum of the roots : α + β = 4/3 ------------(i)

The product of the roots : α × β = 1/3 ------------(ii)

(i) 3α +3 β = 3( α + β ) = 4 [From(i)]

3α × 3 β = 9(1/3) =3 [From (ii)]

Hence the equation whose roots are 3α, 3β is : x2-(sum of the roots)x + product of the roots = 0

x2- 4x + 3 = 0

(ii) α 2 + β 2 = ( α + β )2 –2 α β = (4/3)2 –2(1/3) [From(i) and (ii)]

                                               =10/9

α 2 × β 2 = (1/3)2 = 1/9 [From (ii)]

Hence the equation whose roots are α 2 , β 2 is :

x2-(sum of the roots)x + product of the roots = 0

x2- (10/9)x + (1/9) = 0

9x2-10x+1= 0

Question-110

Write the following in the form of x + iy

          

Solution:
==+0i

Question-111

If α , β are the roots of the quadratic equation x2+px+q =0, then find the equation whose roots are
          (i) α 2 , β 2 (ii) .

Solution:

α , β are the roots of the quadratic equation x2+px+q =0

Then the sum of the roots : α + β = -p ------------(i)

The product of the roots : α × β = q ------------(ii)

(i) α 2 + β 2 = (α + β )2 - 2α β = p2 –2q [From(i) and (ii)]

 

α 2 × β 2 = (α β )2 = q2 [From (ii)]

Hence the equation whose roots are α 2, β 2 is: x2-(sum of the roots)x + product of the roots = 0

x2- (p2-2q)x + (q2) = 0


(ii) = (α + β ) +() = -p - [From(i) and (ii)]

= α β ++2 = = (q+1)2 [From (ii)]

Hence the equation whose roots is is :

x2-(sum of the roots)x + product of the roots = 0

x2- (-p -)x + (q+1)2 = 0

qx2+p(1+q)x+(q+1)2=0,q 0.

Question-112

Find multiplicative inverse of 4 – i3.

Solution:
Let z = 4 – i3

Then = 4 + i3
|z|2 = 16 + 9 = 25

Multiplicative inverse 4 – i3 is 1/z = /|z|2 =

Question-113

Find multiplicative inverse of (+i3).

Solution:
Let z = +i3

Then = - i3

|z|2 = ()2 + (3)2 = 5 + 9 = 14

Multiplicative inverse +i3 is 1/z = /|z|2 =

Question-114

If α , β are the roots of the quadratic equation 2x2-5x+7 =0, then find the equation whose roots are 2α +3 β , 3α +2 β.

Solution:

α , β are the roots of the quadratic equation 2x2-5x+7 =0

Then the sum of the roots : α + β = 5/2 ------------(i)

The product of the roots : α × β = 7/2 ------------(ii)

Sum 2α +3 β + 3α +2 β = 5(α + β ) = 5(5/2) =25/2 [From (i)]

Product (2α +3 β )(3α +2 β ) = 6(α 2 + β 2)+13 α β =6(α + β )2 + α β

                                          = 6(5/2)2+(7/2) [From (i) and (ii)]

                                          = 75/2 + 7/2

                                          = 82/2

Hence the equation whose roots are 2α +3 β , 3α +2 β is :

x2-(sum of the roots)x + product of the roots = 0

x2- (25/2)x + (82/2) = 0

2x2- (25)x + (82) = 0

Question-115

Find multiplicative inverse of –i.

Solution:
Let z = -i
Then = i
|z|2 = 0 + (1)2 = 1
Multiplicative inverse –i is 1/z = /|z|2 = i/1 = 0+i

Question-116

Find the quadratic equation whose roots are the reciprocals of the roots of the equation x2 -px+q =0.

Solution:

Let α , β are the roots of the quadratic equation x2-px+q =0.

Then the sum of the roots : α + β = p ------------(i)

The product of the roots : α × β = q ---------------(ii)

Sum = = [From (i) and (ii)]

Product = 1/αβ = 1/q [From (ii)]

Hence the equation whose roots are is :

x2-(sum of the roots)x + product of the roots = 0

x2- (p/q)x + (1/q) = 0

qx2- px + 1 = 0

Question-117

Change the following complex number in polar form:   1 – i

Solution:
x= rcosθ=1;y = rsinθ=−1
x + iy = 1 – i

x = 1, y = -1
r =

and tan θ = y/x = -1, θ = .
Thus, the polar coordinates of 1 – i are (, ) and its polar form is (cos + i sin).

Question-118

If the sum of the roots of the quadratic equation is 3 and the sum of their cubes is 63, then find the equation.

Solution:

Let α , β be the roots of the quadratic equation.

then α + β =3 -----------------(i)

α 3 + β 3 = 63 ---------------(ii)

From (ii), (α + β )3 –3 α β (α + β ) = 63

Substituting (i), 27 – 3 α β (3) = 63

α β = - 4

Hence the equation is : x2-(sum of the roots)x + product of the roots = 0

x2- 3x - 4 = 0

Question-119

Change the following complex number in polar form:   –1 + i

Solution:
x= rcosθ=−1;y = rsinθ=1
x + iy = -1 + i

x = -1, y = 1

r =

and tan θ = y/x = -1, θ = .
Thus, the polar coordinates of –1+ i are (, ) and its polar form is (cos + i sin).

Question-120

If α , β are the roots of the quadratic equation ax2+bx+c = 0, then find the values of

          (i) (ii)

Solution:

α , β are the roots of the quadratic equation ax2+bx+c = 0

Then the sum of the roots : α + β = -b/a ------------(i)

The product of the roots : α × β = c/a ---------------(ii)

(i) = = [From (i) and (ii)]

                =

(ii) = == [From (i) and (ii)]

                   =

Question-121

Change the following complex number in polar form:     –1 – i

Solution:
x= rcosθ=−1;y = rsinθ=−1
x + iy = -1 - i

x = -1, y = -1

r =

and tan θ = y/x = 1, θ = .
Thus, the polar coordinates of –1+ i are (,) and its polar form is (cos + i sin).

Question-122

If α , β are the roots of the quadratic equation px2+qx+r = 0, then find the equation whose roots are and

Solution:

α , β are the roots of the quadratic equation px2+qx+r =0

Then the sum of the roots : α + β = -q/p ------------(i)

The product of the roots : α × β = r/p ---------------(ii)

Consider = +2=+2

                              = [From (i) and (ii)]

                              =

Sum + =

Product × = 1

Hence the equation is : x2-(sum of the roots)x + product of the roots = 0

x2- ()x + (1) = 0

x2 –qx + = 0, p, r 0

Question-123

Change the following complex number in polar form:–3

Solution:
x + iy = -3
x = -3, y = 0
r =

and tan θ = y/x = 0/3, θ = π .
Thus, the polar coordinates of –1+ i are (3, π ) and its polar form is 3(cos π + i sinπ ).

Question-124

Solve the following equation : .

Solution:



y2 + y –2 =0

(y+2)(y-1) =0

y = -2 or y = 1.

or -2 x = 1 or –8

Solution set :x = 1 , -8

Question-125

Change the following complex number in polar form:  –4 + i4

Solution:
x + iy = –4 + i4
x = -4, y = 4
r =

and tan θ = 4/-4 = -, θ = .
Thus, the polar coordinates of –4 + i4 are (8, ) and its polar form is 8(cos + i sin).

Question-126

Solve the following equation : x4 – 5x2+6 = 0 .

Solution:

x4 – 5x2+6 = 0

Put x2 = y

y2 - 5y + 6 =0

(y-3)(y-2) = 0

y = 3 or y = 2

x2 = 3 or x2 = 2

x = ± or ±

Solution set :x = ± or ±

Question-127

Change the following complex number in polar form: +i

Solution:
x + iy =+ i
x = , y = 1
r =

and tan θ = y/x =, θ = π /6.
Thus, the polar coordinates of + i are (2, π /6) and its polar form is 2(cos + i sin).

Question-128

Solve the following equation (x2 –3x)2-5(x2-3x)+6 = 0.

Solution:

(x2 –3x)2-5(x2-3x)+6 = 0

Put x2 –3x = y

y2 - 5y + 6 =0

(y-3)(y-2) = 0

y = 3 or y = 2

x2 –3x =3 or x2 –3x = 2

x2 –3x -3 = 0 or x2 –3x – 2 = 0

x = or

Solution set :x = ,

Question-129

Change the following complex number in polar form: i

Solution:
x + iy = i
x = 0, y = 1
r =

and tan θ = y/x = , θ = .
Thus, the polar coordinates of –1+ i are (1, ) and its polar form is (cos+ i sin).

Question-130

Solve the following equation : 4x-3.2x+2 +32 = 0.

Solution:

4x - 3.2x+2 +32 = 0

22x- 12× 2x+32 =0

Put 2x = y,

y2-12y+32 =0

(y-8)(y-4)= 0 y = 8 and y = 4 2x = 8, 4 x = 3 , 2.

Solution set :x = 3, 2.

Question-131

Change the following complex number in Cartesian form:   2(cos 0o + isin 0o)

Solution:
2(cos 0o + isin 0o)
r= 2, θ = 0o

Let 2(cos 0o + isin 0o) = x+iy
r = x2 +y2 = 4 -----------(ii)

tanθ = tan 0° =

or y = 0 ------(ii)

Substituting (ii) in (i),

x = 2

Hence the required Cartesian form of 2(cos 0o + isin 0o) = 2+i0

Question-132

Solve the following equation : 4x+1 - 6x - 2.9x +1 = 0.

Solution:

4x+1 - 6x - 2.9x +1 = 0

22x+2 –3x × 2x - 2× 32x+2=0

4- 1 -18=0

Put =y;

4y-1-

4y2-y-18=0

4y2-9y+8y-18=0

y(4y-9)+2(4y-9)=0

(4y-9)(y+2) = 0

y = or -2

i.e = = or x=-2;

= -2 has no solution.

Solution set : x=-2.

Question-133

Change the following complex number in Cartesian form:   5(cos 270o + isin270o)

Solution:
 5(cos 270o + isin 270o) = 5(0-i× 1)
The required Cartesian form is 0 - 5i.

Question-134

Solve the following equation : x4 + x3 - 4x2+x+1 = 0.

Solution:

x4 + x3 - 4x2+x+1 = 0

Divide by x2 ,

x2 + x – 4 +=0

(x2) + (x+) - 4 = 0

Let x+ = y x2= y2 – 2

(y2-2)+y-4 = 0

y2+y-6= 0

(y+3)(y-2) = 0 y = -3 ,2.

Thus, x+= -3 or x+ = 2

x2 +3x+1 = 0 or x2 –2x +1 = 0

or .

Solution set :

Question-135

Change the following complex number in Cartesian form:   4(cos 300o + isin300o)

Solution:
4(cos300o + isin300o) = 4(- i) = 2-2i

The required Cartesian form is 2-2i

Question-136

Solve the following equation : (x+1)(x+2)(x+3)(x+4) +1= 0.

Solution:

(x+1)(x+2)(x+3)(x+4) +1= 0

(x+1)(x+4)(x+2)(x+3) + 1= 0

(x2+5x+4)(x2+5x+6) +1 = 0

Put x2+5x = y

(y+4)(y+6) + 1 = 0

y2 +10y +25 = 0

 

y = -5 , -5

i.e x2 + 5x + 5 = 0

x= ,

x = ,

Solution set : x = ,

Question-137

Find the moduli and arguments of the complex number:  z = -1 - i

Solution:
z = -1 - i
x = -1, y = -
r = === 2
θ =
|z| = 2, arg z = + 2π k, where k is an integer.

Question-138

Solve the following equation : .

Solution:

Cross multiplying,

(2x2-12)(x-1) = (2x-3)(x+2)(x-2)

2x3-2x2-12x+12 = 2x3-3x2-8x+12

x2 - 4x= 0

x(x-4)=0

x=0 or x=4

Solution set : x = 0 or x=4

Question-139

Find the moduli and arguments of the complex number:     z = - + i

Solution:
z = - + i
x = -, y = 1
r = === 2
θ =
|z| = 2, arg z = + 2π k, where k is an integer.

Question-140

Solve the following equation : .

Solution:

Put

y + =

6y2 –13y +6 = 0

i.e

Squaring,

x=

Solution set : x =

Question-141

Find the moduli and arguments of the complex number:     z =

Solution:
z =

z = = ×

1+i = (cos45° +isin45° )

(1+i)20 = [(cos45° +isin45° )] 20

= (cos45° + isin45° )20

            = 23(cos)20

            =23(cos20× ) [Using DeMoivre's formula]

           =23(cos5π +isin5π)

modulus = 8 and argument is 2π k , where k is an integer.

Question-142

Solve the following equation :

Solution:

     (By C and D)

Squaring both sides,

1+x2=4-4x2

5x2 = 3

x = ±

Solution set : x =±

Question-143

Give the following product in polar form:      [2(cos 0o + isin 0o][4(cos90o + isin90o)]

Solution:
z1 = [2(cos 0o + isin 0o)]

z2 =[4(cos90o + isin90o)]

z1z2 =[2(cos 0o + isin 0o)][4(cos90o + isin90o)]

       = 8 [cos (0o + 90o) + isin (0o + 90o)]

       = 8 [cos 90o + isin 90o]

       = 8 [cos + isin ]

Question-144

Solve the following equation : .

Solution:

              (By C and D)

x = ± a or x=0.

x cannot be 0 x = ± a

Solution set :x= ± a

Question-145

Give the following product in polar form:       [2(cos 220o + isin 220o)[4(cos110o + isin110o)]

Solution:
z1 = 2(cos 220o + isin 220o)

z2 = 4(cos110o + isin110o)

z1z2 =[2(cos 220o + isin 220o)][4(cos110o + isin110o)]

       = 8 [cos (220o + 110o) + isin (220o + 110o)]

       = 8 [cos 330o + isin 330o]

       = 8 [cos + isin ]

Question-146

Solve the following equation : .

Solution:

Squaring,

(x+5) +(x+21) +2= 6x+40

= 2x + 7

Squaring,

(x+5)(x+21) = (2x+7)2

x2 +26x+105 = 4x2+28x+49

3x2+2x-56 =0

x=14/3 does not satisfy the equation.

x= 4

Solution set : x= 4

Question-147

Solve the following equation : x2 + .

Solution:

x2 +

Using the formula a2 + b2 = (a - b)2 + 2ab

(x-)2+2x() =3

=3

Put = y
y2 +2y –3 =0

(y+3)(y-1)=0

y= -3 or y =1;

i.e= -3 or 1;

i.e x2+3x+3 = 0 or x2 - x-1 = 0 ;

or

Solution set : x = ,

Question-148

Give the following product in polar form:        [3(cos225o + isin225o)][6(cos 45o + isin 45o)]

Solution:
z1 = [3(cos225o + isin225o)]

z2 = [6(cos 45o + isin 45o)]

z1z2 =[3(cos 225o + isin 225o)][6(cos45o + isin45o)]

       = 18 [cos (225o + 45o) + isin (225o + 45o)]

       = 18 [cos 270o + isin 270o]

       = 18 [cos + isin ]

Question-149


Solution:


(x-6)(x-7) = (x-2)(x-3)

x2-13x+42 = x2-5x+6

8x = 36 x=36/8 = 4.5

Solution set : x= 4.5

Question-150

Give the following quotient in polar form: 

Solution:
   =

                               = 3(cos + isin)

Question-151

Solve the following equation : 4x4-16x3+7x2+16x+4 = 0.

Solution:

4x4-16x3+7x2+16x+4 = 0

Dividing by x2,

4x2 – 16x +7 + = 0

4() –16(x-)+7 =0

Put x-= t

() = t2+2

4(t2+2) –16t +7 = 0

4t2 –16t + 15 = 0

x-=

2x2+5x-2=0 or 2x2+3x-2=0

or =

Solution set : x= , .

Question-152

Give the following quotient in polar form:

Solution:
=

                                =

                               =

                               =

Question-153

Evaluate .

Solution:

Let y =

y =

Squaring,

y2 = 6+y

y2 -y –6 = 0

(y-3)(y+2) = 0

y = 3 or y =-2

y is positive y = 3

Solution set : = 3

Question-154

Find the square root of the following:  –15 – 8i

Solution:
Let x + iy =

Squaring both sides we have:

(x + iy)2 = -15 – 8i

x2 - y2 + 2ixy = -15 – 8i

Comparing both sides

 x2 - y2 = -15…………………………………..(i)

and 2xy = -8    
        
(x2 + y2)2 = (x2 - y2)2 + (2xy)2

               = 225 + (-8)2 = 225 + 64 = 289

x2 + y2 = 17             …………………………………..(ii)

Solving (i) and (ii) we get

2x2 = 2
x2 = 1
x = ± 1

2y2 = 32
y2 = 16
y = ± 4

Since the product xy is negative, we have
x = -1, y = 4 or x = 1, y = -4

Thus, the roots of -15 – 8i are –1 + i4 or 1 – i4.

Question-155

Evaluate 2+ .

Solution:

Let y = 2+ = 2+

y 2 = 2y +1

y 2 - 2y –1 = 0



y is positive y = 1+

Solution set : 1+

Question-156

Find the square root of the following:       –8 – 6i

Solution:
Let x + iy =

Squaring both sides we have:

(x + iy)2 = -8 – 6i

x2 - y2 + 2ixy = -8 – 6i

Comparing both sides

x2 - y2 = -8       …………………………………..(i)

and 2xy = -6        


(x2 + y2)2 = (x2 - y2)2 + (2xy)2

              = (-8)2 + (-6)2= 64 + 36= 100            
              

x2 + y2 = 10        …………………………………..(ii)

Solving (i) and (ii) we get
2x2 = 2
x2 = 1
x = ± 1

-2y2 = -18
y2 = 9
y = ± 3

Since the product xy is negative, we have
x = -1, y = 3 or x = 1, y = -3


Thus, the roots of -8 – 6i are –1 + i3 or 1 - i3.

Question-157

Evaluate 2-.

Solution:

Let y = 2- = 2-

y 2 - 2y +1 = 0

(y-1)2 =0 y = 1,1

2-= 1

Solution set : 1

Question-158

Find the square root of the following :    1 – i

Solution:
Let x + iy =

Squaring both sides we have:

(x + iy)2 = 1 – i

x2 - y2 + 2ixy = 1 – i

Comparing both sides
x2 - y2 = 1         …………………………………..(i)
and 2xy = -1


(x2 + y2) 2 = (x2 - y2) 2 + (2xy)2

                 = 1 + (-1)2= 1 + 1= 2                 
                 
x2 + y2 = …………………………………..(ii)

Solving (i) and (ii) we get

2x2 = 1+
x2 = (1+)/2
x = ±

2y2 = -1+
y2 = (-1+)/2
y = ±

Since the product xy is negative, we have
x = -, y = or x = , y = -

Thus, the roots of 1 – i are + i or -- i.

Question-159

Evaluate .

Solution:

Let y =

y =

y2 = 8-y

y2+y-8=0

y is positive y =

=

Solution set :

Question-160

Find the square root of the following :   –i

Solution:
Let x + iy =

Squaring both sides we have:

(x + iy)2 = -i

x2 - y2 + 2ixy = -i

Comparing both sides

x2 - y2 = 0        …………………………………..(i)

and 2xy = -1


(x2 + y2) 2 = (x2 - y2) 2 + (2xy)2
                = 0 + (-1)2= 0 + 1= 1
                
                 

x2 + y2 = = 1 …………………………………..(ii)

Solving (i) and (ii) we get

2x2 = 1
x2 = 1/2
x = ±

2y2 = 1
y2 = 1/2
y = ±

Since the product xy is negative, we have
x = , y =- or x = -, y =

Thus, the roots of are -+ i and - i.

Question-161

Find the number which exceeds its positive square root by 12.

Solution:

Let the number be x,

x - = 12

x – 12 =

x2 +144 – 24x = x

x2 +144 – 25x = 0

(x-16)(x-9) = 0

x = 16 or x = 9

x 9

So the number is 16 .

Question-162

Find the square root of the following :   i

Solution:
Let x + iy =

Squaring both sides we have:
(x + iy)2 = i

x2 - y2 + 2ixy = i

Comparing both sides

x2 - y2 = 0       …………………………………..(i)

and 2xy = 1

(x2 + y2) 2 = (x2 - y2) 2 + (2xy)2
                
= 0 + (1)2= 0 + 1= 1                
                 

x2 + y2 = = 1 …………………………………..(ii)

Solving (i) and (ii) we get

2x2 = 1
x2 = 1/2
x = ±

2y2 =  1
y2 =  1/2
y = ±

Since the product xy is negative, we have
x = , y = - or x = -, y =

Thus, the roots of are -+ i and - i.

Question-163

Solve x3+y3 = 4914 , x+y = 18.

Solution:

x3+y3 = 4914 --------(i)

x+y = 18 ----------(ii)

From (ii), x+y = 18 (x+y)3 = 183

x3+y3 +3xy(x+y) = 5832

4914 +3xy(18) = 5832 [substituting from (i) and (ii)]

xy = 17    

(x-y)2 = (x+y)2-4xy = 182-4(17)

         = 324 – 68

         = 256

x – y = ± 16---------(iii)

From (ii) and (iii),

2x = 34 or 2 x = 17 or x = 1

y = 1 or 17.

Solution set: x = 17,1 and y = 1,17

Question-164

Find the square root of the following :  1+i

Solution:
Let x + iy =
Squaring both sides we have:

(x + iy)2 = 1 + i
x2 - y2 + 2ixy = 1 + i


Comparing both sides
x2 - y2 = 1   …………………………………..(i)
and 2xy = 1

(x2 + y2) 2 = (x2 - y2) 2 + (2xy)2
                
= 1 + (1)2= 1+1= 2

x2 + y2 = …………………………………..(ii)

Solving (i) and (ii) we get

2x2 = 1 +
x2 = (1 +)/2
x = ±

2y2 = -1+
y2 = (-1 +)/2
y = ±

Since the product xy is negative, we have
x = , y = - or x = -, y =

Thus, the roots of are - i and -+i .

Question-165

Solve x4+y4 = 82 , x+y = 4.

Solution:

x4+y4 = 82 ------------(i)

x+y = 4------------(ii)

From (ii), (x+y)2 = 16

x2+y2+2xy =16

x2+y2= 16-2xy------(iii)

Squaring both sides,

(x2+y2)2= (16-2xy)2

x4+y4+2x2y2= 256+4x2y2-64xy

82+2x2y2= 256+4x2y2-64xy

2 x2y2-64xy+174=0

x2y2-32xy+87=0

29,3

From (iii), (x-y)2=16-4xy= 16-4(29) or 16-4(3)

                       =-100 or 4

(x-y)= ± 10i or x-y = ± 2

             When x-y = ± 2 -------------(iv)

 

From (ii) and (iv), x = 3,1 and y = 1,3

             When x-y = ± 10i -------------(v)

From (ii) and (v), x = 25i; y= 25i

Solution set: 3,1; 1,3; 25i; 25i

Question-166

Prove that Re(z1 z2) = Re z1 Rez2 – Imz1 Imz2

Solution:
Let z1 = r1(cos θ 1 + isinθ 1) and z2 = r2(cos θ 2 + isinθ 2).

z1 z2 = r1 r2 (cos θ 1 + isinθ 1)(cos θ 2 + isinθ 2)

        = r1 r2 [cos θ 1(cos θ 2 + isinθ 2) + isinθ 1(cos θ 2 + isinθ 2)]

        = r1 r2 [cos θ 1cos θ 2 + i cos θ 1sinθ 2 + isinθ 1cos θ 2 + i2 sinθ 1sinθ 2]

        = r1 r2 [(cos θ 1cos θ 2 - sinθ 1sinθ 2 ) + i (cos θ 1sinθ 2 + sinθ 1cos θ 2 )]

        = r1 r2 [cos (θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )]

Re(z1 z2) = r1 r2 cos (θ 1 + θ 2 ) ……………………(i)

Re z1 Rez2 – Imz1 Imz2

        = r1 r2 cos θ 1 cos θ 2 - r1 r2 sin θ 1 sin θ 2
        = r1 r2 cos(θ 1 + θ 2 ) ……………………(ii)

From (i) and (ii) we have

Re(z1 z2) = Re z1 Rez2 – Imz1 Imz2

Question-167

Solve x4+y4 = 257 , x+y = 5.

Solution:

x4+y4 = 257 -----------(i)

x+y = 5--------------(ii)

From (ii), (x+y)2 = 25

x2+y2+2xy =25

Hence, (x2+y2)2= (25-2xy)2

x4+y4+2x2y2= 625+4x2y2-100xy

257+2x2y2= 625+4x2y2-100xy [substituting (i)]

2 x2y2-100xy+368=0

x2y2-50xy+184=0

4,46--------(iii)

From (ii), (x-y)2=25-4xy

             = 25-4(4) or 25-4(46)[substituting (iii)]

             = 9 or -159

(x-y) = ± 3 or ± i

          When x-y = ± 3 ---------------(iv)

From (ii) and (iv), x = 4,1 and y = 1,4

When x-y = ± i ---------------(v)

From (ii) and (v), x =

Solution set : 4,1;1,4;

Question-168

Find the value of

          (i) ω 18
          (ii) ω 21
          (iii) ω -30
          (iv) ω -105

Solution:
(i) ω 18 = (ω 3) 6= (1) 6 = 1 (Since ω 3=1)

(ii) ω 21 = (ω 3) 7 = (1) 7 = 1 (Since ω 3=1)

(iii) ω -30 = 1/(ω 3)10 = 1/(1)10 = 1 (Since ω 3=1)

(iv) ω -105 = 1/(ω 3)35 = 1/(1)35 = 1 (Since ω 3=1)

Question-169

A piece of cloth costs Rs. 35. If the piece were 4 m longer and each metre costs Re.1 less, the cost would have remain unchanged. How long is the piece?

Solution:

Let the piece of cloth is x m. The cost of cloth is Rs.35.

The cost per metre is Rs.35/x

If the piece were 4 m longer and each metre costs Re.1 less the piece of cloth is (x+4) m
and the cost per metre is Rs.35/x – 1.


(x+4)(35/x –1) = 35

35-x+140/x-4 = 35

-x+140/x-4 = 0

x2 +4x –140 =0

x2 +14x-10x –140 =0

(x+14)(x-10)=0

x = -14 or x = 10

x -14,

x = 10.

Hence the piece is 10m.

Question-170

Prove (2 - ω )(2 - ω 2)(2 - ω 10)(2 - ω 11) = 49

Solution:
L.H.S =(2 - ω )(2 - ω 2)(2 - ω 10)(2 - ω 11)
        = (4 - 2ω 2 - 2ω + ω 3)(4 - 2ω 11 - 2ω 10 + ω 21)
        = [4 – 2(ω 2 + ω ) + 1][4 - 2ω 9(ω 2 + ω ) + (ω 3)7]
        = [4 – 2(-1) + 1][4 - 2ω 9(-1) + 17]

        = [4 + 2 + 1][4 + 2(ω 3)3 + 1]

        = 7[5 + 2(1)3 ]

        = 7 × 7
        = 49

        = R.H.S

Question-171

A factory kept increasing its output by the same percentage every year. Find the percentage if its known that the output doubled in the last two years.

Solution:

Let the initial production by P.
Suppose the production increases by r percent every year. Then the P1 after the first year is
P1 = P(1+r/100)
The production at the second year is P1(1+r/100)=P(1+r/100)2
Since we are given that the production doubles in two years,

we have P(1+r/100)2= 2P
(1+r/100)2= 2

(1+r/100) =
r = 100(-1)

Thus, the production increases by 100(-1)% every year.

Question-172

Prove  

Solution:
L.H.S =

         =
         =
         = (Since ω 3=1)
         =
          = ω 2

Question-173

If the roots of the quadratic equation ax2 +bx +c =0, a 0 , are in the ratio p : q then prove that ac(p+q)2 = b2pq.

Solution:

ax2 +bx +c =0

Let the roots be α and ()α .

α + ()α = -b/a α = ---------(i)

α × ()α = c/a α 2 = --------------(ii)

From (i) and (ii),

=

b2pq = ac(p+q)2

Question-174

Prove (1 - ω 2 + ω 4)(1 + ω 2 - ω 4)= 4

Solution:
L.H.S = (1 - ω 2 + ω 4)(1 + ω 2 - ω 4)

        = [1 - ω 2 + ω × ω 3][1 + ω 2 - ω × ω 3]

        = (- ω 2 - ω 2)(- ω - ω ) (Since ω 3=1 and 1 + ω + ω 2 = 0)
        = (- 2ω 2)(- 2ω )
        = 4ω 3
        = 4
        = R.H.S
(1 - ω 2 + ω 4)(1 + ω 2 - ω 4)= 4

Question-175

If α , β are the roots of the quadratic equation x2 - px +q =0, then find the equation whose roots are α β + α + β , α β - α - β.

Solution:

Let α , β are the roots of the quadratic equation x2 - px +q =0

α + β = p----------(i)

α × β = q----------- (ii)

α β +α +β + α β - α - β = 2α β =2q

(α β + α + β )(α β - α - β ) = α 2β 2 – (α +β )2

                                       = q2 – p2

Hence the equation is x2-(sum of the roots)x+(product of the roots) = 0

x2-2qx+( q2 – p2) = 0

Question-176

Prove (1 - ω + ω 2)2 + (1 + ω - ω 2) 2 = -4

Solution:
L.H.S = (1 - ω + ω 2)2 + (1 + ω - ω 2) 2

        = (- ω - ω )2 + (- ω 2- ω 2) 2

        = (- 2ω )2 + (- 2ω 2) 2 (Since ω 3=1 and 1 + ω + ω 2 = 0)

        = 4 ω 2 + 4 ω 4

        = 4 ω 2 + 4 ω

        = -4

        = R.H.S

(1 - ω + ω 2)2 + (1 + ω - ω 2) 2 = -4

Question-177

Prove (1 - ω )(1 - ω 2)(1 - ω 4)(1 - ω 8) = 9

Solution:
L.H.S = (1 - ω )(1 - ω 2)(1 - ω 4)(1 - ω 8)
        = (1 - ω 2 - ω + ω 3)(1 - ω 8 - ω 4 + ω 12)
        = (1 - ω 2 - ω + 1)[1 - ω 8 - ω 4 + (ω 3) 4] (Since ω 3=1 and 1 + ω + ω 2 = 0)
        = (2 - ω 2 - ω )(2 - ω 8 - ω 4)
        = (2 - ω 2 - ω )[2 - ω 2(ω 3) 2 - ω (ω 3)]

        = (2 - ω 2 - ω )(2 - ω 2 - ω )
        = (2 + 1)(2 + 1)

        = 3 × 3 = 9 =R.H.S

(1 - ω )(1 - ω 2)(1 - ω 4)(1 - ω 8) = 9

Question-178

Prove that the roots of the quadratic equations 2x2 – x - (3/2) = 0 are irrational.

Solution:

2x2 – x - =0
D = b2-4ac = (-1)2+ 4× 2× = 1+12 =13
is irrational.
Hence the roots of the quadratic equations 2x2 – x - =0 are irrational.

Question-179

If a, b, and c of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a) = 0 are real, then prove that a2 + b2 + c2 –ab-bc-ca = 0 if and only if a =b = c.

Solution:

(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0

3x2 - 2(a+b+c)x + (ab+bc+ca) = 0

If the a, b, c are equal, D = b2-4ac 0

i.e 4(a+b+c)2-12(ab+bc+ca) 0

i.e (a+b+c)2-3(ab+bc+ca) 0

i.e a2+b2+c2–ab–bc-ca 0

i.e 2[(a-b)2+(b-c)2+(c-a)2] 0

a2 + b2 + c2 –ab-bc-ca = 0 if and only if 2[(a-b)2+(b-c)2+(c-a)2] = 0

i.e iff a = b = c [as (a-b)2,(b-c)2,(c-a)2 are 0]

Question-180

Prove 1 + ω n + ω 2n = 0, when n = 2, 4

Solution:
Put n =2
L.H.S = 1 + ω 2 + ω 4 = 1 + ω 2 + ω × ω 3 = 1 + ω 2 + ω = 0 = R.H.S

Put n = 4
L.H.S = 1 + ω 4 + ω 8 = 1 + ω × ω 3 + ω 2× (ω 3) 2 = 1 + ω + ω 2 = 0 = R.H.S

Question-181

Prove that the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a) = 0 are equal if and only if a =b = c.

Solution:

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a) = 0

3x2 - 2(a+b+c)x +(ab+bc+ca) = 0

If the roots are equal, D = b2-4ac =0

iff 4(a+b+c)2-12(ab+bc+ca) =0

iff 2[(a-b)2+(b-c)2+(c-a)2] = 0

iff a =b = c [as (a-b)2,(b-c)2,(c-a)2 are 0]

Question-182

Prove 1 + ω n + ω 2n = 3, when n is a multiple of 3.

Solution:
L.H.S = 1 + ω n + ω 2n = 1 + 1 + (1) 2 = 1 + 1 + 1 = 3 = R.H.S

1 + ω n + ω 2n = 3

Question-183

What set of points of the complex plane are defined by the condition |z - i| = 1 ?

Solution:
|z - i| = 1
|x + iy – i| = 1

|x + i(y – 1)| = 1


x2 + (y –1 )2 = 1

Circle of radius 1, Centre (0, 1)

Question-184

The sum of the roots of the equation is zero.            
           Prove that the product of the roots is

Solution:

(x2)+x(a+b-2c)+ab-ac-bc=0

The sum of the roots is zero a+b-2c = 0----------(i)

Product of the roots = ab-ac-bc = ab-c(a+b) = ab -[From (i)]

                             =





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