{"id":602129,"date":"2022-05-13T09:16:47","date_gmt":"2022-05-13T09:16:47","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=602129"},"modified":"2022-05-14T05:27:28","modified_gmt":"2022-05-14T05:27:28","slug":"ml-aggarwal-solutions-for-class-8-maths-chapter-10-algebraic-expressions-and-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-8-maths-chapter-10-algebraic-expressions-and-identities\/","title":{"rendered":"ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities"},"content":{"rendered":"\n

Class 8: Maths Chapter 10 solutions. Complete Class 8 Maths Chapter 10 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities<\/h2>\n\n\n\n

ML Aggarwal 8th Maths Chapter 10, Class 8 Maths Chapter 10 solutions<\/p>\n\n\n\n

Exercise 10.1<\/h4>\n\n\n\n

1. Identify the terms, their numerical as well as literal coefficients in each of the following expressions:
(i) 12x2<\/sup>yz \u2013 4xy2<\/sup>
(ii) 8 + mn + nl \u2013 lm<\/strong><\/p>\n\n\n\n

(iii) x2<\/sup>\/3 + y\/6 \u2013 xy2<\/sup><\/strong><\/p>\n\n\n\n

(iv) -4p + 2.3q + 1.7r
Solution:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n


2. Identify monomials, binomials, and trinomials from the following algebraic expressions :
(i) 5p \u00d7 q \u00d7 r2<\/sup>
(ii) 3x2<\/sup> + y \u00f7 2z
(iii) -3 + 7x2<\/sup>
(iv) (5a2 <\/sup>\u2013 3b2<\/sup> + c)\/2
(v) 7x5<\/sup> \u2013 3x\/y
(vi) 5p \u00f7 3q \u2013 3p2<\/sup> \u00d7 q2<\/sup>
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) 5p \u00d7 q \u00d7 r2<\/sup> = 5pqr2<\/sup><\/p>\n\n\n\n

As this algebraic expression has only one term, its therefore a monomial.<\/p>\n\n\n\n

(ii) 3x2<\/sup> + y \u00f7 2z = 3x2<\/sup>\/2z + y\/2z<\/p>\n\n\n\n

As this algebraic expression has two terms, its therefore a binomial.<\/p>\n\n\n\n

(iii) -3 + 7x2<\/sup><\/p>\n\n\n\n

As this algebraic expression has two terms, its therefore a binomial.<\/p>\n\n\n\n

(iv)
\"ML<\/p>\n\n\n\n

As this algebraic expression has three terms, its therefore a trinomial.<\/p>\n\n\n\n

(v) 7x5<\/sup> \u2013 3x\/y<\/p>\n\n\n\n

As this algebraic expression has two terms, its therefore a binomial.<\/p>\n\n\n\n

(vi) 5p \u00f7 3q \u2013 3p2<\/sup> \u00d7 q2<\/sup> = 5p\/3q \u2013 3p2<\/sup>q2<\/sup><\/p>\n\n\n\n

As this algebraic expression has two terms, its therefore a binomial.<\/p>\n\n\n\n


3. Identify which of the following expressions are polynomials. If so, write their degrees.
(i) 2\/5x4<\/sup> \u2013 \u221a3x2<\/sup> + 5x \u2013 1
(ii) 7x3<\/sup> \u2013 3\/x2<\/sup> + \u221a5
(iii) 4a3<\/sup>b2<\/sup> \u2013 3ab4<\/sup> + 5ab + 2\/3
(iv) 2x2<\/sup>y \u2013 3\/xy + 5y3<\/sup> + \u221a3
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) It is a polynomial and the degree of this expression is 4.<\/p>\n\n\n\n

(ii) It is not a polynomial.<\/p>\n\n\n\n

(iii) It is a polynomial and the degree of this expression is 5.<\/p>\n\n\n\n

(iv) It is not a polynomial.<\/p>\n\n\n\n


4. Add the following expressions:
(i) ab \u2013 bv, bv \u2013 ca, ca \u2013 ab
(ii) 5p2<\/sup>q2<\/sup> + 4pq + 7, 3 + 9pq \u2013 2p2<\/sup>q
(iii) l2<\/sup> + m2<\/sup> + n2<\/sup>, lm + mn, mn + nl, nl + lm
(iv) 4x3<\/sup> \u2013 7x2<\/sup> + 9, 3x2<\/sup> \u2013 5x + 4, 7x3<\/sup> \u2013 11x + 1, 6x2<\/sup> \u2013 13x
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) ab \u2013 bc, bc \u2013 ca, ca \u2013 ab<\/p>\n\n\n\n

On adding the expressions, we have<\/p>\n\n\n\n

\u21d2 ab \u2013 bc + bc \u2013 ca + ca \u2013 ab = 0<\/p>\n\n\n\n

(ii) 5p2<\/sup>q2<\/sup> + 4pq + 7,3 + 9pq \u2013 2p2<\/sup>q2<\/sup><\/p>\n\n\n\n

On adding the expressions, we have<\/p>\n\n\n\n

= 5p2<\/sup>q2<\/sup> + 4pq + 7 + 3 + 9pq \u2013 2p2<\/sup>q2<\/sup><\/p>\n\n\n\n

= 5p2<\/sup>q2<\/sup> \u2013 2p2<\/sup>q2<\/sup> + 4pq + 9pq + 7 + 3<\/p>\n\n\n\n

= 3p2<\/sup>q2<\/sup> + 13pq + 10<\/p>\n\n\n\n

(iii) l2<\/sup> + m2<\/sup> + n2<\/sup>, lm + mn, mn + nl, nl + lm<\/p>\n\n\n\n

On adding the expressions, we have<\/p>\n\n\n\n

= l2<\/sup> + m2<\/sup> + n2<\/sup> + lm + mn + mn + nl + nl + lm<\/p>\n\n\n\n

= l2<\/sup> + m2<\/sup> + n2<\/sup> + 2lm + 2mn + 2nl<\/p>\n\n\n\n

(iv) 4x3<\/sup> \u2013 7x2<\/sup> + 9, 3x2<\/sup> \u2013 5x + 4, 7x3<\/sup> \u2013 11x + 1, 6x2<\/sup> \u2013 13x<\/p>\n\n\n\n

On adding the expressions, we have<\/p>\n\n\n\n

= 4x3<\/sup> \u2013 7x2<\/sup> + 9 + 3x2<\/sup> \u2013 5x + 4 + 7x3<\/sup> \u2013 112<\/sup> + 1 + 6x2<\/sup> \u2013 13x<\/p>\n\n\n\n

= 4x2<\/sup> + 7x3<\/sup> \u2013 7x2<\/sup> + 3x2<\/sup> + 6x2<\/sup> \u2013 5x \u2013 11x \u2013 13x + 9 + 4 + 1<\/p>\n\n\n\n

= 11x3<\/sup> \u2013 2x2<\/sup> \u2013 29x + 14<\/p>\n\n\n\n


5. Subtract:
(i) 8a + 3ab \u2013 2b + 7 from 14a \u2013 5ab + 7b \u2013 5
(ii) 8xy + 4yz + 5zx from 12xy \u2013 3yz \u2013 4zx + 5xyz
(iii) 4p2<\/sup>q \u2013 3pq + 5pq2<\/sup> \u2013 8p + 7q -10 from 18 \u2013 3p \u2013 11q + 5pq \u2013 2pq2<\/sup> + 5p2<\/sup>q
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) Subtracting 8a + 3ab \u2013 2b + 7 from 14a \u2013 5ab + 7b \u2013 5, we have<\/p>\n\n\n\n

= (14a \u2013 5ab + 7b \u2013 5) \u2013 (8a + 3ab \u2013 2b + 7)<\/p>\n\n\n\n

= 14a \u2013 5ab + 7b \u2013 5 \u2013 8a \u2013 3ab + 2b \u2013 7<\/p>\n\n\n\n

= 6a \u2013 8ab + 9ab \u2013 12<\/p>\n\n\n\n

(ii) Subtracting 8xy + 4yz + 5zx from 12xy \u2013 3yz \u2013 4zx + 5xyz, we have<\/p>\n\n\n\n

= (12xy \u2013 3yz \u2013 4zx + 5xyz) \u2013 (8xy + 4yz + 5zx)<\/p>\n\n\n\n

= 12xy \u2013 3yz \u2013 4zx + 5xyz \u2013 8xy \u2013 4yz \u2013 5zx<\/p>\n\n\n\n

= 4xy \u2013 7yz \u2013 9zx + 5xyz<\/p>\n\n\n\n

(iii) Subtracting 4p2<\/sup>q \u2013 3pq + 5pq2<\/sup> \u2013 8p + 7q \u2013 10 from 18 \u2013 3p \u2013 11q + 5pq \u2013 2pq2<\/sup> + 5p2<\/sup>q, we have<\/p>\n\n\n\n

= (18 \u2013 3p \u2013 11q + 5pq \u2013 2pq2<\/sup> + 5p2<\/sup>q) \u2013 (4p2<\/sup>q \u2013 3pq + 5pq2<\/sup> \u2013 8p + 7q \u2013 10)<\/p>\n\n\n\n

= 18 \u2013 3p \u2013 11q + 5pq \u2013 2pq2<\/sup> + 5p2<\/sup>q \u2013 7p2<\/sup>q + 3pq \u2013 5pq2<\/sup> + 8p \u2013 7q + 10<\/p>\n\n\n\n

= 28 + 5p \u2013 78q + 8pq \u2013 7pq2<\/sup> + p2<\/sup>q<\/p>\n\n\n\n


6. Subtract the sum of 3x2<\/sup> + 5xy + 7y2<\/sup> + 3 and 2x2<\/sup> \u2013 4xy \u2013 3y2<\/sup> + 7 from 9x2<\/sup> \u2013 8xy + 11y2<\/sup>
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
First, adding 3x2<\/sup> + 5xy + 7y2<\/sup> + 3 and 2x2<\/sup> \u2013 4xy \u2013 3y2<\/sup> + 7, we have<\/p>\n\n\n\n

= 3x2<\/sup> + 5xy + 7y2<\/sup> + 3 + 2x2<\/sup> \u2013 4xy \u2013 3y2<\/sup> + 7<\/p>\n\n\n\n

= 5x2<\/sup> + xy + 4y2<\/sup> + 10<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Subtracting 5x2<\/sup> + xy + 4y2<\/sup> + 10 from 9x2<\/sup> \u2013 8xy + 11y2<\/sup><\/p>\n\n\n\n

= (9x2<\/sup> \u2013 8xy + 11y2<\/sup>) \u2013 (5x2<\/sup> + xy + 4y2<\/sup> + 10)<\/p>\n\n\n\n

= 9x2<\/sup> \u2013 8xy + 11y2<\/sup> \u2013 5x2<\/sup> \u2013 xy \u2013 4y2<\/sup> \u2013 10<\/p>\n\n\n\n

= 4x2<\/sup> \u2013 9xy + 7y2<\/sup> \u2013 10<\/p>\n\n\n\n


7. What must be subtracted from 3a2<\/sup> \u2013 5ab \u2013 2b2<\/sup> \u2013 3 to get 5a2<\/sup> \u2013 7ab \u2013 3b2<\/sup> + 3a?
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
From the question, its understood that we have to subtract 5a2<\/sup> \u2013 7ab \u2013 3b2<\/sup> + 3a from 3a2<\/sup> \u2013 5ab \u2013 2b2<\/sup> \u2013 3<\/p>\n\n\n\n

= 3a2<\/sup> \u2013 5ab \u2013 2b2<\/sup> \u2013 3 \u2013 (5a2<\/sup> \u2013 7ab \u2013 3b2<\/sup> + 3a)<\/p>\n\n\n\n

= 3a2<\/sup> \u2013 5ab \u2013 2b2<\/sup> \u2013 3 \u2013 5a2<\/sup> + 7ab + 3b2<\/sup> \u2013 3a<\/p>\n\n\n\n

= -2a2<\/sup> + 2ab + b2<\/sup> \u2013 3a \u2013 3<\/p>\n\n\n\n


8. The perimeter of a triangle is 7p2<\/sup> \u2013 5p + 11 and two of its sides are p2<\/sup> + 2p \u2013 1 and 3p2<\/sup> \u2013 6p + 3. Find the third side of the triangle.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Perimeter of a triangle = 7p2<\/sup> \u2013 5p + 11<\/p>\n\n\n\n

And, two of its sides are p2<\/sup> + 2p \u2013 1 and 3p2<\/sup> \u2013 6p + 3
<\/strong>
We know that,<\/p>\n\n\n\n

Perimeter of a triangle = Sum of three sides of triangle<\/p>\n\n\n\n

\u21d2 7p2<\/sup> \u2013 5p + 11 = (p2<\/sup> + 2p \u2013 1) + (3p2<\/sup> \u2013 6p + 3) + (Third side of triangle)<\/p>\n\n\n\n

7p2<\/sup> \u2013 5p + 11 = (4p2<\/sup> \u2013 4p + 2) + (Third side of triangle)<\/p>\n\n\n\n

\u21d2 Third side of triangle = (7p2<\/sup> \u2013 5p + 11) \u2013 (4p2<\/sup> \u2013 4p + 2)<\/p>\n\n\n\n

= (7p2<\/sup> \u2013 4p2<\/sup>) + (- 5p + 4p) + (11 \u2013 2)<\/p>\n\n\n\n

= 3p2<\/sup> \u2013 p + 9<\/p>\n\n\n\n

Thus, the third side of the triangle is 3p2<\/sup> \u2013 p + 9.<\/p>\n\n\n\n

Exercise 10.2<\/h4>\n\n\n\n

1. Find the product of:
(i) 4x3<\/sup> and -3xy
(ii) 2xyz and 0
(iii) \u2013(2\/3)p2<\/sup>q, (3\/4)pq2<\/sup> and 5pqr
(iv) -7ab, -3a3<\/sup> and \u2013(2\/7)ab2<\/sup>
(v) \u2013\u00bdx2<\/sup> \u2013 (3\/5)xy, (2\/3)yz and (5\/7)xyz
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
Product of:<\/p>\n\n\n\n

(i) 4x3<\/sup> and -3xy = 4x3<\/sup> \u00d7 (-3xy) = -12x3+1<\/sup> y = -12x4<\/sup>y<\/p>\n\n\n\n

(ii) 2xyz and 0 = 2xyz \u00d7 0 = 0<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

2. Multiply:
(i) (3x \u2013 5y + 7z) by \u2013 3xyz
(ii) (2p2<\/sup> \u2013 3pq + 5q2<\/sup> + 5) by \u2013 2pq
(iii) (2\/3a2<\/sup>b \u2013 4\/5ab2<\/sup> + 2\/7ab + 3) by 35ab
(iv) (4x2<\/sup> \u2013 10xy + 7y2<\/sup> \u2013 8x + 4y + 3) by 3xy
Solution:<\/p>\n\n\n\n


<\/strong>
(i) \u2013 3xyz \u00d7 (3x \u2013 5y + 7z)<\/p>\n\n\n\n

= (- 3xyz) \u00d7 3x + (- 3xyz) \u00d7 (- 5y) + (- 3xyz) \u00d7 (7z)<\/p>\n\n\n\n

= \u2013 9x2<\/sup>yz + 15xyz2<\/sup> \u2013 21xyz2<\/sup><\/p>\n\n\n\n

(ii) -2pq \u00d7 (2p2<\/sup> \u2013 3pq + 5q2<\/sup> + 5)<\/p>\n\n\n\n

= (-2pq) \u00d7 2p2<\/sup> + (-2pq) \u00d7 (-3pq) + (- 2pq) \u00d7 (5q2<\/sup>) + (-2pq) \u00d7 5<\/p>\n\n\n\n

= -4p3<\/sup>q + 6p2<\/sup>q2<\/sup> \u2013 10pq3<\/sup> \u2013 10pq<\/p>\n\n\n\n

(iii)
\"MLby 35ab<\/p>\n\n\n\n

= (2\/3)a2<\/sup>b \u00d7 35ab \u2013 (4\/5)ab2<\/sup> \u00d7 35ab + (2\/7)ab \u00d7 35ab + 3 \u00d7 35ab<\/p>\n\n\n\n

= (70\/3)a3<\/sup>b2<\/sup> \u2013 28a2<\/sup>b3<\/sup> + 10a2<\/sup>b2<\/sup> + 105ab<\/p>\n\n\n\n

(iv) (4x2<\/sup> \u2013 10xy + 7y2<\/sup> \u2013 8x + 4y + 3) by 3xy<\/p>\n\n\n\n

= 4x2<\/sup> \u00d7 3xy \u2013 10xy \u00d7 3xy + 7y2<\/sup> \u00d7 3xy \u2013 8x \u00d7 3xy + 4y \u00d7 3xy + 3 \u00d7 3xy<\/p>\n\n\n\n

= 12x3<\/sup>y \u2013 30x2<\/sup>y2<\/sup> + 21xy3<\/sup> \u2013 24x2<\/sup>y + 12xy2<\/sup> + 9xy<\/p>\n\n\n\n


3. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
(i) (p2<\/sup>q, pq2<\/sup>)
(ii) (5xy, 7xy2<\/sup>)
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) Given, sides of a rectangle are p2<\/sup>q and pq2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area = p2<\/sup>q \u00d7 pq2<\/sup> = p2+1 <\/sup>\u00d7 q2+1<\/sup> = p3<\/sup>q3<\/sup><\/p>\n\n\n\n

(ii) Given, sides are 5xy and 7xy2<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Area = 5xy \u00d7 7xy2<\/sup> = 35x1+1<\/sup> \u00d7 y1+2 <\/sup>= 35x2<\/sup>y3<\/sup><\/p>\n\n\n\n


4. Find the volume of rectangular boxes with the following length, breadth and height respectively:
(i) 5ab, 3a2<\/sup>b, 7a4<\/sup>b2<\/sup>
(ii) 2pq, 4q2<\/sup>, 8rp
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
Given are the length, breadth and height of a rectangular box:<\/p>\n\n\n\n

(i) 5ab, 3a2<\/sup>b, 7a4<\/sup>b2<\/sup><\/p>\n\n\n\n

\u2234 Volume = Length \u00d7 breadth \u00d7 height<\/p>\n\n\n\n

= 5ab \u00d7 3a2<\/sup>b \u00d7 7a4<\/sup>b2<\/sup><\/p>\n\n\n\n

= 5 \u00d7 3 \u00d7 7 \u00d7 a1+2+4<\/sup> \u00d7 b1+1+2<\/sup><\/p>\n\n\n\n

= 105a7<\/sup>b4<\/sup><\/p>\n\n\n\n

(ii) 2pq, 4q2<\/sup>, 8rp<\/p>\n\n\n\n

\u2234 Volume = Length \u00d7 breadth \u00d7 height<\/p>\n\n\n\n

= 2pq \u00d7 4q2<\/sup> \u00d7 8rp<\/p>\n\n\n\n

= 2 \u00d7 4 \u00d7 8 \u00d7 p1+1<\/sup> \u00d7 q1+2<\/sup> \u00d7 r<\/p>\n\n\n\n

= 64p2<\/sup>q3<\/sup>r<\/p>\n\n\n\n

5. Simplify the following expressions and evaluate them as directed:
(i) x2<\/sup>(3 \u2013 2x + x2<\/sup>) for x = 1; x = -1; x = 2\/3 and x = \u20131\/2
(ii) 5xy(3x + 4y \u2013 7) \u2013 3y(xy \u2013 x2<\/sup> + 9) \u2013 8 for x = 2, y = -1
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) x2<\/sup>(3 \u2013 2x + x2<\/sup>)<\/p>\n\n\n\n

For x = 1; x = -1; x = 2\/3 and x = \u20131\/2<\/p>\n\n\n\n

x2<\/sup>(3 \u2013 2x + x2<\/sup>) = 3x2<\/sup> \u2013 2x3<\/sup> + x4<\/sup><\/p>\n\n\n\n

(a) For x = 1<\/p>\n\n\n\n

3x2<\/sup> \u2013 2x3<\/sup> + x4<\/sup> = 3(1)2<\/sup> \u2013 2(1 )3<\/sup> + (1)4<\/sup><\/p>\n\n\n\n

= 3 \u00d7 1 \u2013 2 \u00d7 1 + l<\/p>\n\n\n\n

= 3 \u2013 2 + 1 = 2<\/p>\n\n\n\n

(b) For x = -1<\/p>\n\n\n\n

3x2<\/sup> \u2013 2x3<\/sup> + x4<\/sup> = 3(-1)2<\/sup> \u2013 2(-1)3<\/sup> + (-1)4<\/sup><\/p>\n\n\n\n

= 3 \u00d7 1 \u2013 2 \u00d7 (-1) + 1<\/p>\n\n\n\n

= 3 + 2 + 1 = 6<\/p>\n\n\n\n

(c) For x = 2\/3<\/p>\n\n\n\n

3x2<\/sup> \u2013 2x3<\/sup> + x4<\/sup> = 3(2\/3)2<\/sup> \u2013 2(2\/3)3<\/sup> + (2\/3)4<\/sup><\/p>\n\n\n\n

= 3 \u00d7 (4\/9) \u2013 2 \u00d7 (8\/27) + (16\/81)<\/p>\n\n\n\n

= (4\/3) \u2013 (16\/27) + (16\/81)<\/p>\n\n\n\n

= (108 \u2013 48 + 16)\/81<\/p>\n\n\n\n

= (124 \u2013 48)\/81<\/p>\n\n\n\n

= 76\/81<\/p>\n\n\n\n

(d) For x = -1\/2<\/p>\n\n\n\n

3x2<\/sup> \u2013 2x3<\/sup> + x4<\/sup> = 3(-1\/2)2<\/sup> \u2013 2(-1\/2)3<\/sup> + (-1\/2)4<\/sup><\/p>\n\n\n\n

= 3 \u00d7 (1\/4) \u2013 2 \u00d7 (-1\/8) + (1\/16)<\/p>\n\n\n\n

= (3\/4) + \u00bc + (1\/16)<\/p>\n\n\n\n

= (12 + 4 + 1)\/16<\/p>\n\n\n\n

= 17\/16<\/p>\n\n\n\n

(ii) 5xy(3x + 4y \u2013 7) \u2013 3y(xy \u2013 x2<\/sup> + 9) \u2013 8<\/p>\n\n\n\n

= 15x2<\/sup>y + 20xy2<\/sup> \u2013 35xy \u2013 3xy2<\/sup> + 3 x2<\/sup>y \u2013 21y \u2013 8<\/p>\n\n\n\n

= 18x2<\/sup>y + 17xy2<\/sup> \u2013 35xy \u2013 27y \u2013 8<\/p>\n\n\n\n

When x = 2, y = -1, we have<\/p>\n\n\n\n

= 18(2)2<\/sup> \u00d7 (-1) + 17(2) (-1)2<\/sup> \u2013 35(2) (-1) \u2013 27(-1) \u2013 8<\/p>\n\n\n\n

= 18 \u00d7 4 \u00d7 (-1) + 17 \u00d7 2 \u00d7 1 \u2013 35 \u00d7 2 \u00d7 (-1) \u2013 27 \u00d7 (-1) \u2013 8<\/p>\n\n\n\n

= -74 + 34 + 70 + 27 \u2013 8<\/p>\n\n\n\n

= 131 \u2013 80 = 51<\/p>\n\n\n\n


6. Add the following:
(i) 4p(2 \u2013 p2<\/sup>) and 8p3<\/sup> \u2013 3p
(ii) 7xy(8x + 2y \u2013 3) and 4xy2<\/sup>(3y \u2013 7x + 8)
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
Adding,<\/p>\n\n\n\n

(i) 4p(2 \u2013 p2<\/sup>) and 8p3<\/sup> \u2013 3p<\/p>\n\n\n\n

= 8p \u2013 4p3<\/sup> + 8p3<\/sup> \u2013 3p<\/p>\n\n\n\n

= 5p + 4p3<\/sup><\/p>\n\n\n\n

= 4p3<\/sup> + 5p<\/p>\n\n\n\n

(ii) 7xy(8x + 2y \u2013 3) and 4xy2<\/sup>(3y \u2013 7x + 8)<\/p>\n\n\n\n

= 56x2<\/sup>y + 14xy2<\/sup> \u2013 21xy + 12xy3<\/sup> \u2013 28x2<\/sup>y2<\/sup> + 32xy2<\/sup><\/p>\n\n\n\n

= 12xy3<\/sup> \u2013 28x2<\/sup>y2<\/sup> + 56x2<\/sup>y +46xy2<\/sup> \u2013 21xy<\/p>\n\n\n\n


7. Subtract:
(i) 6x(x \u2013 y + z)- 3y(x + y \u2013 z) from 2z(-x + y + z)
(ii) 7xy(x2<\/sup> -2xy + 3y2<\/sup>) \u2013 8x(x2<\/sup>y \u2013 4xy + 7xy2<\/sup>) from 3y(4x2<\/sup>y \u2013 5xy + 8xy2<\/sup>)
Solution:<\/strong><\/p>\n\n\n\n

Subtracting,
<\/strong>
(i) 6x(x \u2013 y + z) \u2013 3y(x + y \u2013 z) from 2z(-x + y + z)<\/p>\n\n\n\n

\u21d2 6x2<\/sup> \u2013 6xy + 6xz \u2013 3xy \u2013 3y2<\/sup> + 3yz from \u2013 2xz + 2yz + 2z2<\/sup><\/p>\n\n\n\n

= (-2xz + 2yz + 2z2<\/sup>) \u2013 (6x2<\/sup> \u2013 6xy + 6xz \u2013 3xy \u2013 3y2<\/sup> + 3yz)<\/p>\n\n\n\n

= \u2013 2xz + 2yz + 2z2<\/sup> \u2013 6x2<\/sup> + 6xy \u2013 6xz + 3xy + 3y2<\/sup> \u2013 3yz<\/p>\n\n\n\n

= 9xy \u2013 yz \u2013 8zx \u2013 6x2<\/sup> + 3y2<\/sup> + 2z2<\/sup><\/p>\n\n\n\n

(ii) 7xy(x2<\/sup> \u2013 2xy + 3y2<\/sup>) \u2013 8x(x2<\/sup>y \u2013 4xy + 7xy2<\/sup>) from 3y(4x2<\/sup>y \u2013 5xy + 8xy2<\/sup>)<\/p>\n\n\n\n

\u21d2 7x3<\/sup>y \u2013 14x2<\/sup>y2<\/sup> + 21xy3<\/sup> \u2013 8x3<\/sup>y + 32x2<\/sup>y \u2013 56x2<\/sup>y2<\/sup> from 12x2<\/sup>y2<\/sup> \u2013 15xy2<\/sup> + 24xy3<\/sup><\/p>\n\n\n\n

= (12x2<\/sup>y2<\/sup> \u2013 15xy2<\/sup> + 24xy3<\/sup>) \u2013 (7x3<\/sup>y \u2013 14x2<\/sup>y2<\/sup> + 21xy3<\/sup> \u2013 8x3<\/sup>y + 32x2<\/sup>y \u2013 56x2<\/sup>y2<\/sup><\/p>\n\n\n\n

= 12x2<\/sup>y2<\/sup> \u2013 15xy2<\/sup> + 24xy3<\/sup> \u2013 7x3<\/sup>y + 14x2<\/sup>y2<\/sup> \u2013 12xy3<\/sup> + 8x3<\/sup>y \u2013 32x2<\/sup>y + 56x2<\/sup>y2<\/sup><\/p>\n\n\n\n

= 82x2<\/sup>y2<\/sup> + 3xy3<\/sup> + x3<\/sup>y \u2013 15xy2<\/sup> \u2013 32x2<\/sup>y<\/p>\n\n\n\n

Exercise 10.3<\/h4>\n\n\n\n

1. Multiply:
(i) (5x \u2013 2) by (3x + 4)
(ii) (ax + b) by (cx + d)
(iii) (4p \u2013 7) by (2 \u2013 3p)
(iv) (2x2<\/sup> + 3) by (3x \u2013 5)
(v) (1.5a \u2013 2.5b) by (1.5a + 2.56)
(vi) 
Solution:<\/strong><\/p>\n\n\n\n

(i) (5x \u2013 2) by (3x + 4)<\/p>\n\n\n\n

= (5x \u2013 2) \u00d7 (3x + 4)<\/p>\n\n\n\n

= 5x (3x + 4) \u2013 2 (3x + 4)<\/p>\n\n\n\n

= 15x2<\/sup> + 20x \u2013 6x \u2013 8<\/p>\n\n\n\n

= 15x2<\/sup>+ 14x \u2013 8<\/p>\n\n\n\n

(ii) (ax + b) by (cx + d)<\/p>\n\n\n\n

= (ax + b) \u00d7 (cx + d)<\/p>\n\n\n\n

= ax (cx + d) + b (cx + d)<\/p>\n\n\n\n

= acx2<\/sup> + adx + bcx + bd<\/p>\n\n\n\n

(iii) (4p \u2013 7) by (2 \u2013 3p)<\/p>\n\n\n\n

= (4p \u2013 7) \u00d7 (2 \u2013 3p)<\/p>\n\n\n\n

= 4p(2 \u2013 3p) -7(2 \u2013 3p)<\/p>\n\n\n\n

= 8p \u2013 12p2<\/sup> \u2013 14 + 21p<\/p>\n\n\n\n

= 29p \u2013 12p2<\/sup> \u2013 14<\/p>\n\n\n\n

(iv) (2x2<\/sup> + 3) by (3x \u2013 5)<\/p>\n\n\n\n

= (2x2<\/sup> + 3) (3x \u2013 5)<\/p>\n\n\n\n

= 2x2<\/sup>(3x \u2013 5) + 3(3x \u2013 5)<\/p>\n\n\n\n

= 6x3<\/sup> \u2013 10x2<\/sup> + 9x \u2013 15<\/p>\n\n\n\n

(v) (1.5a \u2013 2.5b) by (1.5a + 2.5b)<\/p>\n\n\n\n

= (1.5a \u2013 2.5b) (1.5a + 2.5b)<\/p>\n\n\n\n

= 1.5a(1.5 + 2.5b) \u2013 2.5b(1.5a + 2.5b)<\/p>\n\n\n\n

= 2.25a2<\/sup> + 3.75ab \u2013 3.75a6 \u2013 6.25b2<\/sup><\/p>\n\n\n\n

= 2.25a2<\/sup> \u2013 6.25b2<\/sup><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

2. Multiply:
(i) (x \u2013 2y + 3) by (x + 2y)
(ii) (3 \u2013 5x + 2x2<\/sup>) by (4x \u2013 5)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) (x \u2013 2y + 3) by (x + 2y)<\/p>\n\n\n\n

= (x \u2013 2y + 3) \u00d7 (x + 2y)<\/p>\n\n\n\n

= x (x + 2y) \u2013 2y(x + 2y) + 3 (x + 2y)<\/p>\n\n\n\n

= x2<\/sup> + 2xy \u2013 2xy \u2013 4y2<\/sup> + 3x + 6y<\/p>\n\n\n\n

= x2<\/sup> \u2013 4y2<\/sup> + 3x + 6y<\/p>\n\n\n\n

(ii) (3 \u2013 5x + 2x2<\/sup>) by (4x \u2013 5)<\/p>\n\n\n\n

= (4x \u2013 5) (3 \u2013 5x + 2x2<\/sup>)<\/p>\n\n\n\n

= 4x(3 \u2013 5x + 2x2<\/sup>) \u2013 5(3 \u2013 5x + 2x2<\/sup>)<\/p>\n\n\n\n

= 12x \u2013 20x2<\/sup> + 8x3<\/sup> \u2013 15 + 25x \u2013 10x2<\/sup><\/p>\n\n\n\n

= 8x3<\/sup> \u2013 30x2<\/sup> + 37x \u2013 15<\/p>\n\n\n\n

3. Multiply:
(i) (3x2<\/sup> \u2013 2x \u2013 1) by (2x2<\/sup> + x \u2013 5)
(ii) (2 \u2013 3y \u2013 5y2<\/sup>) by (2y \u2013 1 + 3y2<\/sup>)
Solution:<\/strong><\/p>\n\n\n\n

(i) (3x2<\/sup> \u2013 2x \u2013 1) by (2x2<\/sup> + x \u2013 5)<\/p>\n\n\n\n

= (3x2<\/sup> \u2013 2x \u2013 1) (2x2<\/sup> + x \u2013 5)<\/p>\n\n\n\n

= 3x2<\/sup>(2x2<\/sup> + x \u2013 5) \u2013 2x(2x2<\/sup> + x \u2013 5) -1(2x2<\/sup> + x \u2013 5)<\/p>\n\n\n\n

= 6x4<\/sup> + 3x3<\/sup> \u2013 15x2<\/sup> \u2013 4x3<\/sup> \u2013 2x2<\/sup> + 10x \u2013 2x2<\/sup> \u2013 x + 5<\/p>\n\n\n\n

= 6x4<\/sup> \u2013 x3<\/sup> \u2013 19x2<\/sup> + 9x + 5<\/p>\n\n\n\n

(ii) (2 \u2013 3y \u2013 5y2<\/sup>) by (2y- 1 + 3y2<\/sup>)<\/p>\n\n\n\n

= (2 \u2013 3y \u2013 5y2<\/sup>) \u00d7 (2y- 1 + 3y2<\/sup>)<\/p>\n\n\n\n

= 2(2y \u2013 1 + 3y2<\/sup> ) \u2013 3y (2y \u2013 1 + 3y2<\/sup>) -5y2<\/sup>(2y \u2013 1 + 3y2<\/sup>)<\/p>\n\n\n\n

= 4y \u2013 2 + 6y2<\/sup> \u2013 6y2<\/sup> + 3y \u2013 9y3<\/sup> \u2013 10y3<\/sup> + 5y2<\/sup> \u2013 15y4<\/sup><\/p>\n\n\n\n

= -15y4<\/sup> \u2013 19y3<\/sup> + 5y2<\/sup> + 7y \u2013 2<\/p>\n\n\n\n

4. Simplify:
(i) (x2<\/sup> + 3) (x \u2013 3) + 9
(ii) (x + 3) (x \u2013 3) (x + 4) (x \u2013 4)
(iii) (x + 5) (x + 6) (x + 7)
(iv) (p + q \u2013 2r) (2p \u2013 q + r) \u2013 4qr
(v) (p + q) (r + s) + (p \u2013 q)(r \u2013 s) \u2013 2(pr + qs)
(vi) (x + y + z) (x \u2013 y + z) + (x + y \u2013 z) (-x + y + z) \u2013 4zx
Solution:<\/strong><\/p>\n\n\n\n

(i) (x2<\/sup> + 3) (x \u2013 3) + 9<\/p>\n\n\n\n

= x2<\/sup> (x \u2013 3) + 3(x \u2013 3) + 9<\/p>\n\n\n\n

= x2<\/sup> \u2013 3x2<\/sup> + 3x \u2013 9 + 9<\/p>\n\n\n\n

= x3<\/sup> \u2013 3x2<\/sup> + 3x<\/p>\n\n\n\n

(ii) (x + 3) (x \u2013 3) (x + 4) (x \u2013 4)<\/p>\n\n\n\n

= {(x + 3) (x \u2013 3)} \u00d7 {(x + 4) (x \u2013 4)}<\/p>\n\n\n\n

= {x (x \u2013 3) + 3 (x \u2013 3)} {x (x \u2013 4) + 4 (x \u2013 4)}<\/p>\n\n\n\n

= (x2<\/sup> \u2013 3x + 3x \u2013 9) {x2<\/sup> \u2013 4x + 4x \u2013 16}<\/p>\n\n\n\n

= (x2<\/sup> \u2013 9) (x2<\/sup> \u2013 16)<\/p>\n\n\n\n

= x2<\/sup> (x2<\/sup> \u2013 16) \u2013 9 (x2<\/sup> \u2013 16)<\/p>\n\n\n\n

= x4<\/sup> \u2013 16x2<\/sup> \u2013 9x2<\/sup> + 144<\/p>\n\n\n\n

= x4<\/sup> \u2013 25x2<\/sup> + 144<\/p>\n\n\n\n

(iii) (x + 5) (x + 6) (x + 7)<\/p>\n\n\n\n

= {(x + 5) \u00d7 (x + 6)} (x + 7)<\/p>\n\n\n\n

= (x2<\/sup> + 6x + 5x + 30) (x + 7)<\/p>\n\n\n\n

= (x2<\/sup> + 11x + 30) (x + 7)<\/p>\n\n\n\n

= x(x2<\/sup>+ 11x + 30) + 7(x2<\/sup>+ 11x + 30)<\/p>\n\n\n\n

= x3<\/sup> + 11x2<\/sup> + 30x + 7x2<\/sup> + 77x + 210<\/p>\n\n\n\n

= x3<\/sup> + 18x2<\/sup> + 107x + 210<\/p>\n\n\n\n

(iv) (p + q \u2013 2r)(2p \u2013 q + r) \u2013 4qr<\/p>\n\n\n\n

= p(2p \u2013 q + r) + q(2p \u2013 q + r) \u2013 2r(2p \u2013 q + r) \u2013 4qr<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 pq + pr + 2pq \u2013 q2<\/sup> + qr \u2013 4pr + 2qr \u2013 2r2<\/sup> \u2013 4qr<\/p>\n\n\n\n

= 2p2<\/sup> \u2013 q2<\/sup> \u2013 2r2<\/sup> + pq \u2013 3pr \u2013 2qr<\/p>\n\n\n\n

(v) (p + q)(r + s) + (p \u2013 q) (r \u2013 s) \u2013 2(pr + qs)<\/p>\n\n\n\n

= (pr + ps + qr + qs) + (pr \u2013 ps \u2013 qr + qs) \u2013 2pr \u2013 2qs<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

(vi) (x + y + z)(x \u2013 y + z) + (x + y \u2013 z)(-x + y + z) \u2013 4zx<\/p>\n\n\n\n

= x2<\/sup> \u2013 xy + xz + xy \u2013 y2<\/sup> + yz + xz \u2013 yz + z2<\/sup> \u2013 x2<\/sup> + xy + xz<\/p>\n\n\n\n

\u2013 xy + x2<\/sup> + yx + xz \u2013 yz \u2013 z2<\/sup> \u2013 4zx<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

5. If two adjacent sides of a rectangle are 5x2<\/sup> + 25xy + 4y2<\/sup> and 2x2<\/sup> \u2013 2xy + 3y2<\/sup>, find its area.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

The adjacent sides of a rectangle are 5x2<\/sup> + 25xy + 4y2<\/sup> and 2x2<\/sup> \u2013 2xy + 3y2<\/sup><\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Area of rectangle = Product of two adjacent sides<\/p>\n\n\n\n

= (5x2<\/sup> + 25xy + 4y2<\/sup>) (2x2<\/sup> \u2013 2xy + 3y2<\/sup>)<\/p>\n\n\n\n

= 10x4<\/sup>\u2013 10x3<\/sup>y+ 15x2<\/sup>y2<\/sup> + 50x3<\/sup>y \u2013 50x2<\/sup>y2<\/sup> + 75xy3<\/sup> + 8x2<\/sup>y2<\/sup> \u2013 8xy3<\/sup> + 12y4<\/sup><\/p>\n\n\n\n

= 10x4<\/sup> + 40x3<\/sup>y \u2013 27x2<\/sup>y2<\/sup> + 67xy3<\/sup> + 12y4<\/sup><\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

The area of the rectangle is 10x4<\/sup> + 40x3<\/sup>y \u2013 27x2<\/sup>y2<\/sup> + 67xy3<\/sup> + 12y4<\/sup>.<\/p>\n\n\n\n

Exercise 10.4<\/h4>\n\n\n\n

1. Divide:
(i) \u2013 39pq2<\/sup>r5<\/sup> by \u2013 24p3<\/sup>q3<\/sup>r
(ii) \u20133\/4 a2<\/sup>b3<\/sup> by 6\/7 a3<\/sup>b2<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

(i) \u2013 39pq2<\/sup>r5<\/sup> (\u00f7) \u2013 24p3<\/sup>q3<\/sup>r<\/p>\n\n\n\n

= \u2013 39pq2<\/sup>r5<\/sup>\/ \u2013 24p3<\/sup>q3<\/sup>r<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n


2. Divide:
(i) 9x4<\/sup> \u2013 8x3<\/sup> \u2013 12x + 3 by 3x
(ii) 14p2<\/sup>q3<\/sup> \u2013 32p3<\/sup>q2<\/sup> + 15pq2<\/sup> \u2013 22p + 18q by \u2013 2p2<\/sup>q.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n
\"ML<\/figure>\n\n\n\n


3. Divide:
(i) 6x2<\/sup> + 13x + 5 by 2x + 1
(ii) 1 + y3<\/sup> by 1 + y
(iii) 5 + x \u2013 2x2<\/sup> by x + 1
(iv) x3<\/sup> \u2013 6x2<\/sup> + 12x \u2013 8 by x \u2013 2
Solution:<\/strong><\/p>\n\n\n\n

(i) 6x2<\/sup> + 13x + 5 \u00f7 2x + 1<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u2234 Quotient = 3x + 5 and remainder = 0<\/p>\n\n\n\n


<\/strong>
(ii) 1 + y3<\/sup> \u00f7 1 + y<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u2234 Quotient = y2<\/sup> \u2013 y + 1 and remainder = 0<\/p>\n\n\n\n

(iii) On arranging the terms of dividend in descending order of powers of x and then dividing, we get<\/p>\n\n\n\n

\u2013 2x2<\/sup> + x + 5 \u00f7 x + 1<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u2234 Quotient = \u2013 2x + 3 and remainder = 2<\/p>\n\n\n\n

(iv) x3<\/sup> \u2013 6x2<\/sup> + 12x \u2013 8 \u00f7 x \u2013 2<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u2234 Quotient = x2<\/sup> \u2013 4x + 4 and remainder = 0<\/p>\n\n\n\n


4. Divide:
(i) 6x3<\/sup> + x2<\/sup> \u2013 26x \u2013 25 by 3x \u2013 7
(ii) m3<\/sup> \u2013 6m2<\/sup> + 7 by m \u2013 1
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
(i) 6x3<\/sup> + x2<\/sup> \u2013 26x \u2013 25 \u00f7 3x \u2013 7<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u2234 Quotient = 2x2<\/sup> + 5x + 3 and remainder = \u2013 4<\/p>\n\n\n\n

(ii) m3<\/sup> \u2013 6m2<\/sup> + 7 \u00f7 m \u2013 1<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

\u2234 Quotient = m2<\/sup> \u2013 5m \u2013 5 and remainder = 2.<\/p>\n\n\n\n


5. Divide:
(i) a3<\/sup> + 2a2<\/sup> + 2a + 1 by a2<\/sup> + a + 1
(ii) 12x3<\/sup> \u2013 17x2<\/sup> + 26x \u2013 18 by 3x2<\/sup> \u2013 2x + 5
Solution:<\/strong><\/p>\n\n\n\n

(i) a3<\/sup> + 2a2<\/sup> + 2a + 1 \u00f7 a2<\/sup> + a + 1<\/p>\n\n\n\n


<\/strong>
\u2234 Quotient = a + 1 and remainder = 0.<\/p>\n\n\n\n

(ii) 12x3<\/sup> \u2013 17x2<\/sup> + 26x \u2013 18 \u00f7 3x2<\/sup> \u2013 2x + 5

<\/strong>
\u2234 Quotient = 4x \u2013 3 and remainder = -3<\/p>\n\n\n\n


6. If the area of a rectangle is 8x2<\/sup> \u2013 45y2<\/sup> + 18xy and one of its sides is 4x + 15y, find the length of adjacent side.
Solution:<\/strong><\/p>\n\n\n\n


<\/strong>
Given,<\/p>\n\n\n\n

Area of rectangle = 8x2<\/sup> \u2013 45y2<\/sup> + 18xy<\/p>\n\n\n\n

And, one side = 4x + 15y<\/p>\n\n\n\n

\u2234 Second (adjacent) side = Area of rectangle\/ One side<\/p>\n\n\n\n

= 8x2<\/sup> \u2013 45y2<\/sup> + 18xy \u00f7 4x + 15y<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Thus, length of the adjacent side is 2x \u2013 3y.<\/p>\n\n\n\n

Exercise 10.5<\/h4>\n\n\n\n

1. Using suitable identities, find the following products:
(i) (3x + 5) (3x + 5)
(ii) (9y \u2013 5) (9y \u2013 5)
(iii) (4x + 11y) (4x \u2013 11y)
(iv) (3m\/2 + 2n\/3) (3m\/2 \u2013 2n\/3)
(v) (2\/a + 5\/b) (2a + 5\/b)
(vi) (p2<\/sup>\/2 + 2\/q2<\/sup>) (p2<\/sup>\/2 \u2013 2\/q2<\/sup>)
Solution:<\/strong><\/p>\n\n\n\n

(i) (3x + 5) (3x + 5)<\/p>\n\n\n\n

= (3x + 5)2<\/sup><\/p>\n\n\n\n

= (3x)2<\/sup> + 2 \u00d7 3x \u00d7 5 + (5)2<\/sup> [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 9x2<\/sup> + 30x + 25<\/p>\n\n\n\n

(ii) (9y \u2013 5) (9y \u2013 5)<\/p>\n\n\n\n

= (9y \u2013 5)2<\/sup><\/p>\n\n\n\n

= (9y)2<\/sup> \u2013 2 \u00d7 9y \u00d7 5 + (5)2<\/sup> [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 81y2<\/sup> \u2013 90y + 25<\/p>\n\n\n\n

(iii) (4x + 11y)(4x \u2013 11y)<\/p>\n\n\n\n

= (4x)2<\/sup> \u2013 (11y)2<\/sup><\/p>\n\n\n\n

= 16x2<\/sup> \u2013 121y2<\/sup> [Using, (a + b)(a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

(iv) (3m\/2 + 2n\/3) (3m\/2 \u2013 2n\/3)<\/p>\n\n\n\n

= (3m\/2)2<\/sup> \u2013 (2n\/3)2<\/sup><\/p>\n\n\n\n

= 9m2<\/sup>\/4 \u2013 4n2<\/sup>\/9 [Using, (a + b)(a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

(v) (2\/a + 5\/b) (2a + 5\/b)<\/p>\n\n\n\n

= (2\/a + 5\/b)2<\/sup><\/p>\n\n\n\n

= (2\/a)2<\/sup> + 2(2\/a)(5\/b) + (5\/b)2<\/sup> [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 4\/a2<\/sup> + 20a\/b + 25\/b2<\/sup><\/p>\n\n\n\n

(vi) (p2<\/sup>\/2 + 2\/q2<\/sup>) (p2<\/sup>\/2 \u2013 2\/q2<\/sup>)<\/p>\n\n\n\n

= (p2<\/sup>\/2)2<\/sup> \u2013 (2\/q2<\/sup>)2<\/sup> [Using, (a + b)(a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= p4<\/sup>\/4 \u2013 4\/q4<\/sup><\/p>\n\n\n\n

2. Using the identities, evaluate the following:
(i) 812<\/sup>
(ii) 972<\/sup>
(iii) 1052<\/sup>
(iv) 9972<\/sup>
(v) 6.12<\/sup>
(vi) 496 \u00d7 504
(vii) 20.5 \u00d7 19.5
(viii) 9.62
Solution:<\/strong><\/p>\n\n\n\n

(i) (81)2<\/sup> = (80 + 1)2<\/sup><\/p>\n\n\n\n

= (80)2<\/sup> + 2 \u00d7 80 \u00d7 1 + (1)2<\/sup>  [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 6400 + 160+ 1<\/p>\n\n\n\n

= 6561<\/p>\n\n\n\n

(ii) (97)2<\/sup> = (100 \u2013 3)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> \u2013 2 \u00d7 100 \u00d7 3 + (3)2<\/sup>  [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 10000 \u2013 600 + 9<\/p>\n\n\n\n

= 10009 \u2013 600<\/p>\n\n\n\n

= 9409<\/p>\n\n\n\n

(ii) (105)2<\/sup> = (100 + 5)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> + 2 \u00d7 100 \u00d7 5 + (5)2<\/sup>  [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 10000+ 1000 + 25<\/p>\n\n\n\n

= 11025<\/p>\n\n\n\n

(iv) (997)2<\/sup> = (1000 \u2013 3)2<\/sup><\/p>\n\n\n\n

= (1000)2<\/sup> \u2013 2 \u00d7 1000 \u00d7 3 + (3)2<\/sup>  [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 1000000 \u2013 6000 + 9<\/p>\n\n\n\n

= 1000009 \u2013 6000<\/p>\n\n\n\n

= 994009<\/p>\n\n\n\n

(v) (6.1)2<\/sup> = (6 + 0.1)2<\/sup><\/p>\n\n\n\n

= (6)2<\/sup> + 2 \u00d7 6 \u00d7 0.1 +(0.1)2<\/sup>  [Using, (a + b)2<\/sup> = a2 <\/sup>+ 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 36 + 1.2 + 0.01<\/p>\n\n\n\n

= 37.21<\/p>\n\n\n\n

(vi) 496 \u00d7 504<\/p>\n\n\n\n

= (500 \u2013 4) (500 + 4) [Using, (a + b) (a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= (500)2<\/sup> \u2013 (4)2<\/sup><\/p>\n\n\n\n

= 250000 \u2013 16<\/p>\n\n\n\n

= 249984<\/p>\n\n\n\n

(vii) 20.5 \u00d7 19.5<\/p>\n\n\n\n

= (20 + 0.5) (20 \u2013 0.5) [Using, (a + b) (a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= (20)2<\/sup> \u2013 (0.5)2<\/sup><\/p>\n\n\n\n

= 400 \u2013 0.25<\/p>\n\n\n\n

= 399.75<\/p>\n\n\n\n

(viii) (9.6)2<\/sup> = (10 \u2013 0.4)2<\/sup><\/p>\n\n\n\n

= (10)2<\/sup> \u2013 2 \u00d7 10 \u00d7 0.4 + (0.4)2<\/sup>  [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 100 \u2013 8.0 + 0.16<\/p>\n\n\n\n

= 92.16<\/p>\n\n\n\n

3. Find the following squares, using the identities:
(i) (pq + 5r)2<\/sup> (ii) (5a\/2 \u2013 3b\/5)2<\/sup><\/strong><\/p>\n\n\n\n

(iii) (\u221a2a + \u221a3b)2<\/sup> (iv) (2x\/3y \u2013 3y\/2x)2<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

(i) (pq + 5r)2<\/sup><\/p>\n\n\n\n

= (pq)2<\/sup> + 2 \u00d7 pq \u00d7 5r + (5r)2<\/sup>  [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= p2<\/sup>q2<\/sup> + 10pqr + 25r2<\/sup><\/p>\n\n\n\n

(ii) (5a\/2 \u2013 3b\/5)2<\/sup><\/p>\n\n\n\n

= (5a\/2)2<\/sup> \u2013 2 \u00d7 (5a\/2) \u00d7 (-3b\/5) + (3b\/5)2<\/sup> [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 25a2<\/sup>\/4 \u2013 3ab + 9b2<\/sup>\/25<\/p>\n\n\n\n

(iii) (\u221a2a + \u221a3b)2<\/sup><\/p>\n\n\n\n

= (\u221a2a)2<\/sup> + 2 \u00d7 \u221a2a \u00d7 \u221a3b + (\u221a3b)2<\/sup> [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 2a2<\/sup> + 2\u221a6ab + 3b2<\/sup><\/p>\n\n\n\n

(iv) (2x\/3y \u2013 3y\/2x)2<\/sup><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

4. Using the identity, (x + a) (x + b) = x2<\/sup> + (a + b)x + ab, find the following products:
(i) (x + 7) (x + 3)
(ii) (3x + 4) (3x \u2013 5)
(iii) (p2<\/sup> + 2q) (p2<\/sup> \u2013 3q)
(iv) (abc + 3) (abc \u2013 5)
Solution:<\/strong><\/p>\n\n\n\n

(i) (x + 7) (x + 3)<\/p>\n\n\n\n

= (x)2<\/sup> + (7 + 3)x + 7 \u00d7 3<\/p>\n\n\n\n

= x2<\/sup> + 10x + 21<\/p>\n\n\n\n

(ii) (3x + 4) (3x \u2013 5)<\/p>\n\n\n\n

= (3x)2<\/sup> + (4 \u2013 5) (3x) + 4 \u00d7 (-5)<\/p>\n\n\n\n

= 9x2<\/sup> \u2013 3x \u2013 20<\/p>\n\n\n\n

(iii) (P2<\/sup> + 2q)(p2<\/sup> \u2013 3q)<\/p>\n\n\n\n

= (p2<\/sup>)2<\/sup> + (2q \u2013 3q)p2<\/sup> + 2q \u00d7 (-3q)<\/p>\n\n\n\n

= p4<\/sup> \u2013 p2<\/sup>q \u2013 6pq<\/p>\n\n\n\n

(iv) (abc + 3) (abc \u2013 5)<\/p>\n\n\n\n

= (abc)2<\/sup> + (3 \u2013 5)abc + 3 \u00d7 (-5)<\/p>\n\n\n\n

= a2<\/sup>b2<\/sup>c2<\/sup> \u2013 2abc \u2013 15<\/p>\n\n\n\n

5. Using the identity, (x + a) (x + b) = x2<\/sup> + (a + b)x + ab, evaluate the following:
(i) 203 \u00d7 204
(ii) 8.2 \u00d7 8.7
(iii) 107 \u00d7 93
Solution:<\/strong><\/p>\n\n\n\n

(i) 203 \u00d7 204<\/p>\n\n\n\n

= (200 + 3) (200 + 4)<\/p>\n\n\n\n

= (200)2<\/sup> + (3 + 4) \u00d7 200 + 3 \u00d7 4<\/p>\n\n\n\n

= 40000 + 1400 + 12<\/p>\n\n\n\n

= 41412<\/p>\n\n\n\n

(ii) 8.2 \u00d7 8.7<\/p>\n\n\n\n

= (8 + 0.2) (8 + 0.7)<\/p>\n\n\n\n

= (8)2<\/sup> + (0.2 + 0.7) \u00d7 8 + 0.2 \u00d7 0.7<\/p>\n\n\n\n

= 64 + 8 \u00d7 (0.9) + 0.14<\/p>\n\n\n\n

= 64 + 7.2 + 0.14<\/p>\n\n\n\n

= 71.34<\/p>\n\n\n\n

(iii) 107 \u00d7 93<\/p>\n\n\n\n

= (100 + 7) (100 \u2013 7)<\/p>\n\n\n\n

= (100)2<\/sup> + (7 \u2013 7) \u00d7 100 + 7 \u00d7 (-7)<\/p>\n\n\n\n

= 10000 + 0 \u2013 49<\/p>\n\n\n\n

= 9951<\/p>\n\n\n\n

6. Using the identity a2<\/sup> \u2013 b2<\/sup> = (a + b) (a \u2013 b), find
(i) 532<\/sup> \u2013 472<\/sup>
(ii) (2.05)2<\/sup> \u2013 (0.95)2<\/sup>
(iii) (14.3)2<\/sup> \u2013 (5.7)2<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

(i) 532<\/sup> \u2013 472<\/sup><\/p>\n\n\n\n

= (50 + 3) (50 \u2013 3)<\/p>\n\n\n\n

= (50)2<\/sup> \u2013 (3)2<\/sup><\/p>\n\n\n\n

= 2500 \u2013 9<\/p>\n\n\n\n

= 2491<\/p>\n\n\n\n

(ii) (2.05)2<\/sup> \u2013 (0.95)2<\/sup><\/p>\n\n\n\n

= (2.05 + 0.95) (2.05 \u2013 0.95)<\/p>\n\n\n\n

= 3 \u00d7 1.10<\/p>\n\n\n\n

= 3.3<\/p>\n\n\n\n

(iii) (14.3)2<\/sup> \u2013 (5.7)2<\/sup><\/p>\n\n\n\n

= (14.3 + 5.7) (14.3 \u2013 5.7)<\/p>\n\n\n\n

= 20 \u00d7 8.6<\/p>\n\n\n\n

= 172<\/p>\n\n\n\n

7. Simplify the following:
(i) (2x + 5y)2<\/sup> + (2x \u2013 5y)2<\/sup>
(ii) (7a\/2 \u2013 5b\/2)2<\/sup> \u2013 (5a\/2 \u2013 7b\/2)2<\/sup>
(iii) (p2<\/sup> \u2013 q2<\/sup>r)2<\/sup> + 2p2<\/sup>q2<\/sup>r
Solution:<\/strong><\/p>\n\n\n\n

(i) (2x + 5y)2<\/sup> + (2x \u2013 5y)2<\/sup> [Using, (a \u00b1 b)2<\/sup> = a2<\/sup> \u00b1 2ab + b2<\/sup>]<\/p>\n\n\n\n

= (2x)2<\/sup> + 2 \u00d7 2x \u00d7 5y + (5y)2<\/sup> + (2x)2<\/sup> \u2013 2 \u00d7 2x \u00d7 5y + (5y)2<\/sup><\/p>\n\n\n\n

= 4x2<\/sup> + 20xy + 25y2<\/sup> + 4x2<\/sup> \u2013 20xy + 25y2<\/sup><\/p>\n\n\n\n

= 8x2<\/sup> + 50y2<\/sup><\/p>\n\n\n\n

(ii) (7a\/2 \u2013 5b\/2)2<\/sup> \u2013 (5a\/2 \u2013 7b\/2)2<\/sup><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) (p2<\/sup> \u2013 q2<\/sup>r)2<\/sup> + 2p2<\/sup>q2<\/sup>r [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= (p2<\/sup>)2<\/sup> \u2013 2 \u00d7 p2<\/sup> \u00d7 q2<\/sup>r + (q2<\/sup>r)2<\/sup> + 2p2<\/sup>q2<\/sup>r<\/p>\n\n\n\n

= p4<\/sup> \u2013 2p2<\/sup>q + q4<\/sup>r2<\/sup> + 2p2<\/sup>q2<\/sup>r<\/p>\n\n\n\n

= p4<\/sup> + q4<\/sup>r2<\/sup><\/p>\n\n\n\n

8. Show that:
(i) (4x + 7y)2<\/sup> \u2013 (4x \u2013 7y)2<\/sup> = 112xy
(ii) (3p\/7 \u2013 7q\/6)2<\/sup> + pq = 9p2<\/sup>\/49 + 49q2<\/sup>\/36
(iii) (p \u2013 q)(p + q) + (q \u2013 r)(q + r) + (r \u2013 p) (r + p) = 0
Solution:<\/strong><\/p>\n\n\n\n

(i) Taking LHS, we have<\/p>\n\n\n\n

LHS = (4x + 7y)2<\/sup> \u2013 (4x \u2013 7y)2<\/sup> [Using, (a \u00b1 b)2<\/sup> = a2<\/sup> \u00b1 2ab + b2<\/sup>]<\/p>\n\n\n\n

= [(4x)2<\/sup> + 2 \u00d7 4x \u00d7 7y + (7y)2<\/sup>] \u2013 [(4x)2<\/sup> \u2013 2 \u00d7 4x + 7y + (7y)2<\/sup>]<\/p>\n\n\n\n

= (16x2<\/sup> + 56xy + 49y2<\/sup>) \u2013 (16x2<\/sup> \u2013 56xy + 49y2<\/sup>)<\/p>\n\n\n\n

= l6x2<\/sup> + 56xy + 49y2<\/sup> \u2013 16x2<\/sup> + 56xy \u2013 49y2<\/sup><\/p>\n\n\n\n

= 112xy = RHS<\/p>\n\n\n\n

(ii) Taking LHS, we have<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) Taking LHS, we have<\/p>\n\n\n\n

LHS = (p \u2013 q) (p + q) + (q \u2013 r) (q + r) + (r \u2013 p)(r + p)<\/p>\n\n\n\n

= p2<\/sup> \u2013 q2<\/sup> + q2<\/sup> \u2013 r2<\/sup> + r2<\/sup> \u2013 p2<\/sup> [Using, (a + b) (a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= 0 = RHS<\/p>\n\n\n\n

9. If x + 1\/x = 2, evaluate:
(i) x2<\/sup> + 1\/x2<\/sup> (ii) x4<\/sup> + 1\/x4<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

(i) We have, x + 1\/x = 2<\/p>\n\n\n\n

On squaring on both sides, we get<\/p>\n\n\n\n

(x + 1\/x)2<\/sup> = 22<\/sup><\/p>\n\n\n\n

x2<\/sup> + 2 \u00d7 x \u00d7 1\/x + 1\/x2<\/sup> = 4<\/p>\n\n\n\n

x2<\/sup> + 2 + 1\/x2<\/sup> = 4<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 4 \u2013 2<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 2<\/p>\n\n\n\n

(ii) Again squaring, we get<\/p>\n\n\n\n

(x2<\/sup> + 1\/x2<\/sup>)2<\/sup> = 22<\/sup><\/p>\n\n\n\n

x4<\/sup> + 2 \u00d7 x2<\/sup> \u00d7 1\/x2<\/sup> + 1\/x4<\/sup> = 4<\/p>\n\n\n\n

x4<\/sup> + 2 + 1\/x4<\/sup> = 4<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> = 4 \u2013 2<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> = 2<\/p>\n\n\n\n

10. If x \u2013 1\/x = 7, evaluate:
(i) x2<\/sup> + 1\/x2<\/sup> (ii) x4<\/sup> + 1\/x4<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

We have, x \u2013 1\/x = 7<\/p>\n\n\n\n

On squaring on both sides, we get<\/p>\n\n\n\n

(x \u2013 1\/x)2<\/sup> = 72<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2013 2 \u00d7 x2<\/sup> \u00d7 1\/x2<\/sup> + 1\/x2<\/sup> = 49<\/p>\n\n\n\n

x2<\/sup> \u2013 2 + 1\/x2<\/sup> = 49<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 49 + 2<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 51<\/p>\n\n\n\n

(ii) Again squaring, we get<\/p>\n\n\n\n

(x2<\/sup> + 1\/x2<\/sup>)2<\/sup> = 512<\/sup><\/p>\n\n\n\n

x4 <\/sup>+ 1\/x4<\/sup> + 2 \u00d7 x2<\/sup> \u00d7 1\/x2<\/sup> = 2601<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> + 2 = 2601<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> = 2601 \u2013 2<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> = 2599<\/p>\n\n\n\n

11. If x2<\/sup> + 1\/x2<\/sup> = 23, evaluate:
(i) x + 1\/x (ii) x \u2013 1\/x
Solution:<\/strong><\/p>\n\n\n\n

We have, x2<\/sup> + 1\/x2 <\/sup>= 23<\/p>\n\n\n\n

(i) (x + 1\/x)2<\/sup> = x2<\/sup> + 1\/x2<\/sup> + 2<\/p>\n\n\n\n

= 23 + 2<\/p>\n\n\n\n

= 25<\/p>\n\n\n\n

Taking square root on both sides, we get<\/p>\n\n\n\n

(x + 1\/x) = \u00b15<\/p>\n\n\n\n

Thus, x + 1\/x = 5 or -5<\/p>\n\n\n\n

(ii) (x \u2013 1\/x)2<\/sup> = x2<\/sup> + 1\/x2<\/sup> \u2013 2<\/p>\n\n\n\n

= 23 \u2013 2<\/p>\n\n\n\n

= 21<\/p>\n\n\n\n

Taking square root on both sides, we get<\/p>\n\n\n\n

(x + 1\/x) = \u00b1\u221a21<\/p>\n\n\n\n

Thus, x + 1\/x = \u221a21 or -\u221a21<\/p>\n\n\n\n

12. If a + b = 9 and ab = 10, find the value of a2<\/sup> + b2<\/sup>.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a + b = 9 and ab = 10<\/p>\n\n\n\n

Now, squaring a + b = 9 on both sides, we have<\/p>\n\n\n\n

(a + b)2<\/sup> = (9)<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2ab = 81<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2 \u00d7 10 = 81<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 20 = 81<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> = 81 \u2013 20 = 61<\/p>\n\n\n\n

\u2234 a2<\/sup> + b2<\/sup> = 61<\/p>\n\n\n\n

13. If a \u2013 b = 6 and a2<\/sup> + b2<\/sup> = 42, find the value of
Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

a \u2013 b = 6 and a2<\/sup> + b2<\/sup> = 42<\/p>\n\n\n\n

a \u2013 b = 6<\/p>\n\n\n\n

Now, squaring a \u2013 b = 6 on both sides, we have<\/p>\n\n\n\n

(a \u2013 b)2<\/sup> = (6)2<\/sup><\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> \u2013 2ab = 36<\/p>\n\n\n\n

42 \u2013 2ab = 36<\/p>\n\n\n\n

2ab = 42 \u2013 36 = 6<\/p>\n\n\n\n

ab = 6\/2 = 3<\/p>\n\n\n\n

\u2234 ab = 3<\/p>\n\n\n\n

14. If a2<\/sup> + b2<\/sup> = 41 and ab = 4, find the values of
(i) a + b
(ii) a \u2013 b
Solution:<\/strong><\/p>\n\n\n\n

Given, a2<\/sup> + b2<\/sup> = 41 and ab = 4<\/p>\n\n\n\n

(i) (a + b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab<\/p>\n\n\n\n

= 41 + 2 \u00d7 4<\/p>\n\n\n\n

= 41 + 8<\/p>\n\n\n\n

= 49<\/p>\n\n\n\n

\u2234 a + b = \u00b17<\/p>\n\n\n\n

(ii) (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

= 41 \u2013 2 \u00d7 4<\/p>\n\n\n\n

= 41 \u2013 8<\/p>\n\n\n\n

= 33<\/p>\n\n\n\n

\u2234 a \u2013 b = \u00b1\u221a33<\/p>\n\n\n\n

Check Your Progress<\/p>\n\n\n\n

1. Add the following expressions:
(i) -5x2<\/sup>y + 3xy2<\/sup> \u2013 7xy + 8, 12x2<\/sup>y \u2013 5xy2<\/sup> + 3xy \u2013 2
(ii) 9xy + 3yz \u2013 5zx, 4yz + 9zx \u2013 5y, -5xz + 2x \u2013 5xy
Solution:<\/strong><\/p>\n\n\n\n

(i) (-5x2<\/sup>y + 3xy2<\/sup> \u2013 7xy + 8) + (12x2<\/sup>y \u2013 5xy2<\/sup> + 3xy \u2013 2)<\/p>\n\n\n\n

= 7x2<\/sup>y \u2013 2xy2<\/sup> \u2013 4xy + 6<\/p>\n\n\n\n

(ii) (9xy + 3yz \u2013 5zx) + (4yz + 9zx \u2013 5y, -5xz + 2x \u2013 5xy)<\/p>\n\n\n\n

= 4xy + 7yz \u2013 zx + 2x \u2013 5y<\/p>\n\n\n\n

2. Subtract:
(i) 5a + 3b + 11c \u2013 2 from 3a + 5b \u2013 9c + 3
(ii) 10x2<\/sup> \u2013 8y2<\/sup> + 5y \u2013 3 from 8x2<\/sup> \u2013 5xy + 2y2 <\/sup>+ 5x \u2013 3y
Solution:<\/strong><\/p>\n\n\n\n

(i) 5a \u2013 3b + 11c \u2013 2 from 3a + 5b \u2013 9c + 3<\/p>\n\n\n\n

= (3a + 5b \u2013 9c + 3) \u2013 (5a \u2013 3b + 11c \u2013 2)<\/p>\n\n\n\n

= 3a + 5b \u2013 9c + 3 \u2013 5a + 3b \u2013 11c + 2<\/p>\n\n\n\n

= -2a + 8b \u2013 20c + 5<\/p>\n\n\n\n

(ii) 10x2<\/sup> \u2013 8y2<\/sup> + 5y \u2013 3 from 8x2<\/sup> \u2013 5xy + 2y2<\/sup> + 5x \u2013 3y<\/p>\n\n\n\n

= (8x2<\/sup> \u2013 5xy + 2y2<\/sup> + 5x \u2013 3y) \u2013 (10x2<\/sup> \u2013 8y2<\/sup> + 5y \u2013 3)<\/p>\n\n\n\n

= 8x2<\/sup> \u2013 5xy + 2y2<\/sup> + 5x \u2013 3y \u2013 10x2<\/sup> + 8y2<\/sup> \u2013 5y + 3<\/p>\n\n\n\n

= \u2013 2x2<\/sup> \u2013 5xy + 10y2<\/sup> + 5x \u2013 8y \u2013 3<\/p>\n\n\n\n

3. What must be added to 5x2<\/sup> \u2013 3x + 1 to get 3x3<\/sup> \u2013 7x2<\/sup> + 8?
Solution:<\/strong><\/p>\n\n\n\n

From the question, the required expression is<\/p>\n\n\n\n

= (3x3<\/sup> \u2013 7x2<\/sup> + 8) \u2013 (5x2<\/sup> \u2013 3x + 1)<\/p>\n\n\n\n

= 3x3<\/sup> \u2013 7x2<\/sup> + 8 \u2013 5x2<\/sup> + 3x \u2013 1<\/p>\n\n\n\n

= 3x3<\/sup> \u2013 12x2<\/sup> + 3x + 7<\/p>\n\n\n\n

4. Find the product of
(i) 3x2<\/sup>y and -4xy2<\/sup>
(ii) \u2013(4\/5)xy, (5\/7)yz and \u2013(14\/9)zx
Solution:<\/strong><\/p>\n\n\n\n

Product of:<\/p>\n\n\n\n

(i) 3x2<\/sup>y and -4xy2<\/sup><\/p>\n\n\n\n

= 3x2<\/sup> \u00d7 (-4xy2<\/sup>)<\/p>\n\n\n\n

= -12x2+1<\/sup> y1+2<\/sup><\/p>\n\n\n\n

= 12x3<\/sup>y3<\/sup><\/p>\n\n\n\n

(ii) \u2013(4\/5)xy, (5\/7)yz and \u2013(14\/9)zx<\/p>\n\n\n\n

= \u2013(4\/5)xy \u00d7 (5\/7)yz \u00d7 \u2013(14\/9)zx<\/p>\n\n\n\n

= \u2013(4\/5) \u00d7 (5\/7) \u00d7 \u2013(14\/9) x2<\/sup>y2<\/sup>z2<\/sup><\/p>\n\n\n\n

= (8\/9)x2<\/sup>y2<\/sup>z2<\/sup><\/p>\n\n\n\n

5. Multiply:
(i) (3pq \u2013 4p2<\/sup> + 5q2<\/sup> + 7) by -7pq
(ii) (3\/4x2<\/sup>y \u2013 4\/5xy + 5\/6xy2<\/sup>) by \u2013 15xyz
Solution:<\/strong><\/p>\n\n\n\n

(i) (3pq \u2013 4p2<\/sup> + 5q2<\/sup> + 7) \u00d7 (-7pq)<\/p>\n\n\n\n

= -7pq \u00d7 3pq \u2013 7pq \u00d7 (-4p2<\/sup>) + (-7pq) (5q2<\/sup>) \u2013 7pq \u00d7 7<\/p>\n\n\n\n

= -21p2<\/sup>q2<\/sup> + 28p3<\/sup>q \u2013 35pq3<\/sup> \u2013 49pq<\/p>\n\n\n\n

(ii) (3\/4x2<\/sup>y \u2013 4\/5xy + 5\/6xy2<\/sup>) \u00d7 (\u2013 15xyz)<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

6. Multiply:
(i) (5x2<\/sup> + 4x \u2013 2) by (3 \u2013 x \u2013 4x2<\/sup>)
(ii) (7x2<\/sup> + 12xy \u2013 9y2<\/sup>) by (3x2<\/sup> \u2013 5xy + 3y2<\/sup>)
Solution:<\/strong><\/p>\n\n\n\n

(i) (5x2<\/sup> + 4x \u2013 2) \u00d7 (3 \u2013 x \u2013 4x2<\/sup>)<\/p>\n\n\n\n

= 5x2<\/sup>(3 \u2013 x \u2013 4x2<\/sup>) + 4x(3 \u2013 x \u2013 4x2<\/sup>) \u2013 2(3x \u2013 x \u2013 4x2<\/sup>)<\/p>\n\n\n\n

= 15x2<\/sup> \u2013 5x3<\/sup> \u2013 20x4<\/sup> + 12x \u2013 4x2<\/sup> \u2013 16x3<\/sup> \u2013 6x + 2x + 8x2<\/sup><\/p>\n\n\n\n

= -20x4<\/sup> \u2013 21x3<\/sup> + 19x2<\/sup> + 14x \u2013 6<\/p>\n\n\n\n

(ii) (7x2<\/sup> + 12xy \u2013 9y2<\/sup>) x (3x2<\/sup> \u2013 5xy + 3y2<\/sup>)<\/p>\n\n\n\n

= 7x2<\/sup>(3x2<\/sup> \u2013 5xy + 3y2<\/sup>) + 12xy(3x2<\/sup> \u2013 5xy + 3y2<\/sup>) \u2013 9y2<\/sup>(3x2<\/sup> \u2013 5xy + 3y2<\/sup>)<\/p>\n\n\n\n

= 21x4<\/sup> \u2013 35x3<\/sup>y + 21x2<\/sup>y2<\/sup> + 36x3<\/sup>y \u2013 60x2<\/sup>y2<\/sup> + 36xy3<\/sup> \u2013 27x2<\/sup>y2<\/sup> + 45xy3<\/sup> \u2013 27y4<\/sup><\/p>\n\n\n\n

= 21x4<\/sup> + x3<\/sup>y + 81xy3<\/sup> \u2013 66x2<\/sup>y2<\/sup> \u2013 27y4<\/sup><\/p>\n\n\n\n

7. Simplify the following expressions and evaluate them as directed:
(i) (3ab \u2013 2a2<\/sup> + 5b2<\/sup>) x (2b2<\/sup> \u2013 5ab + 3a2<\/sup>) + 8a3<\/sup>b \u2013 7b4<\/sup> for a = 1, b = -1
(ii) (1.7x \u2013 2.5y) (2y + 3x + 4) \u2013 7.8x2<\/sup> \u2013 10y for x = 0, y = 1.
Solution:<\/strong><\/p>\n\n\n\n

(i) (3ab \u2013 2a2<\/sup> + 5b2<\/sup>) \u00d7 (2b2<\/sup> \u2013 5ab + 3a2<\/sup>) + 8a3<\/sup>b \u2013 7b4<\/sup><\/p>\n\n\n\n

= 3ab(2b2<\/sup> \u2013 5ab + 3a2<\/sup>) \u2013 2a2<\/sup>(2b2<\/sup> \u2013 5ab + 3a2<\/sup>) + 5b2<\/sup>(2b2<\/sup> \u2013 5ab + 3a2<\/sup>) + 8a3<\/sup>b \u2013 7b4<\/sup><\/p>\n\n\n\n

= 6ab32<\/sup> \u2013 15a2<\/sup>b2<\/sup> + 9a3<\/sup>b \u2013 4a2<\/sup>b2<\/sup> + 10a3<\/sup>b \u2013 6a4<\/sup> + 10b4<\/sup> \u2013 25ab3<\/sup> + 15a2<\/sup>b2<\/sup> + 8a3<\/sup>b \u2013 7b4<\/sup><\/p>\n\n\n\n

= 27a3<\/sup>b \u2013 4a2<\/sup>b2<\/sup> \u2013 19ab3<\/sup> \u2013 6a4<\/sup> + 3b4<\/sup><\/p>\n\n\n\n

Putting, a = 1 and b = (-1)<\/p>\n\n\n\n

= 27(1 )3<\/sup> (-1) \u2013 4(1)2<\/sup> (-1)2<\/sup> \u2013 19 (1) (-1)3<\/sup> \u2013 6(1)4<\/sup> + 3(-1)4<\/sup><\/p>\n\n\n\n

= -27 \u2013 4 + 19 \u2013 6 + 3<\/p>\n\n\n\n

= -37 + 22<\/p>\n\n\n\n

= -15<\/p>\n\n\n\n

(ii) (1.7x \u2013 2.5y) (2y + 3x + 4) \u2013 7.8x2<\/sup> \u2013 10y<\/p>\n\n\n\n

1.7x(2y + 3x + 4) \u2013 2.5y(2y + 3x + 4) \u2013 7.8x2<\/sup> \u2013 10y<\/p>\n\n\n\n

= 3.4xy + 5.1x2<\/sup> + 6.8x \u2013 5y2<\/sup> \u2013 7.5xy \u2013 10y \u2013 7.8x2<\/sup> \u2013 10y<\/p>\n\n\n\n

= -2.7x2<\/sup> \u2013 4.1xy \u2013 5y2<\/sup> + 6.8x \u2013 20y<\/p>\n\n\n\n

Putting, x = 0 and y = 1<\/p>\n\n\n\n

= -2.7 \u00d7 0 \u2013 4.1 \u00d7 0 \u00d7 1 \u2013 5(1)2<\/sup> + 6.8 \u00d7 0 \u2013 20 \u00d7 1<\/p>\n\n\n\n

= 0 + 0 \u2013 5 + 0 \u2013 20<\/p>\n\n\n\n

= -25<\/p>\n\n\n\n

8. Carry out the following divisions:
(i) 66pq2<\/sup>r3<\/sup> \u00f7 11qr2<\/sup>
(ii) (x3<\/sup> + 2x2<\/sup> + 3x) \u00f7 2x
Solution:<\/strong><\/p>\n\n\n\n

(i) 66pq2<\/sup>r3<\/sup>\/ 11qr2<\/sup><\/p>\n\n\n\n

= 6pq2-1<\/sup>r3-2<\/sup><\/p>\n\n\n\n

= 6pqr<\/p>\n\n\n\n

(ii) (x3<\/sup> + 2x2<\/sup> + 3x)\/ 2x<\/p>\n\n\n\n

= x3<\/sup>\/2x + 2x2<\/sup>\/2x + 3x\/2x<\/p>\n\n\n\n

= \u00bd x2<\/sup> + x + 3\/2<\/p>\n\n\n\n

9. Divide 10x4<\/sup> \u2013 19x3<\/sup> + 17x2<\/sup> + 15x \u2013 42 by 2x2<\/sup> \u2013 3x + 5.
Solution:<\/strong><\/p>\n\n\n\n

(10x4<\/sup> \u2013 19x3<\/sup> + 17x2<\/sup> + 15x \u2013 42) \u00f7 (2x2<\/sup> \u2013 3x + 5)<\/p>\n\n\n\n

Performing long division, we have<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Thus, Quotient = 5x2<\/sup> \u2013 2x \u2013 7 and Remainder = 4x \u2013 7<\/p>\n\n\n\n

10. Using identities, find the following products:
(i) (3x + 4y) (3x + 4y)
(ii) (5a\/2 \u2013 b) (5a\/2 \u2013 b)
(iii) (3.5m \u2013 1.5n) (3.5m + 1.5n)
(iv) (7xy \u2013 2) (7xy + 7)
Solution:<\/strong><\/p>\n\n\n\n

(i) (3x + 4y) (3x + 4y)<\/p>\n\n\n\n

= (3x + 4y)2<\/sup><\/p>\n\n\n\n

= (3x)2<\/sup> + 2 \u00d7 3x \u00d7 4y + (4y)2<\/sup>  [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 9x2<\/sup> + 24xy + 16y2<\/sup><\/p>\n\n\n\n

(ii) (5a\/2 \u2013 b) (5a\/2 \u2013 b)<\/p>\n\n\n\n

= (5a\/2 \u2013 b)2<\/sup><\/p>\n\n\n\n

= (5a\/2)2<\/sup> + 2 \u00d7 5a\/2 \u00d7 (-b) + (b)2<\/sup>  [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 25a2<\/sup>\/4 \u2013 5ab + b2<\/sup><\/p>\n\n\n\n

(iii) (3.5m \u2013 1.5n) (3.5m + 1.5n)<\/p>\n\n\n\n

= (3.5m)2<\/sup> \u2013 (1.5n)2<\/sup>  [Using, (a \u2013 b)(a + b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= 12.25m2<\/sup> \u2013 2.25n2<\/sup><\/p>\n\n\n\n

(iv) (7xy \u2013 2)(7xy + 7)<\/p>\n\n\n\n

= (7xy)2 <\/sup>+ (-2 + 7) \u00d7 (7xy) + (-2) \u00d7 7 [Using, (x + a)(x + b) = x2<\/sup> + (a + b)x + ab]<\/p>\n\n\n\n

= 49x2<\/sup>y2<\/sup> + 35xy \u2013 14<\/p>\n\n\n\n

11. Using suitable identities, evaluate the following:
(i) 1052<\/sup>
(ii) 972<\/sup>
(iii) 201 \u00d7 199
(iv) 872<\/sup> \u2013 132<\/sup>
(v) 105 \u00d7 107
Solution:<\/strong><\/p>\n\n\n\n

(i) (105)2<\/sup> = (100 + 5)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> + 2 \u00d7 100 \u00d7 5 + (5)2<\/sup> [Using, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 10000 + 1000 + 25<\/p>\n\n\n\n

= 11025<\/p>\n\n\n\n

(ii) (97)2<\/sup> = (100 \u2013 3)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> \u2013 2 \u00d7 100 \u00d7 3 + (3)2<\/sup> [Using, (a \u2013 b)2<\/sup> = a2<\/sup> \u2013 2ab + b2<\/sup>]<\/p>\n\n\n\n

= 10000 \u2013 600 + 9<\/p>\n\n\n\n

= 10009 \u2013 600<\/p>\n\n\n\n

= 9409<\/p>\n\n\n\n

(iii) 201 \u00d7 199 = (200 + 1) (200 \u2013 1)<\/p>\n\n\n\n

= (200)2<\/sup> \u2013 (1)2<\/sup> [Using, (a + b) (a \u2013 b) = a2<\/sup> \u2013 b2<\/sup>]<\/p>\n\n\n\n

= 40000 \u2013 1<\/p>\n\n\n\n

= 39999<\/p>\n\n\n\n

(iv) 872<\/sup> \u2013 132<\/sup><\/p>\n\n\n\n

= (87 + 13) (87- 13) [Using, a2<\/sup> \u2013 b2<\/sup> = (a + b)(a \u2013 b)]<\/p>\n\n\n\n

= 100 \u00d7 74<\/p>\n\n\n\n

= 7400<\/p>\n\n\n\n

(v) 105 \u00d7 107<\/p>\n\n\n\n

= (100 + 5) (100 + 7)<\/p>\n\n\n\n

= (100)2<\/sup> + (5 + 7) \u00d7 100 + 5 \u00d7 7 [Using, (x + a)(x \u2013 b) = x2<\/sup> + (a + b)x + ab]<\/p>\n\n\n\n

= 10000 + 1200 + 35<\/p>\n\n\n\n

= 11235<\/p>\n\n\n\n

12. Prove that following:
(i) (a + b)2<\/sup> \u2013 (a \u2013 b)2<\/sup> + 4ab
(ii) (2a + 3b)2<\/sup> + (2a \u2013 3b)2<\/sup> = 8a2<\/sup> + 18b2<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

(i) Taking the RHS, we have<\/p>\n\n\n\n

RHS = (a \u2013 b)2<\/sup> + 4ab<\/p>\n\n\n\n

= a2<\/sup>\u2013 2ab + b2<\/sup> + 4ab<\/p>\n\n\n\n

= a2<\/sup> + 2ab + b2<\/sup><\/p>\n\n\n\n

= (a + b)2<\/sup> = L.H.S.<\/p>\n\n\n\n

(ii) Taking the LHS, we have<\/p>\n\n\n\n

LHS = (2a + 3b)2<\/sup> + (1a \u2013 3b)2<\/sup><\/p>\n\n\n\n

= (2a)2<\/sup> + 2 \u00d7 2a \u00d7 3b + (3b)2<\/sup> + (2a)2<\/sup> \u2013 2 \u00d7 2a \u00d7 3b + (3b)2<\/sup><\/p>\n\n\n\n

= 4a2<\/sup> + 12ab + 9b2<\/sup> + 4a2<\/sup> \u2013 12ab + 9b2<\/sup><\/p>\n\n\n\n

= 8a2<\/sup> + 18b2<\/sup> = RHS<\/p>\n\n\n\n

13. If x + 1\/x = 5, evaluate
(i) x2<\/sup> + 1\/x2 <\/sup>(ii) x4<\/sup> + 1\/x4<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

(i) We have, x + 1\/x = 5<\/p>\n\n\n\n

On squaring on both sides, we get<\/p>\n\n\n\n

(x + 1\/x)2<\/sup> = 52<\/sup><\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> + 2 \u00d7 x \u00d7 1\/x = 25<\/p>\n\n\n\n

x2<\/sup> + 2 + 1\/x2<\/sup> = 25<\/p>\n\n\n\n

x2<\/sup> + 1\/x2<\/sup> = 25 \u2013 2<\/p>\n\n\n\n

Hence, x2<\/sup> + 1\/x2<\/sup> = 23<\/p>\n\n\n\n

(ii) Again, squaring x2<\/sup> + 1\/x2<\/sup> = 23 on both sides, we get<\/p>\n\n\n\n

(x2<\/sup> + 1\/x2<\/sup>)2<\/sup> = 232<\/sup><\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> + 2 \u00d7 x4<\/sup> \u00d7 1\/x4<\/sup> = 529<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> + 2 = 529<\/p>\n\n\n\n

x4<\/sup>+ 1\/x4<\/sup> = 529 \u2013 2<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

x4<\/sup> + 1\/x4<\/sup> = 527<\/p>\n\n\n\n

14. If a + b = 5 and a2<\/sup> + b2<\/sup> = 13, find ab.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a + b = 5 and a2<\/sup> + b2<\/sup> = 13<\/p>\n\n\n\n

On squaring a + b = 5 both sides, we get<\/p>\n\n\n\n

(a + b)2<\/sup> = (5)2<\/sup><\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2ab = 25<\/p>\n\n\n\n

13 + 2ab = 25 \u21d2 2ab = 25 \u2013 13 = 12<\/p>\n\n\n\n

\u21d2 ab = 12\/2 = 6<\/p>\n\n\n\n

\u2234 ab = 6<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n