Question1
Solution:
Let z_{1} = x_{1} + iy_{1}
z^{2} = x_{2} + iy_{2}
(z_{1}+ z_{2})^{2} =
=
= (x_{1}+ x_{2})^{2} + 2i (x_{1}+ x_{2}) (y_{1}+y_{2})  (y_{1}+y_{2})^{2}
= x_{1}^{2} + 2x_{1}x_{2}+ x_{2}^{2} + 2ix_{1}y_{1} + 2ix_{1}y_{2} + 2ix_{2}y_{1} + 2ix_{2}y_{2}  y_{1}^{2}2y_{1}y_{2}  y_{2}^{2}
= x_{1}^{2} + 2ix_{1}y_{1}  y_{1}^{2} + x_{2}^{2} + 2ix_{2}y_{2}  y_{2}^{2} + 2(x_{1}+iy_{1}) (x_{2} +iy_{2})
= x_{1}^{2} +2ix_{1}y_{1}+(iy_{1})^{2} + x_{2}^{2} +2ix_{2}y_{2} + (iy_{2})^{2} + 2(x_{1}+iy_{1}) (x_{2} +iy_{2})
= (x_{1}+iy_{1})^{2} + (x_{2}+iy_{2})^{2} + 2 z_{1}z_{2}
= z_{1}^{2} + 2z_{1}z_{2} + z_{2}^{2}
Question2
i.
ii. 1 +
iii. â€“ 1 
iv
v. , (x>0)
vi. â€“ b + , (a, c > 0)
Solution:
i. = = = i
ii. 1 + = 1 + i
iii. â€“ 1  = 1  = 1  i
iv. ==
v. = + i0
vi. â€“ b + = â€“ b + = b + 2i
Question3
Solution:
Let Î±, Î² be the roots of the equation.
Sum of the roots Î± +Î² = 2+3 = 5
Product of the roots Î±x Î²= 2x3 = 6
âˆ´ The equation is given by
x^{2}  (sum of roots)x + product of roots = 0
âˆ´ Equation is x^{2}  5x + 6 = 0.
Question4
Solution:
25x^{2}â€“30x+9 = 0
D = b^{2}4ac = 900  4Ã— 25Ã— 9 =900 â€“ 900=0
Hence the two real equal roots of the equation are :
i.e
Question5
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
Question6
Solution:
If Î± and Î² an the root of the equation 3x^{2} +2x+6 = 0
Sum of the roots Î± +Î² =
Product of roots Î±Î² = = 2
(i) =
(ii) Î±^{2} + Î²^{2} = (Î± + Î² )^{2}  2 Î±Î² =
(iii .
Question7
Solution:
(1i)^{2} = 1^{2}  2(i) (1) +(i)^{2} = 12(i) +(i)^{2} = 12i1 = 2i
Question8
Solution:
2x^{2}2âˆš 3x + 1 =0
D = b^{2}4ac = 12  4Ã— 2Ã— 1 =12 â€“ 8= 4 > 0
= 2
Hence the two real and unequal roots are :
i.e
Question9
Solution:
Squaring both sides
=(x2)^{2}
x = x^{2}  2(x)(2)+4
0 = x^{2}  4x+ 4  x
= x^{2}  5x + 4
x^{2}  5x + 4 = 0
x^{2 } 4x  x + 4 = 0
x(x  4)  (x  4) = 0
x = 4 or x = 1
x=1 doesn't satisfy the equation
âˆ´ x = 4 .
Question10
i. 3 + i
ii. 3 â€“ i
iii.
iv. i
v. 4/5
vi. 49 â€“ i/7
Solution:
i. Conjugate of 3 + i is 3 â€“ i.
ii. Conjugate of 3 â€“ 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 â€“ i/7 is 49 + i/7.
Question11
Solution:
Squaring,
(3x+1)+(x1)  2= 4
4x  2= 4
= 2x2
Squaring,
3x^{2} â€“2x1 = (2x2)^{2 }
3x^{2}2x1= 4x^{2} â€“8x+4
x^{2}6x+5=0
D = b^{2}4ac = 36  4Ã— 1Ã— 5 =16 > 0
= 4
Hence the two real and unequal roots are :
i.e 5,1
Question12
Solution:
=.
âˆ´Conjugate of = i
Question13
Solution:
(a+ib) (c+id) = ac + iad + ibc +i^{2}bd = ac + i(ad+bc)  bd = (acbd) + i(ad+bc)
âˆ´ac  bd + i(ad + bc)
= (acbd) i (ad+bc)  (1)
(aib) (cid) = (acbd) i (bc+ad) (2)
From (1) and (2)
Question14
Solution:
4x + i(3x  y) = 3 â€“ i6
Equating the real and imaginary, we have
4x = 3
x = Â¾
3x â€“ y = 6
3(3/4) â€“ y = 6
9/4 â€“ y = 6
â€“ y = 6 â€“ 9/4
y = 33/4
Question15
Solution:
2x^{2}+1 =0 â‡’ x^{2} =1/2
Hence the complex roots of the equation are +i,i.
Question16
Solution:
The given equation is 2x^{2} â€“ 4x + 3 = 0.
Comparing with ax^{2} + bx + c = 0
a = 2, b = 4, c = 3
b^{2} â€“ 4ac = (4)^{2} â€“ 4Ã—2Ã— 3 = 16 â€“ 24 = 8 < 0
The roots are not equal.
Hence the roots of the given equation is =
Question17
Solution:
(3y  2) + i(7 â€“ 2x) = 0
Equating the real and imaginary, we have
3y â€“ 2 = 0
y = 2/3
7 â€“ 2x = 0
2x = 7
x = 7/2
The value of x = 7/2 and y = 2/3.
Question18
Solution:
x_{1}^{2} + y_{1}^{2} = x2^{2} + y_{2}^{2}
x_{1}^{2} = x_{2}^{2} and y_{1}^{2} = y_{2}^{2}
x_{1}= Â±x_{2} y_{1} = Â±y
âˆ´ z_{1} need not be z_{2}
Question19
Solution:
x^{2}â€“4x+7 = 0
D = b^{2}4ac = 16  4Ã— 1Ã— 7 =16 28= 12 <0
= 2i
Hence the two complex roots are :
i.e 2+i , 2+i
Question20
Solution:
Let the roots be Î±, 2Î±.
Î± + 2Î± =
â‡’ 3Î± =  2Î±  1
â‡’ Î± =
Î± .2Î± =
2Î± ^{2} =
= Î±^{2} +2
2(4Î±^{2} + 4Î± + 1) = 9(Î±^{2} +2)
8Î±^{2} +8Î± + 2 = 9Î±^{2} + 18
Î±^{2} + 8Î±  16 = 0
Î±^{2}  8Î± + 16 = 0
Î±^{2} 4Î±  4Î± + 16 = 0
Î±(Î±  4)  4(Î±  4) = 0
âˆ´Î± = 4 or Î± = 4
Question21
Solution:
Let Î±,Î² be the root of the equation x^{2}bx+c = 0; Î³, Î´and be the roots of the equation x^{2}â€“ cx + b = 0.
Then a + b = b, Î±Î²=c , Î³+ Î´=c and Î³ Î´ = b
Given that a  b = g  Î´
(
b^{2 } 4c = c^{2 } 4b
b^{2}c^{2}+4b4c = 0
(bc)(b+c)+4(bc) = 0
(bc) (b+c+4) = 0
Hence bc = 0 or b+c+4 = 0
(ie) b+c+4 = 0 or b = c
Question22
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.
Question23
Solution:
Let z_{1}, z_{2}, z_{3} be x_{1} + iy_{1}, x_{2} + iy_{2} and x_{3} + iy_{3} 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 z_{1}, z_{2}, z_{3} lie on a circle.
Question24
Solution:
x^{2}+x+1=0
D = b^{2}4ac = 1  4 =3 <0
= i
Hence the two complex roots are :
Question25
Solution:
Conjugate of 4 â€“ 3i is 4 + 3i. The absolute value of 4 â€“ 3i = = = = 5

Question26
Solution:
Let the price of the tape recorder be Rs. x
Let no. of student be n.
At the last moment
No. of students = (n2)
Increased contribution =
Original contribution =
According to the question
= +1
=
nx = (n2)(x+n) = nx + n^{2}  2x  2n
n^{2} 2n = 2x
x = , Also 170 < x < 195
170 < < 195
â‡’340 < n^{2}  2n < 390
Either 340 < n^{2} 2n
n^{2}  2n 340 â‰¥ 0
Roots are given by
n =
n â‰¥1+ or n < 1 
n = 20 or n^{2} 2n 390 < 0  (1)
n= n =
âˆ´ 1 < n < 1 +
Since n is a natural no. n = 1, 2, 3 â€¦â€¦..20  (2)
From (1) and (2),
n = 20
Cost of tape  recorder x = = = Rs. 180.
Question27
Solution:
x^{2} +2x +2 =0
D = b^{2}4ac = 4  4Ã— 2 =4 <0
= 2i
Hence the two complex roots are :
i.e 1i , 1+i
Question28
Solution:
Conjugate of â€“3 + i5 is â€“3  i5. The absolute value of â€“3 + i5 = = =

Question29
Solution:
(x^{2}  5x + 7)^{2}  (x  2) (x  3) = 1
(x^{2}  5x + 7)^{2}  [x^{2}  (2 + 3)x +2 Ã— 3] = 1
(x^{2}5x+7)^{2}  [x^{2} 5x+6] 1 = 0
Let x^{2}  5x = y (1)
(y+7)^{2}  (y+6) 1 = 0
y^{2} + 14y + 49  y  6  1 = 0
y^{2} + 13y + 42 =0
y =
=
=
=
=
= 6 or 7
âˆ´ y = 6 or 7
Substituting y = 6 in (1)
x^{2}  5x = 6
x^{2}  5x + 6=0
x = 3 or x = 2
Substituting y = 7 in (1)
x^{2}  5x = 7
x^{2}  5x + 7=0
x = .
Question30
Solution:
25x^{2}â€“30x +11=0
D = b^{2}4ac = 900  4Ã— 25Ã— 11 =200 <0
= 10i
Hence the two complex roots are :
i.e
Question31
Solution:
(x+1+i) (x+1i) (x1+i) (x1i) = [(x+1)^{2}i^{2}] [(x1)^{2} â€“i^{2}]
= (x^{2}+2x+1+1) (x^{2}2x+1+1)
= [(x^{2}+2)+2x] [(x+2)2x]
= (x^{2}+2)^{2} 4x^{2}
= x^{4}+4x^{2}+44x^{2}
= x^{4}+4
Question32
Solution:
5x^{2 } 6x + 2 = 0
D = b^{2}4ac = 36 4Ã— 5Ã— 2 = 4 <0
= 2i
Hence the two complex roots are :
i.e
Question33
Solution:
Conjugate of 5 is 5. The absolute value of 5 = = = 5

Question34
Solution:
(1+x)^{n} = p_{o}+p_{1} x + p_{2}x^{2}+ ............. p_{n}x^{n} â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)
Put x = 1,w,w^{2} in (1) and add
[1+w = w^{2} and 1+w^{2} = w]
3(p_{o}+p_{3}+p_{6} ............) = 2^{n}+(w^{2})^{n} + (w)^{n} ............... (2)
Now w =
âˆ´ w = i= cos isin, (._{.}. r = 1, Î¸ = )
w^{2} = cos+isin
âˆ´(w)^{n} +(w^{2})^{n} = 2cos (Demoivreâ€™ s Theorem)
Substituting in (2),3(p_{o}+p_{3}+p_{6}..............) = 2^{n} + 2cos
or p_{o}+p_{3}+p_{6}..............=.
Question35
2x^{2} + 2(m + n)x + m^{2} + n^{2} = 0.
Solution:
Let Î± , Î² be the roots of the equation 2x^{2} + 2(m + n)x + m^{2} + n^{2} = 0.
Then Î± + Î² = 2(m + n)/2 = (m + n)
Î±Î² = (m^{2} + n^{2 })/2
The roots of the required equation are (Î± +Î² )^{2} and (Î±  Î² )^{2}
Sum of the roots = (Î± + Î² )^{2} + (Î±  Î² )^{2 }= (Î± + Î² )^{2} + [(Î± + Î² )^{2 } 4Î±Î² ]
= (m + n)^{2} + [(m + n)^{2 } ]
= 4mn
Product of the roots = (Î± + Î² )^{2} (Î±  Î² )^{2 }= (Î± + Î² )^{2} [(Î± + Î² )^{2 } 4Î±Î² ]
= (m + n)^{ 2} [(m + n)^{2 } ]
= (m + n)^{ 2} [2mn â€“ m^{2} â€“ n^{2}]
The required equation is x^{2 } 4mnx + (m + n)^{2}[2mn â€“ m^{2}  n^{2}] = 0
or x^{2 } 4mnx + (m + n)^{2}[â€“ (m â€“ n)^{2}] = 0
or x^{2 } 4mnx  (m^{2}  n^{2})^{ 2 }= 0
Question36
Solution:
1i = isin
=
= sin
= 2^{1/4}[cos(8n+1)
= 2^{1/4} (cosisin for n = 0
= 2^{1/4}(cos(isin for n = 1
= 2^{1/4}(cosisin where
cos, sin
Question37
Solution:
3x^{2 } 7x + 5 = 0
D = b^{2}4ac = 49 4Ã— 3Ã— 5 = 11 <0
= i
Hence the two complex roots are :
Question38
Solution:
x = = =
x = ,
Question39
Solution:
Conjugate of 2i is 2i. The absolute value of 2i = = = 2

Question40
Solution:
Conjugate of is The absolute value of = = =

Question41
Solution:
Let a , b be the roots of the equation x^{2} â€“ lx + m = 0.
Î± + Î² = l
Î±Î² = m
Î±  Î² = 1
(Î± +Î² )^{2 }= (Î±  Î² )^{2 }+ 4Î±Î²
l^{2} = 1+ 4m
Question42
Solution:
13 x^{2}â€“7x + 1=0
D = b^{2}4ac = 49 4Ã— 13Ã— 1 = 3 <0
= i
Hence the two complex roots are :
Question43
Solution:
z = x+iy and 2^{1/3} = aib
(x+iy)^{1/3} = aib
Cubing both sides,
x+iy = (aib)^{3}
= a^{3}+b^{3}i3abi(aib)
= a^{3}+b^{3}i3a^{2}bi3ab^{2}
Equating the real and imaginary,
x = a^{3}3ab^{2}
y = b^{3}3a^{2}b
= 4(a^{2}b^{2})
Question44
Solution:
(1w+w^{2}) (1w^{2}+w^{4}) (1w^{4}+w^{8}) ................. 2n factors.
= (1w+w^{2}) (1w^{2}+w) (1w+w^{2}) ................... 2n factors.
( since w^{4}=w, w^{8} = w^{2} ........)
= (2w)(2w^{2})(2w)(2w^{2}) ....................... 2n factors.
= (2^{2}w^{3}) (2^{2}w^{3}) ................. n factors.
= (2^{2})^{n} = 2^{2n}
Question45
Solution:
= 3i Conjugate of 3i is â€“3i. The absolute value of 3i = = = 3

Question46
Solution:
9x^{2}+10x+3 =0
D = b^{2}4ac = 100 4Ã— 9Ã— 3 = 8 <0
= 2i
Hence the two complex roots are :
i.e
Question47
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 (n1) point leaving itself. But these line segments include those also which are in the reverse order.
Hence actual number of line segments = = 10
= 10
n^{2}  n = 20
n^{2}  n  20 = 0
n^{2}  5n + 4n 20 =0
n(n  5) + 4(n  5) = 0
n = 5 or n = 4
Since line segment cannot be negative.
âˆ´ Number of points on the plane = 5 .
Question48
Solution:
8x^{2}+9x+3 =0
D = b^{2}4ac = 81 4Ã— 8Ã— 3 = 15<0
= i
Hence the two complex roots are :
Question49
Solution:
Let Î±, Î² be the two root of the equation,
x^{2} +px+45 = 0
Sum of roots Î±+Î² = p
Product of roots Î± Î² = 45
(Î± Î²)^{2} = 144 (given)
We know
(Î±+Î²)^{2}  (Î± Î² )^{2} = 4Î±Î²
(p)^{2}  (Î± Î²)^{2} = 4 x 45
p^{2}  144 = 180
p^{2} = 144 + 180 = 324
p = =
Case (i) p = 18
Then the given equation is
x^{2} + 18x+45 = 0
x^{2} +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
x^{2}  18x+45 =0
x^{2} 15x3x+45 = 0
x(x15) 3 (x15) = 0
(x  15)(x  3)= 0
x = 15 or 3
Hence the roots are 3,15, or 3,15.
Question50
Solution:
âˆ´ 2cosÎ¸ = x+ and 2cosÏ† = y +
âˆ´ x = cosÎ¸ + isinÎ¸
y = cosÏ†+ i sinÏ†
= (cosÎ¸ +isinÎ¸ ) (cosÏ†  isinÏ† ) = cos(Î¸ Ï† ) + i sin(Î¸ Ï† )
Similarly =cossin
Question51
Solution:
Conjugate of â€“ 4i/3 is + 4i/3. The absolute value of â€“ 4i/3 = = = 4/3

Question52
Solution:
Conjugate of is . The absolute value of = = = = 1 
Question53
Solution:
x^{2}  6x+a = 0
Sum of the root Î±+Î² = 6  1
Product of the root Î±Î² = a  2
Also 3Î± +2Î² = 20  3
(1) Ã— 2 â‡’ 2Î± +2Î² = 12  4
(3)  (4)
Î± = 8
Substituting in (1)
Now 8 + Î² = 6
âˆ´ Î² = 2
8(2) = a
âˆ´ a = 16.
Question54
Solution:
17x^{2}â€“28x +12=0
D = b^{2}4ac = 784  4Ã— 17Ã— 12 = 32<0
= 4i
Hence the two complex roots are :
i.e
Question55
Solution:
cosÎ±+ cosÎ² +cosÎ³ = 0  (i)
sinÎ± + sin Î² +sinÎ³= 0  (ii)
Multiply (ii) by i and add to (i),
(cosÎ± + isinÎ± ) + (cosÎ² +isinÎ² ) + (cosÎ³ + sinÎ³ ) = 0  (iii)
Let z_{1} = cosÎ± + isinÎ±
z_{2} = cosÎ² + isinÎ²
z_{3} = cosÎ³ + i sinÎ³
From( iii),
z_{1}+z_{2}+z_{3} = 0  (iv)
âˆ´ z_{1}^{3} +z_{2}^{3}+z_{3}^{3} = 3z_{1}z_{2}z_{3}
âˆ´ (cosÎ± + isinÎ± )^{3} + (cosÎ² +isinÎ²)^{3} + (cosÎ³ + sinÎ³ )^{3}= 3(cosÎ± + isinÎ± )( cosÎ² +isinÎ² )( cosÎ³ +isinÎ³ )
or (cos3Î± + isin3Î± ) +(cos3Î² +isin3Î² )+(cos3Î³ + isin3Î³ )= 3[cos(Î± +Î² +Î³ )+i sin(Î± +Î² +Î³ )] ............(v)
Equation of the real parts,
cos3Î± + cos3Î² + cos3Î³ = 3cos(Î± +Î² +Î³ )
Question56
Solution:
Number of lines given is 15.
(x/2)(x  1) = 15
x(x  1) = 30
x^{2}  x â€“ 30 = 0
x^{2}  6x + 5x â€“ 30 = 0
x(x â€“ 6) + 5(x â€“ 6) = 0
(x + 5)(x â€“ 6) = 0
x = 5 or x = 6
âˆ´ A figure has 6 points if only 15 lines can be drawn connecting its vertices.
Question57
Solution:
âˆ´
âˆ´
Question58
Solution:
21x^{2}+9x+1=0
D = b^{2}4ac = 81  4Ã— 21Ã— 1 =  3 <0
= i
Hence the two complex roots are :
Question59
Solution:
Conjugate of 1 is 1. = = 1

Question60
Solution:
Put 5^{x} = y
y Ã—5^{1}+ = 125 + 1
5y + = 126
5y^{2} + 25  126y = 0
5y^{2}  126y + 25 = 0
5y^{2}  125y  1y + 25 =0
5y(y25)  1 (y25) = 0
y = 25 or 5y 1 = 0
âˆ´ y = 25 or 1/5
Put y = 25
Then 5^{x} = 5^{2}
âˆ´ x = 2
Put y = 5^{1}
Then 5^{x} = 5^{1}
âˆ´ x = 1
âˆ´ the solution is 2 and 1.
Question61
Solution:
Conjugate of i is i. The absolute value of i = = = 1 
Question62
Solution:
17x^{2}8x+1=0
D = b^{2}4ac = 64  4Ã— 17Ã— 1 =  4 <0
= 2i
Hence the two complex roots are :
i.e
Question63
Solution:
21x^{2}29x+11=0
D = b^{2}4ac = 841 4Ã— 21Ã— 11 =  83 <0
= i
Hence the two complex roots are :
Question64
Solution:
=
=
=
=
= A+iB
When A = 0 and B =
Question65
Solution:
The complex number whose absolute value is 2 are 2, 2, 2i, 2i,+i, i, +i
andi
Question66
Solution:
For a perfect square b^{2}  4ac = 0
(2k +4)^{2}  4(8k+1)(4  k) = 0
4k^{2} + 2 x 2k x 4 + 16  4(32k  8k^{2} + 4  k) = 0
4k^{2} + 16k + 16  4(31k  8k^{2} + 4) = 0
4(k^{2} + 4k + 4)  4(31k  8k^{2} + 4) = 0
k^{2} + 4k + 4  31k +8k^{2}  4 = 0
9k^{2}  27 k = 0
k^{2} = 3k
k(k  3) = 0
k = 0 or k =3 .
Question67
Solution:
The roots of the equation are real if D > 0.
b^{2}  4ac > 0.
2^{2} (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.
Question68
Solution:
21x^{2}+28x+10=0
D = b^{2}4ac = 784 4Ã— 21Ã— 10 =  56 <0
= i
Hence the two complex roots are :
i.e .
Question69
i. (2i)^{3}
ii. (8i)(i/8)
iii. i^{6} + i^{8}
iv. i^{ }+i^{2}+i^{3}+i^{4 }
v. i^{9} + i^{10} + i^{11} + i^{12}
Solution:
i. (2i)^{3}= 08i
ii. (8i)= i^{2} = 1+0i
iii. i^{6} + i^{8} = i^{6 }(1 + i^{2}) = i^{6 }(1 â€“ 1) = 0+0i
iv. i^{ }+i^{2}+i^{3}+i^{4 }= i(1^{ }+ i + i^{2 }+ i^{3})^{ }= i(1^{ }+ i  1^{ } i) = 0+0i
v. i^{9}(1 + i + i^{2} + i^{3}) = i^{9} (1 + i  1  i) = 0+0i
Question70
Solution:
x+i = x+i
=
=
=
Hence .
Question71
Solution:
27x^{2}+10x+1=0
D = b^{2}4ac = 100 4Ã— 27Ã— 1 =  8 <0
= i
Hence the two complex roots are :
i.e .
Question72
i. i^{4} + i^{8} + i^{12 }+ i^{16}
ii. i + i^{5} + i^{9} + i^{13}
iii. i^{38}
Solution:
i. (i^{2})^{ 2} + (i^{2})^{ 4} + (i^{2})^{ 6} + (i^{2})^{ 8} = (1)^{ 2} + (1)^{ 4} + (1)^{ 6} + (1)^{ 8 }= 1 + 1 + 1 + 1 = 4+0i
ii.i(1 + i^{4} + i^{8} + i^{12}) = i[1 + (i^{2})^{ 2} + (i^{2})^{4} + (i^{2})^{6}]
= i[1 + (1)^{ 2} + (1)^{4} + (1)^{6}]
= i[1 + 1 + 1 + 1]
= 0+4i
iii. i^{38 }= (i^{2})^{19}= (1)^{19 }=(1)^{1 }= 1+0i
Question73
Solution:
Let one of the numbers be x.
Then the other number = x + 2
As given x^{2} + (x + 2)^{2} = 100
x^{2} + x^{2} + 4x + 4 = 100
2x^{2} + 4x  96 = 0
x^{2} + 2x  48 = 0
x = ===6, 8
Since as given x is + ve.
x = 6, x + 2 = 8
Hence the numbers are 6, 8.
Question74
Solution:
=  (i)
Hence (i) is real if sinÎ¸ = 0
This gives Î¸ = nÏ€ , where n is an integer.
If (i) is purely imaginary,
34sin2Î¸^{ }= 0 â‡’ sin2Î¸^{ }=
or sinÎ¸ = Â±
This gives Î¸ = n where n is an integer.
Question75
Solution:
x^{2}(3i)x  6=0
D = b^{2}4ac = (3i)^{2} 4Ã— 1Ã— (6) =18412+24 = 14+12
Let = a+bi
14+12= a^{2}b^{2}+2abi
â‡’ a^{2}b^{2}=14 (i)
ab = 6
(a^{2}+b^{2})^{2} = (14)^{2}+4(72)= 196 + 288 = 484
a^{2}+b^{2 }= 22 (ii)
From (i) and (ii),
a^{2} = 18 and b^{2} = 4 â‡’ a = Â± 3 and b = Â± 2
a+bi = 3+2i or 32i
Hence the two complex roots are :
i.e 3,2i
Question76
Solution:
x = a+b, y = aÎ± + bÎ²and z = aÎ² +bÎ±
xyz = (a+b)(aÎ± +bÎ² ) (aÎ² +bÎ± ) where Î± = w and Î²= w^{2}
= (a+b)(aw+bw^{2})(aw^{2}+bw)
= (a+b) (a^{2}w^{3}+abw^{2}+abw^{4}+b^{2}w^{3})
= (a+b) [a^{2}+b^{2}+ab(w^{2}+w^{4})]
= (a+b) [a^{2}+b^{2}+ab(1)] = (a+b)(a^{2}ab+b^{2})
= a^{3}+b^{3}
Question77
Solution:
Put = y â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)
8y  = 2
8y^{2} â€“ 1 = 2y
8y^{2} â€“ 2y â€“ 1 = 0
8y^{2} + 2y  4y â€“ 1 = 0
2y(4y + 1)  (4y + 1) = 0
(2y  1)(4y + 1) = 0
y = 1/2 or y = Â¼
Substitute y = 1/2 in (i)
Squaring both sides we have
=
4x = x + 3
3x = 3
x = 1
= Â¼ which is impossible since the expression on LHS is not negative x Â¹ Â¼ and y = Â¼ gives
x = 1 which is the required solution of the given solution.
Question78
i. (5+i4)+(5i4)
ii. â€“(1 + i) + i7 â€“5
iii. 3(7 + i7) + i(7 + i7)
iv. (1  i) â€“ (1 + i6)
Solution:
i. (5+i4)+(5i4) =10+0i
ii. â€“(1 + i) + i7 â€“5 = 1  i + i7 â€“5 = â€“4 + 6i
iii. 3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i^{2} = 21 + 21i + 7i â€“ 7 = 14 + 28 i
iv. (1  i) â€“ (1 + i6) = 1 â€“ i + 1 â€“ 6i = 2 â€“ 7i
Question79
(i) 2x^{2}3x+5=0
(ii) (3k1)x^{2}mx+(ab) = 0, k â‰ .
(iii) x+ =7
(iv) â‰ 0.
Solution:
(i) 2x^{2}3x+5=0
The sum of the roots : =
The product of the roots : c/a = 5/2
(ii) (3k1)x^{2}mx+(ab) = 0, k â‰ .
The sum of the roots : =
The product of the roots : c/a = (a  b)/(3k  1)
(iii) x+ =7
x^{2}7x+1=0
The sum of the roots : = 7
The product of the roots : c/a = 1
(iv) â‰ 0.
x^{2}+kx k^{2} = 0
The sum of the roots : = k
The product of the roots : c/a = k^{2}
Question80
i.
ii. (7 â€“ i2) â€“ (4 + i) + (3 + i5)
Solution:
i.= =
ii. (7 â€“ i2) â€“ (4 + i) + (3 + i5) =(7 4 â€“ 3) + (â€“ i2 â€“ i + i5) = 0+2i
Question81
Solution:
=
=
=
Question82
(i)
(ii) 7i,2i
(iii)
(iv) 34i,2+3i
(v)
(vi) 3i,1+2i
Solution:
(i)
The sum of the roots : = 3
The product of the roots :
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2 } (3)x + 7/4 = 0
4x^{2} â€“12x +7 = 0
(ii) 7i,2i
The sum of the roots : 7i+ 2i = 9i
The product of the roots : 7i Ã— 2i = 14
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2 } (9i)x â€“14 =0
(iii)
The sum of the roots : = 0
The product of the roots : =
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2 } (0)x + 1/16 = 0
16x^{2} +1 = 0
(iv) 34i,2+3i
The sum of the roots : 34i+2+3i =5i
The product of the roots : (34i)Ã— (2+3i) = 18+i
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2 } (5+i)x +18+i =0
(v)
The sum of the roots :
The product of the roots : =
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2 } ()x + =0
2x^{2} (3+7i) +(3+9i)= 0
(vi) 3i,1+2i
The sum of the roots : (3i)+(1+2i) = 2 + i
The product of the roots : (3i)Ã— (1+2i) = 3+2 +i(6+1)= 1+7i
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2 }â€“ (2+i)x +(1+7i) =0
Question83
Solution:
2x^{2}16x+k=0
Let the roots be Î± and 2Î± .
The sum of the roots : Î± + 2Î± = 16/2
3Î±=8 (i)
The product of the roots : Î± Ã— 2Î± = k/2
2Î±2 = k/2 (ii)
From (i) and (ii) ,
2(8/3)^{2} = k/2
k = 256/9
Question84
x^{2} + y^{2} + z^{2} = 6ab.
Solution:
x^{2} + y^{2} + z^{2}= (a + b)^{2} + (aÏ‰ + bÏ‰ ^{2})^{2} + (aÏ‰ ^{2} + bÏ‰ )^{2}
= a^{2} + b^{2} + 2ab + a^{2}Ï‰ ^{2} + b^{2}Ï‰ ^{4} + 2abÏ‰ ^{3} + a^{2}Ï‰^{ 4} + b^{2}Ï‰ ^{2} + 2abÏ‰ ^{3}
= a^{2} + b^{2} + 2ab + a^{2}Ï‰ ^{2} + b^{2}Ï‰+ 2ab + a^{2}Ï‰ + b^{2}Ï‰^{2 }+ 2ab [Ï‰ ^{3} = 1]
= a^{2} + a^{2}Ï‰ + a^{2}Ï‰ ^{2 }+ b^{2} + b^{2}Ï‰+ b^{2}Ï‰ ^{2}+ 2ab + 2ab + 2ab
= a^{2}(1 + Ï‰ + Ï‰ ^{2})^{ }+ b^{2}(1 + Ï‰ + Ï‰ ^{2}) + 6ab
= a^{2}(0)^{ }+ b^{2}(0) + 6ab [1 + Ï‰ + Ï‰ ^{2} = 0]
= 6ab
âˆ´ x^{2} + y^{2} + z^{2} = 6ab
Question85
i^{3} + (6 + i3) â€“ (20 + i5) + (14 + i3)
Solution:
i^{3} + (6 + i3) â€“ (20 + i5) + (14 + i3) = i + (6 + i3) â€“ (20 + i5) + (14 + i3)
= (6  20 + 14) + i(1 + 3  5 +3)
= 0 + 0i
Question86
Solution:
x^{2}+(2m+1)x+m^{2}+2=0
Let the roots be Î± and 2Î± .
The sum of the roots : Î± + 2Î± = (2m+1)
3Î± = (2m + 1)(i)
The product of the roots : Î± Ã— 2Î± = m^{2}+2
2Î±^{2} = m^{2}+2 (ii)
From (i) and (ii) ,
2( )^{2} = m^{2}+2
8m^{2}+8m+2 = 9m^{2} +18
m^{2} â€“8m + 16 = 0
m = 4.
Question87
Solution:
Let Î± ,Î² be the roots of the equation x^{2 }â€“ lx + m = 0.
Then Î± + Î² = l and Î± Î² = m.
Î±  Î² = 1
â‡’ (Î±  Î² )^{2} = 1
â‡’ (Î± + Î² )^{2}  4Î± Î² = 1
â‡’ l^{2}  4m = 1
â‡’ l^{2} = 1 + 4m
Question88
Solution:
=  3
Question89
Solution:
2x^{2}+3x+k = 0
Let the roots be Î± and Î± .
The sum of the roots : Î± + Î± = (3/2) (i)
The product of the roots : Î± Ã— Î± = k/2 (ii)
From (i) and (ii) ,
k =
Question90
(1 + i)^{4}
Solution:
(1 + i)^{4}= (1 + i)^{2 }(1 + i)^{2}
= (1^{2} + i^{2} + 2i)^{ }(1^{2} + i^{2 }+ 2i)
= (1^{2}  1 + 2i)^{ }(1^{2}  1^{ }+ 2i)
= (2i)^{ }(2i)
= 4i^{2 }
= 4
Question91
Solution:
ax^{2}+bx+a= 0
Let the roots be Î± and Î² .
The sum of the roots : Î± + Î² = (b/a) (i)
The product of the roots : Î± Ã— Î² = 1 (ii)
From (ii),
Î± =
or the roots of the quadratic equation ax^{2}+bx+a= 0 are reciprocals of each other.
Question92
Solution:
Since Î± , Î² are the complex roots of unity,
Î± = Ï‰ and Î² = Ï‰ ^{2}
Hence Î± ^{4} + Î² ^{4} + Î± ^{1}Î² ^{1 }= Ï‰ ^{4} + Ï‰ ^{8} + Ï‰ ^{1}Ï‰ ^{2 }
= Ï‰ ^{3}Ï‰ + Ï‰ ^{6}Ï‰ ^{2} + (Ï‰ ^{3})^{2} [Ï‰ ^{3} = 1]
= Ï‰ + Ï‰ ^{2} + 1
= 0
Question93
(7 + i5)(7 â€“ i5)
Solution:
(7 + i5)(7 â€“ i5) = 7(7 â€“ i5) + i5(7 â€“ i5)
= 49 â€“ i35 + i35 + 25
= 74
Question94
Solution:
x^{2} 4x+k = 0
Let the roots be Î± and Î² .
The sum of the roots : Î± + Î² = 4 (i)
The product of the roots : Î± Ã— Î² = k (ii)
Î±  Î² = 2 (given) (iii)
From (i) and (iii) ,
Î± = 3 and Î² = 1
âˆ´ k = Î± Ã— Î² = 3.
Question95
(Î±  Î³ ) (Î²  Î³ ) (Î± + Î´ ) (Î² + Î´ ) = q^{2} â€“ p^{2 }
Solution:
Î± +Î² = p ; Î± Î² = 1
Î³ + Î´ = q ; Î³ Î´ = 1
L.H.S. = (Î±  Î³ ) (Î²  Î³ ) (Î± + Î´ ) (Î² + Î´ )
= (Î± Î²  Î³ Î±  Î³ Î² + Î³ ^{2}) (Î± Î² + Î± Î´ + Î² Î´ + Î´ ^{2})= (1  Î³ p + Î³ ^{2}) (1 + pÎ´ + Î´ ^{2})
= 1 + pÎ´ + Î´ ^{2}  Î³ p  Î³ Î´ p^{2}  Î³ Î´ ^{2}p + Î³ ^{2} + pÎ³ ^{2}Î´ + Î³ ^{2}Î´ ^{2 }
= 1 + pÎ´ + Î´ ^{2}  Î³ p  p^{2}  Î´ p + Î³ ^{2} + pÎ³ +1
= 1 + Î´ ^{2} + Î³ ^{2} + 1 â€“ p^{2 }
= 2 + Î´ ^{2} + Î³ ^{2}  P^{2 }
= 2Î´ Î³ + Î´ ^{2} + Î³ ^{2}  P^{2 }
= (Î´ + Î³ )^{2} â€“ P^{2 }
= q^{2} â€“ p^{2 }
Question96
3i^{3} (15i^{6})
Solution:
3i^{3} (15i^{6}) = 3(i^{2})i(15(i^{2})^{ 3})
= 3(1)i(15(1)^{ 3})
= 3(1)i15(1)
= 45i
Question97
Solution:
2kx^{2}20x+21 = 0
Let the roots be Î± and Î± 2.
The sum of the roots : Î± + Î± 2 = 10/k (i)
The product of the roots : Î± Ã— (Î± 2) = 21/2k (ii)
From (i) ,
Î± = 5/k + 1
( + 1)( â€“ 1) = 21/2k
() =
2(25 â€“ k^{2}) = 21k
2k^{2} +21k â€“50 = 0
2k^{2} +25k â€“4k â€“50 = 0
(k2)(2k+25) = 0
â‡’ k =2 or k =
Question98
Solution:
(a + b + c)(a + bÏ‰ + cÏ‰ ^{2})(a + bÏ‰ ^{2} + cÏ‰ )
= (a + b + c)[a^{2} + abÏ‰ ^{2} + acÏ‰ + abÏ‰ + b^{2}Ï‰ ^{3} + bcÏ‰ ^{2} + acÏ‰ ^{2} + bcÏ‰ ^{4} + c^{2}Ï‰ ^{3}]
= (a + b + c)[(a^{2} + b^{2 }+ c^{2}) +^{ }ab(Ï‰ + Ï‰ ^{2}) + ac(Ï‰ + Ï‰ ^{2}) + bc(Ï‰ ^{2} + Ï‰ )]
= (a + b + c)[(a^{2} + b^{2 }+ c^{2}) +^{ }ab(1) + ac(1) + bc(1)]
= (a + b + c)(a^{2} + b^{2 }+ c^{2} â€“^{ }ab â€“ ac â€“ bc)
= a^{3} + b^{3} + c^{3}  3abc
âˆ´ (a + b + c)(a + bÏ‰ + cÏ‰ ^{2})(a + bÏ‰ ^{2} + cÏ‰ ) = a^{3} + b^{3} + c^{3}  3abc
Question99
Solution:
[âˆš 2 (cos 56^{o} 15â€™ + i sin56^{o} 15â€™)]^{8}
= 16[cos (8Ã— 56^{o} 15â€™) + i sin (8 Ã— 56^{o} 15â€™)]
= 16[cos 450^{o} + i sin 450^{o}]
= 16[cos 90^{o} + i sin 90^{o}]
= 16[0 + i]
= 16i
âˆ´ [âˆš 2 (cos 56^{o} 15â€™ + i sin56^{o} 15â€™)]^{8} = 16i.
Question100
Solution:
x^{2 } 3x+2 =0
If the roots be Î± and Î² .
Then the sum of the roots : Î± + Î² = 3 (i)
The product of the roots : Î± Ã— Î² = 2
The equation whose roots are larger by 2 are Î± +2 and Î² +2.
The sum of the roots : Î± +2 + Î² +2 = Î± + Î² + 4 =7 [From (i)]
The product of the roots :(Î± +2) (Î² +2 ) = Î± Î² +2(Î± + Î² ) +4
= 2+2(3)+4
= 12[From (i) and (ii)]
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2}(7)x + 12 = 0
Question101
Solution:
=
=
=
=
Question102
Solution:
x^{2} +px+q =0
If the roots be Î± and Î² .
Then the sum of the roots : Î± + Î² = p (i)
The product of the roots : Î± Ã— Î² = q (ii)
The equation whose roots are n times are nÎ± and nÎ² .
The sum of the roots : nÎ± + nÎ² = n(Î± + Î² ) =np [From (i)]
The product of the roots :(nÎ± ) ( nÎ² ) = n^{2} Î± Î² = n^{2} q [From (ii)]
The equation is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2}+npx + n^{2} q = 0
Question103
(i) Î± ^{2} + Î² ^{2} (ii) Î± ^{3} + Î² ^{3} (iii) (iv) (v) Î± Î² ^{2} + Î± ^{2} Î²
Solution:
Î± , Î² are the roots of the quadratic equation 3x^{2}5x8 =0;
Then the sum of the roots : Î± + Î² = 5/3 (i)
The product of the roots : Î± Ã— Î² = 8/3 (ii)
(i) Î± ^{2} + Î² ^{2} = (Î± + Î² )^{2}  2Î± Î²
= (5/3)^{2} â€“2(8/3) [From (i) and (ii)]
==
(ii) Î± ^{3} + Î² ^{3} = (Î± + Î² )^{3}  3Î± Î² (Î± +Î² )
= (5/3)^{3} 3(8/3)(5/3) [From (i) and (ii)]
= =
(iii) = =[From (i) and (ii)]
=
(iv) = = [From (i) and (ii)]
=
(v) Î± Î² ^{2} + Î± ^{2} Î² = Î± Î² (Î± + Î² ) = (5/3)(8/3) [From (i) and (ii)]
= 40/9
Question104
Solution:
(5 + i9) Ã· (3 + i4) =
=
=
=
Question105
(i) (ii) Î± ^{2} + Î² ^{2}.
Solution:
Î± , Î² are the roots of the quadratic equation x^{2}+3x+6 = 0;
Then the sum of the roots : Î± + Î² = 3 (i)
The product of the roots : Î± Ã— Î² = 6 (ii)
(i) = =[From (i) and (ii)]
==
(ii) Î± ^{2} + Î² ^{2} = (Î± + Î² )^{2} 2Î± Î² = (3)^{2}2(6) [From (i) and (ii)]
= 3
Question106
Solution:
(2 â€“ i5) Ã· (3 â€“ i6) =
=
=
=
=
Question107
(i) Î± ^{3} + Î² ^{3} (ii) Î± ^{4} + Î² ^{4}
Solution:
Î± + Î² = 1 and Î± ^{2} + Î² ^{2} = 2 (given) (i)
Ãž 2Î± Î² = (Î± + Î² )^{2} â€“(Î± ^{2} + Î² ^{2}) = 12 = 1;
Ãž a Î² = 1/2 (ii)
(i) Î± ^{3} + Î² ^{3} = (Î± + Î² )^{3}  3Î± Î² (Î± +Î² ) [From(i) and (ii)]
= 1 3(1/2)(1) = 1+3/2 =5/2
(ii) Î± ^{4} + Î² ^{4} = (Î± ^{2} + Î² ^{2})^{2}2(Î± Î² )^{2 }
= (2)^{2}2(1/2)^{2 }[From(i) and (ii)]
= 4 â€“ (1/2) = 7/2
Question108
Solution:
=
=
=
=
=
=
=
=
Question109
(i) 3Î± ,3Î² (ii) Î± ^{2}, Î² ^{2}.
Solution:
Î± , Î² are the roots of the quadratic equation 3x^{2}4x+1 =0
Then the sum of the roots : Î± + Î² = 4/3 (i)
The product of the roots : Î± Ã— Î² = 1/3 (ii)
(i) 3Î± +3 Î² = 3( Î± + Î² ) = 4 [From(i)]
3Î± Ã— 3 Î² = 9(1/3) =3 [From (ii)]
Hence the equation whose roots are 3Î±, 3Î² is : x^{2}(sum of the roots)x + product of the roots = 0
x^{2} 4x + 3 = 0
(ii) Î± ^{2} + Î² ^{2} = ( Î± + Î² )^{2} â€“2 Î± Î² = (4/3)^{2} â€“2(1/3) [From(i) and (ii)]
=10/9
Î± ^{2 Ã— }Î² ^{2} = (1/3)^{2} = 1/9 [From (ii)]
Hence the equation whose roots are Î± ^{2} , Î² ^{2} is :
x^{2}(sum of the roots)x + product of the roots = 0
x^{2} (10/9)x + (1/9) = 0
9x^{2}10x+1= 0
Question110
Solution:
==+0i
Question111
(i) Î± ^{2} , Î² ^{2} (ii) .
Solution:
Î± , Î² are the roots of the quadratic equation x^{2}+px+q =0
Then the sum of the roots : Î± + Î² = p (i)
The product of the roots : Î± Ã— Î² = q (ii)
(i) Î± ^{2} + Î² ^{2} = (Î± + Î² )^{2}  2Î± Î² = p^{2} â€“2q [From(i) and (ii)]
Î± ^{2} Ã— Î² ^{2 }= (Î± Î² )^{2 }= q^{2 }[From (ii)]
Hence the equation whose roots are Î± ^{2}, Î² ^{2} is: x^{2}(sum of the roots)x + product of the roots = 0
x^{2} (p^{2}2q)x + (q^{2}) = 0
(ii) = (Î± + Î² ) +() = p  [From(i) and (ii)]
= Î± Î² ++2^{ }= = (q+1)^{2 }[From (ii)]
Hence the equation whose roots is is :
x^{2}(sum of the roots)x + product of the roots = 0
x^{2} (p )x + (q+1)^{2 }= 0
qx^{2}+p(1+q)x+(q+1)^{2}=0,q â‰ 0.
Question112
Solution:
Let z = 4 â€“ i3
Then = 4 + i3
z^{2 }= 16 + 9 = 25
Multiplicative inverse 4 â€“ i3 is 1/z = /z^{2} =
Question113
Solution:
Let z = +i3
Then =  i3
z^{2 }= ()^{2} + (3)^{2 }= 5 + 9 = 14
Multiplicative inverse +i3 is 1/z = /z^{2} =
Question114
Solution:
Î± , Î² are the roots of the quadratic equation 2x^{2}5x+7 =0
Then the sum of the roots : Î± + Î² = 5/2 (i)
The product of the roots : Î± Ã— Î² = 7/2 (ii)
Sum 2Î± +3 Î² + 3Î± +2 Î² = 5(Î± + Î² ) = 5(5/2) =25/2 [From (i)]
Product (2Î± +3 Î² )(3Î± +2 Î² ) = 6(Î± ^{2} + Î² ^{2})+13 Î± Î² =6(Î± + Î² )^{2 }+ Î± Î²
= 6(5/2)^{2}+(7/2) [From (i) and (ii)]
= 75/2 + 7/2
= 82/2
Hence the equation whose roots are 2Î± +3 Î² , 3Î± +2 Î² is :
x^{2}(sum of the roots)x + product of the roots = 0
x^{2} (25/2)x + (82/2) = 0
2x^{2} (25)x + (82) = 0
Question115
Solution:
Let z = i
Then = i
z^{2 }= 0 + (1)^{2 }= 1
Multiplicative inverse â€“i is 1/z = /z^{2} = i/1 = 0+i
Question116
Solution:
Let Î± , Î² are the roots of the quadratic equation x^{2}px+q =0.
Then the sum of the roots : Î± + Î² = p (i)
The product of the roots : Î± Ã— Î² = q (ii)
Sum = = [From (i) and (ii)]
Product = 1/Î±Î² = 1/q [From (ii)]
Hence the equation whose roots are is :
x^{2}(sum of the roots)x + product of the roots = 0
x^{2} (p/q)x + (1/q) = 0
qx^{2} px + 1 = 0
Question117
Solution:
x= rcosÎ¸=1;y = rsinÎ¸=âˆ’1
x + iy = 1 â€“ i
x = 1, y = 1
r =
and tan Î¸ = y/x = 1, Î¸ = .
Thus, the polar coordinates of 1 â€“ i are (, ) and its polar form is (cos + i sin).
Question118
Solution:
Let Î± , Î² be the roots of the quadratic equation.
then Î± + Î² =3 (i)
Î± ^{3} + Î² ^{3 }= 63 (ii)
From (ii), (Î± + Î² )^{3} â€“3 Î± Î² (Î± + Î² ) = 63
Substituting (i), 27 â€“ 3 Î± Î² (3) = 63
Î± Î² =  4
Hence the equation is : x^{2}(sum of the roots)x + product of the roots = 0
x^{2} 3x  4 = 0
Question119
Solution:
x= rcosÎ¸=âˆ’1;y = rsinÎ¸=1
x + iy = 1 + i
x = 1, y = 1
r =
and tan Î¸ = y/x = 1, Î¸ = .
Thus, the polar coordinates of â€“1+ i are (, ) and its polar form is (cos + i sin).
Question120
(i) (ii)
Solution:
Î± , Î² are the roots of the quadratic equation ax^{2}+bx+c = 0
Then the sum of the roots : Î± + Î² = b/a (i)
The product of the roots : Î± Ã— Î² = c/a (ii)
(i) = = [From (i) and (ii)]
=
(ii) = == [From (i) and (ii)]
=
Question121
Solution:
x= rcosÎ¸=âˆ’1;y = rsinÎ¸=âˆ’1
x + iy = 1  i
x = 1, y = 1
r =
and tan Î¸ = y/x = 1, Î¸ = .
Thus, the polar coordinates of â€“1+ i are (,) and its polar form is (cos + i sin).
Question122
Solution:
Î± , Î² are the roots of the quadratic equation px^{2}+qx+r =0
Then the sum of the roots : Î± + Î² = q/p (i)
The product of the roots : Î± Ã— Î² = r/p (ii)
Consider = +2=+2
= [From (i) and (ii)]
=
Sum + =
Product Ã— = 1
Hence the equation is : x^{2}(sum of the roots)x + product of the roots = 0
x^{2} ()x + (1) = 0
x^{2} â€“qx + = 0, p, r â‰ 0
Question123
Solution:
x + iy = 3
x = 3, y = 0
r =
and tan Î¸ = y/x = 0/3, Î¸ = Ï€ .
Thus, the polar coordinates of â€“1+ i are (3, Ï€ ) and its polar form is 3(cos Ï€ + i sinÏ€ ).
Question124
Solution:
y^{2} + y â€“2 =0
(y+2)(y1) =0
y = 2 or y = 1.
âˆ´ or 2 â‡’ x = 1 or â€“8
Solution set :x = 1 , 8
Question125
Solution:
x + iy = â€“4 + i4
x = 4, y = 4
r =
and tan Î¸ = 4/4 = , Î¸ = .
Thus, the polar coordinates of â€“4 + i4 are (8, ) and its polar form is 8(cos + i sin).
Question126
Solution:
x^{4} â€“ 5x^{2}+6 = 0
Put x^{2} = y
y^{2}  5y + 6 =0
(y3)(y2) = 0
y = 3 or y = 2
x^{2} = 3 or x^{2 }= 2
â‡’ x = Â± or Â±
Solution set :x = Â± or Â±
Question127
Solution:
x + iy =+ i
x = , y = 1
r =
and tan Î¸ = y/x =, Î¸ = Ï€ /6.
Thus, the polar coordinates of + i are (2, Ï€ /6) and its polar form is 2(cos + i sin).
Question128
Solution:
(x^{2} â€“3x)^{2}5(x^{2}3x)+6 = 0
Put x^{2} â€“3x = y
y^{2}  5y + 6 =0
(y3)(y2) = 0
y = 3 or y = 2
x^{2} â€“3x =3 or x^{2} â€“3x = 2
x^{2} â€“3x 3 = 0 or x^{2} â€“3x â€“ 2 = 0
x = or
Solution set :x = ,
Question129
Solution:
x + iy = i
x = 0, y = 1
r =
and tan Î¸ = y/x = âˆž , Î¸ = .
Thus, the polar coordinates of â€“1+ i are (1, ) and its polar form is (cos+ i sin).
Question130
Solution:
4^{x } 3.2^{x+2} +32 = 0
2^{2x} 12Ã— 2^{x}+32 =0
Put 2^{x} = y,
y^{2}12y+32 =0
(y8)(y4)= 0 â‡’ y = 8 and y = 4âˆ´ 2^{x} = 8, 4 â‡’ x = 3 , 2.
Solution set :x = 3, 2.
Question131
Solution:
2(cos 0^{o} + isin 0^{o})
r= 2, Î¸ = 0^{o }
Let 2(cos 0^{o} + isin 0^{o}) = x+iy
r = â‡’ x^{2} +y^{2} = 4 (ii)
tanÎ¸ = â‡’ tan 0Â° =
or y = 0 (ii)
Substituting (ii) in (i),
x = 2
Hence the required Cartesian form of 2(cos 0^{o} + isin 0^{o}) = 2+i0
Question132
Solution:
4^{x+1 } 6^{x } 2.9^{x +1} = 0
2^{2x+2} â€“3^{x} Ã— 2^{x}  2Ã— 3^{2x+2}=0
4 1 18=0
Put =y;
4y1
4y^{2}y18=0
4y^{2}9y+8y18=0
y(4y9)+2(4y9)=0
(4y9)(y+2) = 0
y = or 2
i.e = â‡’ = or x=2;
= 2 has no solution.
Solution set : x=2.
Question133
Solution:
5(cos 270^{o} + isin 270^{o}) = 5(0iÃ— 1)
The required Cartesian form is 0  5i.
Question134
Solution:
x^{4} + x^{3 } 4x^{2}+x+1 = 0
Divide by x^{2} ,
x^{2 }+ x â€“ 4 +=0
(x^{2}) + (x+)  4 = 0
Let x+ = y â‡’ x^{2}= y^{2 }â€“ 2
âˆ´ (y^{2}2)+y4 = 0
y^{2}+y6= 0
(y+3)(y2) = 0 â‡’ y = 3 ,2.
Thus, x+= 3 or x+ = 2
x^{2} +3x+1 = 0 or x^{2} â€“2x +1 = 0
or .
Solution set :
Question135
Solution:
4(cos300^{o} + isin300^{o}) = 4( i) = 22i
The required Cartesian form is 22i
Question136
Solution:
(x+1)(x+2)(x+3)(x+4) +1= 0
(x+1)(x+4)(x+2)(x+3) + 1= 0
(x^{2}+5x+4)(x^{2}+5x+6) +1 = 0
Put x^{2}+5x = y
(y+4)(y+6) + 1 = 0
y^{2} +10y +25 = 0
y = 5 , 5
i.e x^{2 }+ 5x + 5 = 0
x= ,
x = ,
Solution set : x = ,
Question137
Solution:
z = 1  i
x = 1, y = 
r = === 2
Î¸ =
z = 2, arg z = + 2Ï€ k, where k is an integer.
Question138
Solution:
Cross multiplying,
(2x^{2}12)(x1) = (2x3)(x+2)(x2)
2x^{3}2x^{2}12x+12 = 2x^{3}3x^{2}8x+12
x^{2 } 4x= 0
x(x4)=0
x=0 or x=4
Solution set : x = 0 or x=4
Question139
Solution:
z =  + i
x = , y = 1
r = === 2
Î¸ =
z = 2, arg z = + 2Ï€ k, where k is an integer.
Question140
Solution:
Put
y + =
6y^{2} â€“13y +6 = 0
i.e
Squaring,
x=
Solution set : x =
Question141
Solution:
z =
z = = Ã—
1+i = (cos45Â° +isin45Â° )
âˆ´ (1+i)^{20} = [(cos45Â° +isin45Â° )] ^{20}
= (cos45Â° + isin45Â° )^{20 }
= 2^{3}(cos)^{20 }
=2^{3}(cos20Ã— ) [Using DeMoivre's formula]
=2^{3}(cos5Ï€ +isin5Ï€)
â‡’ modulus = 8 and argument is 2Ï€ k , where k is an integer.
Question142
Solution:
(By C and D)
Squaring both sides,
1+x^{2}=44x^{2 }
5x^{2} = 3
x = Â±
Solution set : x =Â±
Question143
Solution:
z_{1} = [2(cos 0^{o} + isin 0^{o})]
z_{2} =[4(cos90^{o }+ isin90^{o})]
z_{1}z_{2} =[2(cos 0^{o} + isin 0^{o})][4(cos90^{o }+ isin90^{o})]
= 8 [cos (0^{o} + 90^{o}) + isin (0^{o} + 90^{o})]
= 8 [cos 90^{o} + isin 90^{o}]
= 8 [cos + isin ]
Question144
Solution:
(By C and D)
x = Â± a or x=0.
x cannot be 0 â‡’ x = Â± a
Solution set :x= Â± a
Question145
Solution:
z_{1} = 2(cos 220^{o} + isin 220^{o})
z_{2} = 4(cos110^{o} + isin110^{o})
z_{1}z_{2} =[2(cos 220^{o} + isin 220^{o})][4(cos110^{o }+ isin110^{o})]
= 8 [cos (220^{o} + 110^{o}) + isin (220^{o} + 110^{o})]
= 8 [cos 330^{o} + isin 330^{o}]
= 8 [cos + isin ]
Question146
Solution:
Squaring,
(x+5) +(x+21) +2= 6x+40
= 2x + 7
Squaring,
(x+5)(x+21) = (2x+7)^{2 }
x^{2 }+26x+105 = 4x^{2}+28x+49
3x^{2}+2x56 =0
x=14/3 does not satisfy the equation.
âˆ´ x= 4
Solution set : x= 4
Question147
Solution:
x^{2} +
Using the formula a^{2} + b^{2} = (a  b)^{2} + 2ab
(x)^{2}+2x() =3
=3
Put = y
y^{2} +2y â€“3 =0
(y+3)(y1)=0
y= 3 or y =1;
i.e= 3 or 1;
i.e x^{2}+3x+3 = 0 or x^{2 } x1 = 0 ;
or
Solution set : x = ,
Question148
Solution:
z_{1} = [3(cos225^{o} + isin225^{o})]
z_{2} = [6(cos 45^{o} + isin 45^{o})]
z_{1}z_{2} =[3(cos 225^{o} + isin 225^{o})][6(cos45^{o }+ isin45^{o})]
= 18 [cos (225^{o} + 45^{o}) + isin (225^{o} + 45^{o})]
= 18 [cos 270^{o} + isin 270^{o}]
= 18 [cos + isin ]
Question149
Solution:
(x6)(x7) = (x2)(x3)
x^{2}13x+42 = x^{2}5x+6
8x = 36 â‡’ x=36/8 = 4.5
Solution set : x= 4.5
Question150
Solution:
=
= 3(cos + isin)
Question151
Solution:
4x^{4}16x^{3}+7x^{2}+16x+4 = 0
Dividing by x^{2},
4x^{2} â€“ 16x +7 + = 0
4() â€“16(x)+7 =0
Put x= t
() = t^{2}+2
4(t^{2}+2) â€“16t +7 = 0
4t^{2} â€“16t + 15 = 0
âˆ´ x=
â‡’ 2x^{2}+5x2=0 or 2x^{2}+3x2=0
or =
Solution set : x= , .
Question152
Solution:
=
=
=
=
Question153
Solution:
Let y =
y =
Squaring,
y^{2} = 6+y
y^{2} y â€“6 = 0
(y3)(y+2) = 0
y = 3 or y =2
y is positive â‡’ y = 3
Solution set : = 3
Question154
Solution:
Let x + iy =
Squaring both sides we have:
(x + iy)^{2 }= 15 â€“ 8i
x^{2}  y^{2} + 2ixy = 15 â€“ 8i
Comparing both sides
x^{2}  y^{2} = 15â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)
and 2xy = 8
(x^{2} + y^{2})^{2} = (x^{2}  y^{2})^{2} + (2xy)^{2 }
= 225 + (8)^{2 }= 225 + 64 = 289
x^{2} + y^{2} = 17 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)
Solving (i) and (ii) we get
2x^{2} = 2
x^{2} = 1
x = Â± 1
2y^{2} = 32
y^{2} = 16
y = Â± 4
Since the product xy is negative, we have
x = 1, y = 4 or x = 1, y = 4
Thus, the roots of 15 â€“ 8i are â€“1 + i4 or 1 â€“ i4.
Question155
Solution:
Let y = 2+ = 2+
y ^{2 }= 2y +1
y ^{2 } 2y â€“1 = 0
y is positive â‡’ y = 1+
Solution set : 1+
Question156
Solution:
Let x + iy =
Squaring both sides we have:
(x + iy)^{2 }= 8 â€“ 6i
x^{2}  y^{2} + 2ixy = 8 â€“ 6i
Comparing both sides
x^{2}  y^{2} = 8 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)
and 2xy = 6
(x^{2} + y^{2})^{2} = (x^{2}  y^{2})^{2} + (2xy)^{2 }
= (8)^{2} + (6)^{2}= 64 + 36= 100^{ }
x^{2} + y^{2} = 10 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)
Solving (i) and (ii) we get
2x^{2} = 2
x^{2} = 1
x = Â± 1
2y^{2} = 18
y^{2} = 9
y = Â± 3
Since the product xy is negative, we have
x = 1, y = 3 or x = 1, y = 3
Thus, the roots of 8 â€“ 6i are â€“1 + i3 or 1  i3.
Question157
Solution:
Let y = 2 = 2
y ^{2 } 2y +1 = 0
(y1)^{2} =0 â‡’ y = 1,1
âˆ´ 2= 1
Solution set : 1
Question158
Solution:
Let x + iy =
Squaring both sides we have:
(x + iy)^{2 }= 1 â€“ i
x^{2}  y^{2} + 2ixy = 1 â€“ i
Comparing both sides
x^{2}  y^{2} = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)
and 2xy = 1
(x^{2} + y^{2})^{ 2} = (x^{2}  y^{2})^{ 2} + (2xy)^{2 }
= 1 + (1)^{2}= 1 + 1= 2^{ }
x^{2} + y^{2} = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)
Solving (i) and (ii) we get
2x^{2} = 1+
x^{2} = (1+)/2
x = Â±
2y^{2} = 1+
y^{2} = (1+)/2
y = Â±
Since the product xy is negative, we have
x = , y = or x = , y = 
Thus, the roots of 1 â€“ i are + i or  i.
Question159
Solution:
Let y =
y =
y^{2 }= 8y
y^{2}+y8=0
y is positive â‡’ y =
âˆ´ =
Solution set :
Question160
Solution:
Let x + iy =
Squaring both sides we have:
(x + iy)^{2 }= i
x^{2}  y^{2} + 2ixy = i
Comparing both sides
x^{2}  y^{2} = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)
and 2xy = 1
(x^{2} + y^{2})^{ 2} = (x^{2}  y^{2})^{ 2} + (2xy)^{2}
= 0 + (1)^{2}= 0 + 1= 1
^{ }
x^{2} + y^{2} = = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)
Solving (i) and (ii) we get
2x^{2} = 1
x^{2} = 1/2
x = Â±
2y^{2} = 1
y^{2} = 1/2
y = Â±
Since the product xy is negative, we have
x = , y = or x = , y =
Thus, the roots of are + i and  i.
Question161
Solution:
Let the number be x,
x  = 12
x â€“ 12 =
x^{2 }+144 â€“ 24x = x
x^{2 }+144 â€“ 25x = 0
(x16)(x9) = 0
x = 16 or x = 9
x â‰ 9
So the number is 16 .
Question162
Solution:
Let x + iy =
Squaring both sides we have:
(x + iy)^{2 }= i
x^{2}  y^{2} + 2ixy = i
Comparing both sides
x^{2}  y^{2} = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)
and 2xy = 1
(x^{2} + y^{2})^{ 2} = (x^{2}  y^{2})^{ 2} + (2xy)^{2
}= 0 + (1)^{2}= 0 + 1= 1^{ }
x^{2} + y^{2} = = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)
Solving (i) and (ii) we get
2x^{2} = 1
x^{2} = 1/2
x = Â±
2y^{2} = 1
y^{2} = 1/2
y = Â±
Since the product xy is negative, we have
x = , y =  or x = , y =
Thus, the roots of are + i and  i.
Question163
Solution:
x^{3}+y^{3} = 4914 (i)
x+y = 18 (ii)
From (ii), x+y = 18 â‡’ (x+y)^{3} = 18^{3}
x^{3}+y^{3} +3xy(x+y) = 5832
4914 +3xy(18) = 5832 [substituting from (i) and (ii)]
xy = 17
(xy)^{2} = (x+y)^{2}4xy = 18^{2}4(17)
= 324 â€“ 68
= 256
x â€“ y = Â± 16(iii)
From (ii) and (iii),
âˆ´ 2x = 34 or 2 â‡’ x = 17 or x = 1
âˆ´ y = 1 or 17.
Solution set: x = 17,1 and y = 1,17
Question164
Solution:
Let x + iy =
Squaring both sides we have:
(x + iy)^{2 }= 1 + i
x^{2}  y^{2} + 2ixy = 1 + i
Comparing both sides
x^{2}  y^{2} = 1 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(i)
and 2xy = 1
(x^{2} + y^{2})^{ 2} = (x^{2}  y^{2})^{ 2} + (2xy)^{2
}= 1 + (1)^{2}= 1+1= 2
x^{2} + y^{2} = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)
Solving (i) and (ii) we get
2x^{2} = 1 +
x^{2} = (1 +)/2
x = Â±
2y^{2} = 1+
y^{2} = (1 +)/2
y = Â±
Since the product xy is negative, we have
x = , y =  or x = , y =
Thus, the roots of are  i and +i .
Question165
Solution:
x^{4}+y^{4} = 82 (i)
x+y = 4(ii)
From (ii), (x+y)^{2} = 16
x^{2}+y^{2}+2xy =16
x^{2}+y^{2}= 162xy(iii)
Squaring both sides,
(x^{2}+y^{2})^{2}= (162xy)^{2 }
x^{4}+y^{4}+2x^{2}y^{2}= 256+4x^{2}y^{2}64xy
82+2x^{2}y^{2}= 256+4x^{2}y^{2}64xy
2 x^{2}y^{2}64xy+174=0
x^{2}y^{2}32xy+87=0
29,3
From (iii), (xy)^{2}=164xy= 164(29) or 164(3)
=100 or 4
(xy)= Â± 10i or xy = Â± 2
When xy = Â± 2 (iv)
From (ii) and (iv), x = 3,1 and y = 1,3
When xy = Â± 10i (v)
From (ii) and (v), x = 25i; y= 25i
Solution set: 3,1; 1,3; 25i; 25i
Question166
Solution:
Let z_{1} = r_{1}(cos Î¸ _{1} + isinÎ¸ _{1}) and z_{2 }= r_{2}(cos Î¸ _{2} + isinÎ¸ _{2}).
z_{1} z_{2} = r_{1} r_{2} (cos Î¸ _{1} + isinÎ¸ _{1})(cos Î¸ _{2} + isinÎ¸ _{2})
= r_{1} r_{2} [cos Î¸ _{1}(cos Î¸ _{2} + isinÎ¸ _{2}) + isinÎ¸ _{1}(cos Î¸ _{2} + isinÎ¸ _{2})]
= r_{1} r_{2} [cos Î¸ _{1}cos Î¸ _{2} + i cos Î¸ _{1}sinÎ¸ _{2} + isinÎ¸ _{1}cos Î¸ _{2} + i^{2} sinÎ¸ _{1}sinÎ¸ _{2}]
= r_{1} r_{2} [(cos Î¸ _{1}cos Î¸ _{2}  sinÎ¸ _{1}sinÎ¸ _{2 }) + i (cos Î¸ _{1}sinÎ¸ _{2} + sinÎ¸ _{1}cos Î¸ _{2 })]
= r_{1} r_{2} [cos (Î¸ _{1} + Î¸ _{2 }) + i sin(Î¸ _{1} + Î¸ _{2 })]
Re(z_{1} z_{2}) = r_{1} r_{2} cos (Î¸ _{1} + Î¸ _{2 }) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)
Re z_{1} Rez_{2} â€“ Imz_{1} Imz_{2 }
= r_{1} r_{2} cos Î¸ _{1} cos Î¸ _{2}^{ } r_{1} r_{2} sin Î¸ _{1} sin Î¸ _{2}
= r_{1} r_{2} cos(Î¸ _{1} + Î¸ _{2}^{ }) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)
From (i) and (ii) we have
Re(z_{1} z_{2}) = Re z_{1} Rez_{2} â€“ Imz_{1} Imz_{2 }
Question167
Solution:
x^{4}+y^{4} = 257 (i)
x+y = 5(ii)
From (ii), (x+y)^{2} = 25
x^{2}+y^{2}+2xy =25
Hence, (x^{2}+y^{2})^{2}= (252xy)^{2 }
x^{4}+y^{4}+2x^{2}y^{2}= 625+4x^{2}y^{2}100xy
257+2x^{2}y^{2}= 625+4x^{2}y^{2}100xy [substituting (i)]
2 x^{2}y^{2}100xy+368=0
x^{2}y^{2}50xy+184=0
4,46(iii)
From (ii), (xy)^{2}=254xy
= 254(4) or 254(46)[substituting (iii)]
= 9 or 159
(xy)^{ }= Â± 3 or Â± i
When xy = Â± 3 (iv)
From (ii) and (iv), x = 4,1 and y = 1,4
When xy = Â± i (v)
From (ii) and (v), x =
Solution set : 4,1;1,4;
Question168
(i) Ï‰ ^{18}
(ii) Ï‰ ^{21}
(iii)^{ }Ï‰ ^{30}
(iv) Ï‰ ^{105 }
Solution:
(i) Ï‰ ^{18} = (Ï‰ ^{3})^{ 6}= (1)^{ 6} = 1 (Since Ï‰ ^{3}=1)
(ii) Ï‰ ^{21} = (Ï‰ ^{3})^{ 7} = (1)^{ 7 }= 1 (Since Ï‰ ^{3}=1)
(iii)^{ }Ï‰ ^{30} = 1/(Ï‰ ^{3})^{10} = 1/(1)^{10 }= 1 (Since Ï‰ ^{3}=1)
(iv) Ï‰ ^{105} = 1/(Ï‰ ^{3})^{35} = 1/(1)^{35 }= 1 (Since Ï‰ ^{3}=1)
Question169
Solution:
Let the piece of cloth is x m. The cost of cloth is Rs.35.
âˆ´ The cost per metre is Rs.35/x
If the piece were 4 m longer and each metre costs Re.1 less â‡’ the piece of cloth is (x+4) m
and the cost per metre is Rs.35/x â€“ 1.
âˆ´(x+4)(35/x â€“1) = 35
35x+140/x4 = 35
x+140/x4 = 0
x^{2} +4x â€“140 =0
x^{2} +14x10x â€“140 =0
(x+14)(x10)=0
x = 14 or x = 10
x â‰ 14,
âˆ´ x = 10.
Hence the piece is 10m.
Question170
Solution:
L.H.S =(2  Ï‰ )(2  Ï‰ ^{2})(2  Ï‰ ^{10})(2  Ï‰ ^{11})
= (4  2Ï‰ ^{2}  2Ï‰ + Ï‰ ^{3})(4  2Ï‰ ^{11}  2Ï‰ ^{10} + Ï‰ ^{21})
= [4 â€“ 2(Ï‰ ^{2} + Ï‰ ) + 1][4  2Ï‰ ^{9}(Ï‰ ^{2} + Ï‰ ) + (Ï‰ ^{3})^{7}]
= [4 â€“ 2(1) + 1][4  2Ï‰ ^{9}(1) + 1^{7}]
= [4 + 2 + 1][4 + 2(Ï‰ ^{3})^{3} + 1]
= 7[5 + 2(1)^{3} ]
= 7 Ã— 7
= 49
= R.H.S
Question171
Solution:
Let the initial production by P.
Suppose the production increases by r percent every year. Then the P_{1} after the first year is
P_{1} = P(1+r/100)
The production at the second year is P_{1}(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.
Question172
Solution:
L.H.S =
=
=
= (Since Ï‰ ^{3}=1)
=
= Ï‰ ^{2}
Question173
Solution:
ax^{2} +bx +c =0
Let the roots be Î± and ()Î± .
Î± + ()Î± = b/a â‡’ Î± = (i)
Î± Ã— ()Î± = c/a â‡’ Î± ^{2} = (ii)
From (i) and (ii),
=
â‡’ b^{2}pq = ac(p+q)^{2 }
Question174
Solution:
L.H.S = (1  Ï‰ ^{2} + Ï‰ ^{4})(1 + Ï‰ ^{2}  Ï‰ ^{4})
= [1  Ï‰ ^{2} + Ï‰ Ã— Ï‰ ^{3}][1 + Ï‰ ^{2}  Ï‰ Ã— Ï‰ ^{3}]
= ( Ï‰ ^{2}  Ï‰ ^{2})( Ï‰  Ï‰ ) (Since Ï‰ ^{3}=1 and 1 + Ï‰ + Ï‰ ^{2 }= 0)
= ( 2Ï‰ ^{2})( 2Ï‰ )
= 4Ï‰ ^{3}
= 4
= R.H.Sâˆ´ (1  Ï‰ ^{2} + Ï‰ ^{4})(1 + Ï‰ ^{2}  Ï‰ ^{4})= 4
Question175
Solution:
Let Î± , Î² are the roots of the quadratic equation x^{2}  px +q =0
Î± + Î² = p(i)
Î± Ã— Î² = q (ii)
Î± Î² +Î± +Î² + Î± Î²  Î±  Î² = 2Î± Î² =2q
(Î± Î² + Î± + Î² )(Î± Î²  Î±  Î² ) = Î± ^{2}Î² ^{2} â€“ (Î± +Î² )^{2 }
= q^{2 }â€“ p^{2 }
Hence the equation is x^{2}(sum of the roots)x+(product of the roots) = 0
x^{2}2qx+( q^{2 }â€“ p^{2}) = 0
Question176
Solution:
L.H.S = (1  Ï‰ + Ï‰ ^{2})^{2} + (1 + Ï‰  Ï‰ ^{2})^{ 2}
= ( Ï‰  Ï‰ )^{2} + ( Ï‰ ^{2} Ï‰ ^{2})^{ 2}
= ( 2Ï‰ )^{2} + ( 2Ï‰ ^{2})^{ 2 }(Since Ï‰ ^{3}=1 and 1 + Ï‰ + Ï‰ ^{2 }= 0)
= 4 Ï‰ ^{2} + 4 Ï‰ ^{4}
= 4 Ï‰ ^{2} + 4 Ï‰
= 4
= R.H.S
âˆ´ (1  Ï‰ + Ï‰ ^{2})^{2} + (1 + Ï‰  Ï‰ ^{2})^{ 2} = 4
Question177
Solution:
L.H.S = (1  Ï‰ )(1  Ï‰ ^{2})(1  Ï‰ ^{4})(1  Ï‰ ^{8})
= (1  Ï‰ ^{2}  Ï‰ + Ï‰ ^{3})(1  Ï‰ ^{8}  Ï‰ ^{4} + Ï‰ ^{12})
= (1  Ï‰ ^{2}  Ï‰ + 1)[1  Ï‰ ^{8}  Ï‰ ^{4} + (Ï‰ ^{3})^{ 4}] (Since Ï‰ ^{3}=1 and 1 + Ï‰ + Ï‰ ^{2 }= 0)
= (2  Ï‰ ^{2}  Ï‰ )(2  Ï‰ ^{8}  Ï‰ ^{4})
= (2  Ï‰ ^{2}  Ï‰ )[2  Ï‰ ^{2}(Ï‰ ^{3})^{ 2}  Ï‰ (Ï‰ ^{3})]
= (2  Ï‰ ^{2}  Ï‰ )(2  Ï‰ ^{2}  Ï‰ )
= (2 + 1)(2 + 1)
= 3 Ã— 3 = 9 =R.H.S
âˆ´ (1  Ï‰ )(1  Ï‰ ^{2})(1  Ï‰ ^{4})(1  Ï‰ ^{8}) = 9
Question178
Solution:
2x^{2 }â€“ x  =0
D = b^{2}4ac = (1)^{2}+ 4Ã— 2Ã— = 1+12 =13
is irrational.
Hence the roots of the quadratic equations 2x^{2 }â€“ x  =0 are irrational.
Question179
Solution:
(xa)(xb) + (xb)(xc) + (xc)(xa) = 0
3x^{2}  2(a+b+c)x + (ab+bc+ca) = 0
If the a, b, c are equal, D = b^{2}4ac â‰¥ 0
i.e 4(a+b+c)^{2}12(ab+bc+ca) â‰¥ 0
i.e (a+b+c)^{2}3(ab+bc+ca) â‰¥ 0
i.e a^{2}+b^{2}+c^{2}â€“abâ€“bcca â‰¥ 0
i.e 2[(ab)^{2}+(bc)^{2}+(ca)^{2}] â‰¥ 0
a^{2} + b^{2} + c^{2} â€“abbcca = 0 if and only if 2[(ab)^{2}+(bc)^{2}+(ca)^{2}] = 0
i.e iff a = b = c [as (ab)^{2},(bc)^{2},(ca)^{2 }are â‰¥ 0]
Question180
Solution:
Put n =2
L.H.S = 1 + Ï‰ ^{2} + Ï‰ ^{4} = 1 + Ï‰ ^{2} + Ï‰ Ã— Ï‰ ^{3} = 1 + Ï‰ ^{2} + Ï‰ = 0 = R.H.S
Put n = 4
L.H.S = 1 + Ï‰ ^{4} + Ï‰ ^{8} = 1 + Ï‰ Ã— Ï‰ ^{3} + Ï‰ ^{2Ã— }(Ï‰ ^{3})^{ 2 }= 1 + Ï‰ + Ï‰ ^{2} = 0 = R.H.S
Question181
Solution:
(xa)(xb)+(xb)(xc)+(xc)(xa) = 0
3x^{2}  2(a+b+c)x +(ab+bc+ca) = 0
If the roots are equal, D = b^{2}4ac =0
iff 4(a+b+c)^{2}12(ab+bc+ca) =0
iff 2[(ab)^{2}+(bc)^{2}+(ca)^{2}] = 0
iff a =b = c [as (ab)^{2},(bc)^{2},(ca)^{2 }are â‰¥ 0]
Question182
Solution:
L.H.S = 1 + Ï‰ ^{n} + Ï‰ ^{2n} = 1 + 1 + (1)^{ 2 }= 1 + 1 + 1 = 3 = R.H.S
âˆ´ 1 + Ï‰ ^{n} + Ï‰ ^{2n} = 3
Question183
Solution:
z  i = 1
x + iy â€“ i = 1
x + i(y â€“ 1) = 1
x^{2} + (y â€“1 )^{2} = 1
Circle of radius 1, Centre (0, 1)
Question184
Prove that the product of the roots is
Solution:
(x^{2})+x(a+b2c)+abacbc=0
The sum of the roots is zero â‡’ a+b2c = 0(i)
Product of the roots = abacbc = abc(a+b) = ab [From (i)]
=