Exercise 4(A) Page: 57 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 4- Expansions Exercise 4(A) Page: 57 1. Find the square of:

Concise Selina Solutions for Class 9 Maths Chapter 4-

Expansions

Exercise 4(A) Page: 57 1. Find the square of:

(i) 2a+b

(ii) 3a+7b

(iii) 3a-4b

(iv) 𝟑𝐚

𝟐𝐛−

𝟐𝐛

𝟑𝐚

Solution:

(i)

(ii)

(iii)

(iv)

2. Use identities to evaluate:

(i) 1012

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Expansions

(ii) 5022

(iii) 972

(iv) 9982

Solution: (i)

(ii)

(iii)

(iv)




Expansions

3. Evaluate: (i)

(ii)

Solution:

(i)

(ii)




Expansions

4. Evaluate: (i)

(ii)

Solution:

(i)Consider the given expression:




Expansions

(ii)Consider the given expression:

5. If a+b=7 and ab = 10; find a-b.

Solution:

6. If a-b=7 and ab = 18; find a+b.




Expansions

Solution:

7. If 𝒙 + 𝒚 =𝟕

𝟐 and 𝒙𝒚 =

𝟓

𝟐; find:

(i) x-y

(ii) x2-y2

Solution: (i)




Expansions

(ii)

8. If a-b=0.9 and ab = 0.36; find

(i) a+b.

(ii) a2-b2

Solution: (i)

(ii)




Expansions

9. If a-b=4 and a+b=6; find :

(i) a2+b2

(ii) ab

Solution: (i)




Expansions

(ii)

10. If 𝒂 +𝟏

𝒂= 𝟔 and 𝒂 ≠ 0; find:

(i) 𝒂 −𝟏

𝒂

(ii) 𝒂𝟐 −𝟏

𝒂𝟐

Solution: (i)




Expansions

(ii)

11. If 𝒂 −𝟏

𝒂= 𝟖 and 𝒂 ≠ 0; find:

(i) 𝒂 +𝟏

𝒂

(ii) 𝒂𝟐 −𝟏

𝒂𝟐

Solution:

(i)




Expansions

(ii)




Expansions

12. If a2-3a+1=0 and a≠0; find:

(i) 𝒂 +𝟏

𝒂

(ii) 𝒂𝟐 +𝟏

𝒂𝟐

Solution: (i)

(ii)

13. If a2-5a-1=0 and a≠0; find:

(i) 𝒂 −𝟏

𝒂

(ii) 𝒂 +𝟏

𝒂




Expansions

(iii) 𝒂𝟐 −𝟏

𝒂𝟐

Solution: (i)

(ii)

(iii)




Expansions

14. If 3x+4y=16 and xy=4; find the value of 9x2+16y2

Solution:




Expansions

15. The number x is 2 more than the number y. If the sum of the squares of x and

y is 34; find the product of x and y.

Solution: Given x is 2 more than y, so x = y + 2

Sum of squares of x and y is 34, so x2 + y2 = 34.

Replace x = y + 2 in the above equation and solve for y.

We get (y + 2)2 + y2 = 34

2y2 + 4y - 30 = 0

y2 + 2y - 15 = 0

(y + 5)(y - 3) = 0

So y = -5 or 3

For y = -5, x =-3

For y = 3, x = 5

Product of x and y is 15 in both cases.

16. The difference between two positive numbers is 5 and the sum of their

squares is 73. Find the product of these numbers.

Solution: Let the two positive numbers be a and b.

Given difference between them is 5 and sum of squares is 73.

So, a - b = 5, a2 + b2 = 73

Squaring on both sides gives

(a - b)2 = 52

a2 + b2 - 2ab = 25

but a2 + b2 = 73

so 2ab = 73 - 25 = 48

ab = 24

So, the product of numbers is 24.




Expansions

Exercise 4(B) Page: 60

1. Find the cube of:

(i) 3a-2b

(ii) 5a+3b

(iii) 𝟐𝒂 +𝟏

𝟐𝒂 ; (𝒂 ≠ 𝟎)

(iv) 𝟑𝒂 −𝟏

𝒂 ; (𝒂 ≠ 𝟎)

Solution: (i)

(ii)

(iii)

(iv)




Expansions

2. If 𝒂𝟐 +𝟏

𝒂𝟐= 𝟒𝟕 and 𝒂 ≠ 0; find:

(i) 𝒂 +𝟏

𝒂

(ii) 𝒂𝟑 +𝟏

𝒂𝟑

Solution: (i)

(ii)




Expansions

3. If 𝒂𝟐 +𝟏

𝒂𝟐= 𝟏𝟖 and 𝒂 ≠ 0; find:

(i) 𝒂 −𝟏

𝒂

(ii) 𝒂𝟑 −𝟏

𝒂𝟑

Solution:

(i)

(ii)

4. If 𝒂 +𝟏

𝒂= 𝒑 and 𝒂 ≠ 0; then show that:

𝒂𝟑 +𝟏

𝒂𝟑= 𝒑(𝒑𝟐 − 𝟑)

Solution:




Expansions

5. If a + 2b = 5; then show that:

a3+8b3+30ab=125

Solution:

6. If (𝒂 +𝟏

𝒂)

𝟐= 𝟑 𝒂𝒏𝒅 𝒂 ≠ 𝟎; then show that:

𝒂𝟑 +𝟏

𝒂𝟑= 𝟎

Solution:




Expansions

7. If a+2b+c=0; then show that:

a3+8b3+c3=6abc.

Solution:

8. Use property to evaluate:

(i) 133 + (-8)3 + (-5)3

(ii) 73 + 33 + (-10)3

(iii) 93 - 53 - 43

(iv) 383 + (-26)3 + (-12)3




Expansions

Solution: Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc

(i) a = 13, b = -8 and c = -5

133 + (-8)3 + (-5)3 = 3(13)(-8)(-5) = 1560

(ii) a = 7, b = 3, c = -10

73 + 33 + (-10)3 = 3(7)(3)(-10) = -630

(iii)a = 9, b = -5, c = -4

93 - 53 - 43 = 93 + (-5)3 + (-4)3 = 3(9)(-5)(-4) = 540

(iv) a = 38, b = -26, c = -12

383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568

9. If 𝒂 ≠ 0 and 𝒂 −𝟏

𝒂= 𝟑; find:

(i) 𝒂𝟐 +𝟏

𝒂𝟐

(ii) 𝒂𝟑 −𝟏

𝒂𝟑

Solution: (i)

(ii)

10. If 𝒂 ≠ 0 and 𝒂 −𝟏

𝒂= 𝟒; find:




Expansions

(i) 𝒂𝟐 +𝟏

𝒂𝟐

(ii) 𝒂𝟒 +𝟏

𝒂𝟒

(iii) 𝒂𝟑 −𝟏

𝒂𝟑

Solution: (i)

(ii)

(iii)

11. If 𝒙 ≠ 0 and 𝒙 −𝟏

𝒙= 𝟐; then show that:

Solution:




Expansions

Thus from equations (1), (2) and (3), we have

12. If 2x-3y=10 and xy=16; find the value of 8x3-27y3.

Solution: Given that 2x - 3y = 10, xy = 16

:. (2x - 3y)3 = (10)3

Þ 8x3 - 27y3 - 3 (2x) (3y) (2x - 3y) = 1000 Þ 8x3 - 27 y3 -18xy (2x - 3y) = 1000

Þ 8x3 - 27 y3 - 18 (16) (10) = 1000

Þ 8x3 - 27 y3 - 2880 = 1000

Þ 8x3 - 27 y3 = 1000 + 2880

Þ8x3 - 27 y3 =3880




Expansions

13. Expand:

(i) (3x + 5y + 2z) (3x - 5y + 2z)

(ii) (3x - 5y - 2z) (3x - 5y + 2z)

Solution: (i)

(3x + 5y + 2z) (3x - 5y + 2z)

= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}

= (3x + 2z)2 - (5y)2

{since (a + b) (a - b) = a2 - b2}

= 9x2 + 4z2 + 2 × 3x × 2z - 25y2

= 9x2 + 4z2 + 12xz - 25y2

= 9x2 + 4z2 - 25y2 + 12xz

(ii)

(3x - 5y - 2z) (3x - 5y + 2z)

= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}

= (3x - 5y)2 - (2z)2{since(a + b) (a - b) = a2 - b2}

= 9x2 + 25y2 - 2 × 3x × 5y - 4z2

= 9x2 + 25y2- 30xy - 4z2

= 9x2 +25y2 - 4z2 - 30xy

14. The sum of two numbers is 9 and their product is 20. Find the sum of their:

(i) Squares

(ii) Cubes

Solution: Given sum of two numbers is 9 and their product is 20.

Let the numbers be a and b.

a + b = 9

ab = 20

(i) Squaring on both sides gives

(a+b)2 = 92

a2 + b2 + 2ab = 81

a2 + b2 + 40 = 81

So sum of squares is 81 - 40 = 41

(ii) Cubing on both sides gives

(a + b)3 = 93

a3 + b3 + 3ab(a + b) = 729

a3 + b3 + 60(9) = 729

a3 + b3 = 729 - 540 = 189




Expansions

So the sum of cubes is 189.

15. Two positive numbers x and y are such that x>y. if the difference of these

numbers is 5 and their product is 24, find:

(i) Sum of these numbers.

(ii) Difference of their cubes.

(iii) Sum of their cubes.

Solution: (i) Given x - y = 5 and xy = 24 (x>y)

(x + y)2 = (x - y)2 + 4xy = 25 + 96 = 121

So, x + y = 11; sum of these numbers is 11.

(ii) Difference of their Cubes

(x - y)3 = 53

x3 - y3 - 3xy(x - y) = 125

x3 - y3 - 72(5) = 125

x3 - y3= 125 + 360 = 485

So, difference of their cubes is 485.

(iii) Difference of their Cubes

(x + y)3 = 113

x3 + y3 + 3xy(x + y) = 1331

x3 + y3 = 1331 - 72(11) = 1331 - 792 = 539

So, sum of their cubes is 539.

16. If 4x2+y2=a and xy=b, find the value of 2x+y

Solution: xy = ab ….(i)

4x2 + y2 = a ….(ii)

Now, (2x + y)2 = (2x)2 + 4xy + y2

= 4x2 + y2 + 4xy

= a + 4b ….[From (i) and (ii)]




Expansions

Exercise 4(C) Page: 62 1. Expand:

(i) (x+8)(x+10)

(ii) (x+8)(x-10)

(iii) (x-8)(x+10)

(iv) (x-8)(x-10)

Solution:

2. Expand:

(i) (𝟐𝒙 −𝟏

𝒙) (𝟑𝒙 +

𝟐

𝒙)

(ii) (𝟑𝒂 +𝟐

𝒃) (𝟐𝒂 −

𝟑

𝒃)

Solution:




Expansions

3. Expand:

Solution:

4. If a+b+c=12 and a2+b2+c2=50; find ab+bc+ca.

Solution:




Expansions

5. If a2+b2+c2=35 and ab+bc+ca=23; find a+b+c.

Solution:

6. If a+b+c=p and ab+bc+ca=q; find a2+b2+c2.

Solution:




Expansions

7. If a2+b2+c2=50 and ab+bc+ca=47; find a+b+c.

Solution:

8. If x+y-z=4 and x2+y2+z2=30, then find the value of xy-yz-zx.

Solution:




Expansions

Exercise 4(D) Page: 64

1. If x+2y+3z=0 and

x3+4y3+9z3=18xyz; evaluate;

Solution:

Given that x3 + 4y3 + 9z3 = 18xyzand x + 2y + 3z = 0

x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y

Now

2. If 𝒂 +𝟏

𝒂= 𝒎 and 𝒂 ≠ 𝟎; find in terms of ‘m’; the value of

(i) 𝒂 −𝟏

𝒂

(ii) 𝒂𝟐 −𝟏

𝒂𝟐

Solution: (i)




Expansions

(ii)

3. In the expansion of (2x2-8)(x-4)2; find the value of:

(i) Coefficient of x3

(ii) Coefficient of x2

(iii) Constant term.

Solution:

4. If x>0 and 𝒙𝟐 +𝟏

𝟗𝒙𝟐=

𝟐𝟓

𝟑𝟔, find: 𝒙𝟑 +

𝟏

𝟐𝟕𝒙𝟑.

Solution: Given that




Expansions

5. If 2(x2+1)=5x, find:




Expansions

(i) 𝒙 −𝟏

𝒙

(ii) 𝒙𝟑 −𝟏

𝒙𝟑

Solution: (i)

2(x2 + 1} = 5x

Dividing by x, we have




Expansions

(ii)

6. If a2 + b2 = 34 and ab= 12; find:

(i) 3(a + b)2 + 5(a - b)2

(ii) 7(a - b)2 - 2(a + b)2

Solution: a2 + b2 = 34, ab= 12

(a + b)2 = a2 + b2 + 2ab

= 34 + 2 x 12 = 34 + 24 = 58

(a - b)2 = a2 + b2 - 2ab

= 34 - 2 x 12 = 34- 24 = 10

(i) 3(a + b)2 + 5(a - b)2

= 3 x 58 + 5 x 10 = 174 + 50

= 224

(ii) 7(a - b)2 - 2(a + b)2

= 7 x 10 - 2 x 58 = 70 - 116 = -46

7. If 𝟑𝒙 −𝟒

𝒙 =4 and 𝒙 ≠ 𝟎; find:

Solution:

Given 3x -

We need to find




Expansions

8. If and x ≠0; find the value of:

Solution:

Given that

We need to find the value of

Consider the given equation:




Expansions

9. If and x ≠5, find: 𝒙𝟐 −𝟏

𝒙𝟐

Solution:

By cross multiplication,

=> x (x - 5) = 1 => x2 - 5x = 1 => x2 - 1 = 5x

Dividing both sides by x,




Expansions

10. If and x ≠5, find: 𝒙𝟑 −𝟏

𝒙𝟑

Solution:

By cross multiplication,

=> x (5 - x) = 1 => x2 - 5x =-1 => x2 + 1 = 5x

Dividing both sides by x,

11. If 3a + 5b + 4c = 0, show that:

27a3 + 125b3 + 64c3 = 180abc

Solution: Given that 3a + 5b + 4c = 0

3a + 5b = -4c

Cubing both sides,

(3a + 5b)3 = (-4c)3

=>(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3

=>27a3 + 125b3 + 45ab x (-4c) = -64c3

=>27a3 + 125b3 - 180abc = -64c3

=>27a3 + 125b3 + 64c3 = 180abc

Hence proved.




Expansions

12. The sum of two numbers is 7 and the sum of their cubes is 133, find the sum

of their squares.

Solution: Let a, b be the two numbers

.'. a + b = 7 and a3 + b3 = 133

(a + b)3 = a3 + b3 + 3ab (a + b)

=> (7)3 = 133 + 3ab (7)

=> 343 = 133 + 21ab => 21ab = 343 - 133 = 210

=> 21ab = 210 => ab= 2I

Now a2 + b2 = (a + b)2 - 2ab

=72 - 2 x 10 = 49 - 20 = 29

13. In each of the following, find the value of ‘a’:

(i) 4x2 + ax + 9 = (2x + 3)2

(ii) 4x2 + ax + 9 = (2x - 3)2

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2

Solution: (i) 4x2 + ax + 9 = (2x + 3)2

Comparing coefficients of x terms, we get

ax = 12x

so, a = 12

(ii) 4x2 + ax + 9 = (2x - 3)2


ax = -12x

so, a = -12

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2


(7a - 5)x = 30x

7a - 5 = 30

7a = 35

a = 5

14. If 𝒙𝟐+𝟏

𝒙= 𝟑

𝟏

𝟑 and x>1; find

(i) 𝒙 −𝟏

𝒙

(ii) 𝒙𝟑 −𝟏

𝒙𝟑




Expansions

Solution: Given

15. The difference between two positive numbers is 4 and the difference between

their cubes is 316. Find:

(i) their product

(ii) the sum of their squares.

Solution:

Given difference between two positive numbers is 4 and difference

between their cubes is 316.

Let the positive numbers be a and b

a - b = 4

a3 - b3 = 316

Cubing both sides,

(a - b)3 = 64

a3 - b3 - 3ab(a - b) = 64




Expansions

Given a3 - b3 = 316

So 316 - 64 = 3ab(4)

252 = 12ab

So ab = 21; product of numbers is 21

Squaring both sides, we get

(a - b)2 = 16

a2 + b2 - 2ab = 16

a2 + b2 = 16 + 42 = 58

Sum of their squares is 58.




Expansions

Exercise 4(E) Page: 66 1. Simplify:

(i) (x + 6)(x + 4)(x - 2)

(ii) (x - 6)(x - 4)(x + 2)

(iii) (x - 6)(x - 4)(x - 2)

(iv) (x + 6)(x - 4)(x - 2)

Solution: Using identity:

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

(i) (x + 6)(x + 4)(x - 2)

= x3 + (6 + 4 - 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)

= x3 + 8x2 + (24 - 8 - 12)x - 48

= x3 + 8x2 + 4x - 48

(ii) (x - 6)(x - 4)(x + 2)

= x3 + (-6 - 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2

= x3 - 8x2 + (24 - 8 - 12)x + 48

= x3 - 8x2 + 4x + 48

(iii) (x - 6)(x - 4)(x - 2)

= x3 + (-6 - 4 - 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)

= x3 - 12x2 + (24 + 8 + 12)x - 48

= x3 - 12x2 + 44x - 48

(iv) (x + 6)(x - 4)(x - 2)

= x3 + (6 - 4 - 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)

= x3 - 0x2 + (-24 + 8 - 12)x + 48

= x3 - 28x + 48

2. Simplify using following identity: (𝑎 ± 𝑏)(𝑎2 ± 𝑎𝑏 + 𝑏2) = 𝑎3 ± 𝑏3

Solution:




Expansions

3. Using suitable identity, evaluate:

(i) (104)3

(ii) (97)3

Solution: Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)

(i) (104)3 = (100 + 4)3

= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)

= 1000000 + 64 + 1200 × 104

= 1000000 + 64 + 124800

= 1124864

(ii) (97)3 = (100 - 3)3

= (100)3 - (3)3 - 3 × 100 × 3(100 - 3)

= 1000000 - 27 - 900 × 97

= 1000000 - 27 - 87300

= 912673

4. Simplify:




Expansions

Solution:

5. Evaluate:

Solution:




Expansions

6. If a - 2b + 3c = 0, state the value of a3 - 8b3 + 27c3

Solution: a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

Since a - 2b + 3c = 0, we have

a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

= 3(a)( -2b)(3c)

= -18abc

7. If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.

Solution: x + 5y = 10

⇒ (x + 5y)3 = 103

⇒ x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000

⇒ x3 + (5y)3 + 3(x)(5y)(10) = 1000




Expansions

= x3 + (5y)3 + 150xy = 1000

= x3 + (5y)3 + 150xy - 1000 = 0

8. If , find:

Solution:




Expansions

9. If ; find a3+b3.

Solution:

10. Prove that:

x2 + y2 + z2 - xy - yz – zx is always positive.

Solution: x2 + y2 + z2 - xy - yz - zx

= 2(x2 + y2 + z2 - xy - yz - zx)

= 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx

= x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx

= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)

= (x - y)2 + (z - x)2 + (y - z)2

Since square of any number is positive, the given equation is always positive.

11. Find:

(i) (a + b)(a + b) = (a + b)2

(ii) (a + b)(a + b)(a + b)

(iii) (a - b)(a - b)(a - b) by using the result of part (ii)

Solution: (i) (a + b)(a + b) = (a + b)2

= a × a + a × b + b × a + b × b




Expansions

= a2 + ab + ab + b2

= a2 + b2 + 2ab

(ii) (a + b)(a + b)(a + b)

= (a × a + a × b + b × a + b × b)(a + b)

= (a2 + ab + ab + b2)(a + b)

= (a2 + b2 + 2ab)(a + b)

= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b

= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2

= a3 + b3 + 3a2b + 3ab2

(iii) (a - b)(a - b)(a - b)

In result (ii), replacing b by -b, we get

(a - b)(a - b)(a - b)

= a3 + (-b)3 + 3a2(-b) + 3a(-b)2

= a3 - b3 - 3a2b + 3ab2



Exercise 4(A) Page: 57 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 4- Expansions Exercise 4(A) Page: 57 1. Find the square of:

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