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## Karnataka State Syllabus Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 1.

Choose the correct answer

(a) Terms having the same literal factor with the same exponents are called

(A) exponents

(B) like terms

(C) factors

(D) unlike term

Answer:

(B) like terms

(b) The coefficient of ab in 2ab is

(A) ab

(B) 2

(C) 2a

(D) 2b

Answer:

(B) 2

(c) The exponential form of a x a x a is

(A) 3a

(B) 3 + a

(C) a^{3}

(D) 3 – a

Answer:

(C) a^{3}

(d) Sum of two negative integers is

(A) negative

(B) positive

(C) zero

(D) infinite

Answer:

(A) negative

(e) What should be added to a^{2} + 2ab to make it a complete square?

(A) b^{2}

(B) 2ab

(C) ab

(D) 2a

Answer:

(A) b^{2}

(f) What is the product of (x + 2) (x – 3)?

(A) 2x – 6

(B) 3x – 2

(C) x^{2} – x – 6

(D) x^{2} – 6x

Answer:

(C) x^{2} – x – 6

(g) The value of (7.2)^{2} is (use an iden-tity to expand)

(A) 49.4

(B) 14.4

(C) 51.84

(D) 49.04

Answer:

(C) 51.84

(h) The expansion of (2x – 3y)^{2}is

(A) 2x^{2} + 3y^{2}+ 6xy

(B) 4x^{2} + 9y^{2} – 12xy

(C) 2x^{2} +3y^{2} – 6xy

(D) 4x^{2} +9y^{2} + 12xy

Answer:

(B) 4x^{2} + 9y^{2} – 12xy

(i) The product of 58 × 62 is (use an identity)

(A) 4596

(B) 2596

(C) 3596

(D) 6596

Answer:

(C) 3596

Question 2.

Take away 8x – 7y – 8p + 10q from 10x + 10y – 7p + 9q

Answer:

Question 3.

Expand

(i) (4x + 3)^{2}

Answer:

(a + b)^{2} = a^{2} + 2ab + b^{2} ‘

Here a = 4x, b = 3

(4x + 3)^{2} = (4x)^{2} + 2.4×3 .+ 3^{2}

= 16x^{2} + 24x + 9

(ii) (x + 2y)^{2}

Answer:

(a + b)^{2} = a^{2} + 2ab + b^{2}

Here a = x, b = 2y

(x + 2y)^{2} = x^{2} + 2.x.2y + (2y)^{2}

= x^{2} + 4xy + 4y^{2}

(iii)

Answer:

(a + b)^{2} = a^{2} + 2ab + b^{2}

Here a = x, b = 1/x

Question 4.

Expand

(i) (2t + 5) (2t – 5).

Answer:

(a + b) (a – b) = a^{2} – b^{2}

Here = a = 2t, b = 5 ]

(2t + 5) (2t – 5) = (2t)^{2} – 5^{2}

= 4t^{2} – 25

(ii) (xy + 8) (xy – 8)

Answer:

(a + b) (a -b) = a^{2} – b^{2}

Here a = xy, b = 8

(xy + 8) (xy – 8) = (xy)^{2} – 8^{2}

= x^{2}y^{2} – 64

(ii) (2x + 3y) (2x – 3y)

Answer:

(a + b) (a -b) = a^{2} – b^{2}

Here = a = 2x, b = 3y

(2x + 3y) (2x – 3y) = (2X)^{2} – (3y)^{2} = 4x^{2} – 9y^{2}

Question 5.

Expand

(i) (n – 1) (n + 1) (n^{2} + 1)

Answer:

(n – 1) (n + 1) (n^{2} + 1)

= (n^{2} – l^{2}) (n^{2} + 1)

= (n^{2} – 1) (n^{2} + 1)

= (n^{2})^{2} – l2

= n^{4} – 1

(ii)

Answer:

(iii) (x – 1) (x + 1) (x^{2} + 1) (x^{4} + 1)

Answer:

(x – 1) (x + 1) (x^{2} + 1) (x^{4} + 1)

= (x^{2} – l^{2}) (x^{2} + 1) (x^{4} + 1)

= (x^{2} – 1) (x^{2} + 1) (x^{4} + 1)

= [(x^{2})^{2} – l^{2})] (x^{4} + 1)

= (x^{4} – 1) (x^{4} + 1)

= (x^{4})^{2} – l^{2}

= X^{8} – 1

(iv) (2x – y) (2x + y) (4x^{2} + y^{2})

Answer:

(2x – y) (2x + y) (4x^{2} + y^{2})

= [(2x)^{2} – y^{2}) (4x^{2} + y^{2})

= (4x^{2} – y^{2}) (4x^{2} + y^{2})

= [(4x^{2})^{2} – (y^{2})^{2})]

=16x^{4} – y^{4}

Question 6.

Use appropriate formulae and compute

(i) (103)^{2}

Answer:

103^{2} = (100 + 3)^{2}

(a + b)^{2} = a^{2} + 2ab + b^{2}

Here a = 100, b = 3

(100 + 3)^{2} = 100^{2} + 2.100.3 + 3^{2}

= 10000 + 600 + 9

103^{2} = 10609

(ii) (96)^{3}

Answer:

96^{2} = (100 – 4)^{2}

(a – b)^{2} = a^{2} – 2ab + b^{2}

Here a = 100, b = 4

(100 + 3)^{2} = 100^{2} + 2.100.4 + 42

= 10000 – 800 + 16

103^{2} = 9216

(iii) 107 x 93

Answer:

107 x 93 = (100 + 7)(100 – 7)

(a + b) (a – b) = a^{2} – b^{2}

Here a = 100, b = 7

(100 + 7) (100 – 7) = 100^{2} – 7^{2}

= 10000 – 49 = 9951

(iv) 1008 x 992

Answer:

1008 x 992 = (1000 + 8)(100 – 8)

(a + b) (a – b) = a^{2} – b^{2}

Here a = 1000, b = 8

(1000 + 8) (100 – 8) = (1000)^{2} – 8^{2}

= 1000000 – 64

1008 x 992 = 999936

(v) 185^{2} – 115^{2}

Answer:

185^{2} – 115^{2}

(a – b)^{2} = (a + b) (a – b)

Here a = 185, b = 115

185^{2} – 115^{2} = (185 + 115) (185 – 115)

= (300) (70)

185^{2} – 115^{2} = 21000

Question 7.

If x + y = 7 and xy = 12 find x^{2} + y^{2}

Answer:

(x + y)^{2} = x^{2} + 2xy + y^{2}

(x + y)^{2} = x^{2} + y^{2} + 2xy

7^{2} = x^{2} + y^{2} + 2(12)

[Substituting for x +y and xy]

49 = x^{2} + y^{2} +24

x^{2} + y^{2} = 49 24

x^{2} + y^{2} = 25

Question 8.

If x + y = 12 and xy = 32 find x^{2} + y^{2}

Answer:

(x + y)^{2} = x^{2} + y^{2} + 2xy

12^{2} = x^{2} + y^{2} + 2(32)

[Substituting for x +y and xy]

144 = x^{2} + y^{2} + 64

x^{2} + y^{2} = 144 – 64

x^{2} + y^{2} = 80

Question 9.

If 4x^{2} + y^{2} = 40 and xy = 6 find 2x + y

Answer:

4x^{2} + y^{2} = 40

(2x)^{2} + y^{2} = 40

(2x + y)^{2} = (2x)^{2} + y^{2} + 2.2x.y

(2x + y)^{2} = 4x^{2} + y^{2} + 4xy

= 40 + 4 x 6

= 40 + 24

(2x + y)^{2} = 64

2x + y = ±√64

2x + y = ±√8

Question 10.

If x – y = 3 and xy = 10 find x^{2} + y^{2}

Answer:

(x – y)^{2} = x2 + y2 – 2xy

3^{2} = x^{2} + y^{2} + 2(10)

[Substituting for x – y and xy]

9 = x^{2} + y^{2} – 20

9 + 20 = x^{2} + y^{2}

∴ 29 = x^{2} + y^{2}

x^{2} + y^{2} = 29

Question 11.

If = 3 find and

Answer:

Question 12.

find and

Answer:

Question 13.

Simplify (i) (x + y)^{2} + (x – y)^{2}

Answer:

(x + y)^{2} + (x – y)^{2}

= x^{2} + y^{2} + 2xy + x^{2} + y^{2} – 2xy

= x^{2} + y^{2} + x^{2} + y^{2} = 2x^{2} + 2y^{2} ]

= 2(x^{2} + y^{2})

(ii) (x + y)^{2} x (x – y)^{2}

Answer:

(x + y)^{2} x (x – y)^{2}

= (x^{2} + y^{2} + 2xy) (x^{2} + y^{2} – 2xy)

= x^{2} (x^{2} + y^{2} – 2xy) + y^{2} (x^{2} + y^{2} – 2xy) + 2xy (x^{2} + y^{2} – 2xy)

= x^{4} + x^{2}y^{2} – 2x^{3}y + x^{2}y^{2} + y^{4} – 2xy^{3} + 2x^{3}y + 2xy^{3} – 4x^{2}y^{2}

= X^{4} + 2x^{2}y^{2} – 4x^{2}y^{2} + y^{4}

= x^{4} – 2x^{2}y^{2} + y^{4}

Question 14.

Express the following as difference of two squares

(i) (x + 2z) (2x + z)

Answer:

x(2x + z) + 2z (2x + z)

= 2x^{2} + xz + 4xz + 2z^{2}

= 2x^{2} + 2z^{2} + 4xz + xz

= 2 (x^{2} + z^{2} + 2xz) + xz

= 2(x + z)^{2} + xz

(iii) 4(x + 2y) (2x + y)

Answer:

4(x + 2y) (2x + y)

= 4(2x^{2} + xy + 4xy + 2y^{2})

= 4(2x^{2} + 2y^{2} + 4xy + xy)

=4 (2 (x^{2} + y^{2} + 2xy) + xy)

= 4 (2 (x + y)^{2} + xy)

= 8 (x + y)^{2} + 4xy

= 8 (x + y)^{2} + (x + y)^{2} – (x – y)^{2}

[∵ 4xy = (x + y)^{2} – (x – y)^{2}] = 9 (x + y)x^{4}+\frac{1}{x^{4}} – (x – y)2 = [3 (x + y)]^{2} – (x – y)^{2}

(iii) (x + 98)( (x+ 102)

Answer:

(x + 98)( (x + 102)

= (x + 100 – 2) (x + 100 + 2)

= (x + 100)2 – 22

[∵(a + b) (a – b) = a^{2} – b^{2}]

(iv) 505 x 495

Answer:

505 x 495 = (500 + 5) (500 – 5)

= (500)^{2} – 5^{2}

Question 15.

If a = 3x – 5y, b = 6x + 3y and c = 2y – 4x find

(i) a + b – c

Answer:

a + b – c = 3x – 5y + 6x + 3y – (2y – 4x)

= 9x – 2y – 2y + 4x

= 13x – 4y

(ii) 2a – 3b + 4c = 2(3x – 5y) – 3 (6x + 3y) + 4 (2y – 4x)

= 6x – 10y – 18x – 9y + 8y – 16 x

= 6x – 18x – 16 x 10y – 9y + 8y

= 6x – 34x – 19y + 8y = – 28x – 11y

= – (28x + 11y)

Question 16.

The perimeter of a triangle is 15×2 – 23x + 9 and two of its sides are 5×2 + 8x – 1 and 6×2 – 9x + 4. Find the third side

Answer:

Let the sides of the triangle be a, b and c

Let a = 5x^{2} + 8x – 1,

b = 6x^{2} – 9x + 4

Perimeter = 15x^{2} – 23x + 9

a + b + c = 15x^{2} – -23x + 9 5x^{2} + 8x – 1 + 6x^{2} – 9x + 4 + c

= 15x^{2} – 23x + 9

llx^{2} – x + 3 + c <= 15x^{2} – 23x + 9

c = 15x^{2} – 23x + 9 – (llx^{2} – x + 3)

= 15x^{2} – 23x + 9 – llx^{2} + x – 3

c = 4x^{2} 22x + 6

∴ The third side is 4x^{2} – 22x + 6

Question 17.

Two adjacent sides of a rectangle are 2x^{2} – 5xy + 3z^{2} and 4xy – x^{2} – z^{2}. Find its perimeter.

Answer:

Let l = 2x^{2} – 5xy + 3z^{2} and b = 4xy – x^{2} – z^{2}

Perimeter of a rectangel = 2(l + b)

= 2 [2x^{2} – 5xy + 3z^{2} + 4xy – x^{2} – z^{2}]

= 4x^{2} – lOxy + 6z^{2} + 8xy – 2x^{2}– 2z^{2}]

= 2x^{2} + 4z^{2} – 2xy

Question 18.

The base and altitutde of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.

Answer:

Area of a triangle = x base x height

= x(3x-4y)(5x + 5y)

= [(3x(6x + 5y) – 4y(6x + 5y) ]

= [18x^{2} +15y)-24y-20y^{2}]

= [18x^{2} – 9xy – 20y^{2}]

Question 19.

The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of side length x + y is removed. What is the area of the remaining region.

Answer:

Area of the remaining region

= Area of rectangle – Area of sqûare

= length x breadth – (side)^{2}

= (2x + 3y) (3x + 2y) – (x + y)^{2}

=2x(3x + 2y) + 3y(3x + 2y) – [x^{2} + y^{2} + 2xy]

= 6x^{2} + 4xy + 9xy + 6y^{2} – [x^{2} + y^{2} + 2xy]

= 6x^{2} + 13xy + 6y^{2} – x^{2} – y^{2} – 2xy = 5x^{2} + 11xy + 5y^{2}

Question 20.

If a, b, c are rotational numbers such that a^{2} + b^{2} + c^{2} – ab – be – ca = 0 prove that a = b = c

Answer:

a^{2} +b^{2} + c^{2} – ab – bc – ca = 0 Multiplying both the sides by 2

2(a^{2} + b^{2} + c^{2} – ab – be – ca) = 2 x 0

2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ca = 0

a^{2} + a^{2} + b^{2} + b^{2} + c^{2} + c^{2} – 2ab – 2bc – 2ca = 0

a^{2} – 2ab + b^{2} + b^{2} – 2bc + c^{2} + c^{2} – 2ca + a^{2} = 0

(a – b)^{2} + (b – c)^{2} + (c – a)^{2} =0

(a – b)^{2} = 0, (b – c)^{2} = 0, (c – a)^{2} = 0 [v If the sum of positive quantities is equal to zero then each quantity is equal to zero]

∴ (a-b)^{2} = 0 ⇒ a – b = 0 ⇒ a = b

(b-c)^{2} = 0 ⇒ b – c = 0 ⇒ b = c

(c-a)^{2} = 0 ⇒ c – a = 0 ⇒ c = a

∴ a = b = c.