NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities | Class 8th Maths
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths. Here We learn what is in ncert maths class 8 chapter 9 Algebraic Expressions and Identities and how to solve questions with easiest method. ┬аIn this chapter we solve the question of NCERT 8th class maths chapter 9. NCERT class 8 maths solutions Algebraic Expressions and Identities┬аare part of NCERT Solutions for class 8 maths chapter 9 solution PDF. Ncert solutions for class 8 chapter 9 Algebraic Expressions and Identities with formula and solution. class 8th maths
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8 maths Here we solve class 8th maths ncert solutions chapter 9 Algebraic Expressions and Identities┬аconcepts all questions with easy method with expert solutions. It help students in their study, home work and preparing for exam. Soon we provide ncert solutions for class 8 maths chapter 9 Algebraic Expressions and Identities question and answers. Soon we provided ncert class 8 class 8th maths chapter 9 Algebraic Expressions and Identities in free PDF here. ncert solutions for class 8 maths chapter 9 pdf will be provide soon. 8th class maths chapter 9 NCERT Solution and ncert solutions for class 8 maths chapter 9 pdf download book PDF.
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities class 8th maths
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.1
Exercise 9.1
1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 тИТ 3zy | (ii) 1 + x + x2 | (iii) 4x2y2 тИТ 4x2y2z2 + z2 |
(iv) 3 тИТ pq + qr тИТ rp | (v) \(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\) | (vi) 0.3a тИТ 0.6ab + 0.5b |
Solutions :
┬а | ┬а | Terms | Coefficients |
(i) | 5xyz2 тИТ 3zy | There are two terms in this expressions. 5xyz2 and 3zy |
5 and тИТ 3 |
(ii) | 1 + x + x2 | There are two terms in this expressions. 1, x and x2 |
1, 1 and 1 |
(iii) | 4x2y2 тИТ 4x2y2z2 + z2 | There are three terms in this expressions. 4x2y2 , тИТ 4x2y2z2 and z2 |
4, тИТ 4 and 1 |
(vi) | 3 тИТ pq + qr – rp | There are four terms in this expressions. 3, тИТ pq, qr and тИТ rp |
3, тИТ 1, 1 and тИТ 1 |
(v) | \(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\) | There are three terms in this expressions. \(\displaystyle \frac{x}{2}\) , \(\displaystyle \frac{y}{2}\) and тИТ xy |
\(\displaystyle \frac{1}{2}\) , \(\displaystyle \frac{1}{2}\) and тИТ 1 |
(vi) | 0.3a тИТ 0.6ab + 0.5b | There are three terms in this expressions. 0.3, тИТ 0.6ab and 0.5b |
0.3, тИТ 0.6 and 0.5 |
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3, 7 + y + 5x, 2y тИТ 3y2, 2y тИТ 3y2 + 4y3, 5x тИТ 4y + 3xy, 4z тИТ 15z2, ab + bc + cd + da, pqr, p2q, 2p + 2q
Solutions :
Monomials :- 1000, pqr
Binomials :- x + y, 2y тИТ 3y2, 4z тИТ 15z2, p2q + pq2, 2p + 2q
Trinomials :- 7 + y + 5x, 2y тИТ 3y2 + 4y3, 5x тИТ 4y + 3xy
These polynomials do not fit in any of these three categories :- x + x2 + x3 + x4, ab + bc + cd + da
3. Add the following :
(i) ab тИТ bc, bc тИТ ca, ca тИТ ab | (ii) a тИТ b + ab, b тИТ c + bc, c тИТ a + ac |
(iii) 2p2q2 тИТ 3pq + 4, 5 + 7pq тИТ 3p2q2 | (iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl |
Solutions :
(i) ab тИТ bc, bc тИТ ca, ca тИТ ab
on writing similar terms together –
= ab тИТ bc + bc тИТ ca + ca тИТ ab
= ab тИТ ab + bc тИТ bc + ca тИТ ca
= 0
(ii) a тИТ b + ab, b тИТ c + bc, c тИТ a + ac
= a тИТ b + ab + b тИТ c + bc + c тИТ a + ac
on writing similar terms together –
= a тИТ a + b тИТ b + c тИТ c + ab + bc + ac
by adding and substracting same terms –
= ab + bc + ac
(iii) 2p2q2 тИТ 3pq + 4, 5 + 7pq тИТ 3p2q2
= 2p2q2 тИТ 3pq + 4 + 5 + 7pq тИТ 3p2q2
on writing similar terms together –
= 2p2q2 тИТ 3p2q2 + 7pq тИТ 3pq + 4 + 5
= тИТ p2q2 + 4pq + 9
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
= l2 + m2┬а+ m2 + n2┬а+ n2 + l2┬а+ 2lm + 2mn + 2nl
on writing similar terms together –
= l2 + l2 + m2┬а+ m2 + n2┬а+ n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
= 2(l2 + m2 + n2 + lm + mn + nl)
4. (a) Subtract 4a тИТ 7ab + 3b + 12 from 12a тИТ 9ab + 5b тИТ 3
(b) Subtract 3xy + 5yz тИТ 7zx from 5xy тИТ 2yz тИТ 2zx + 10xyz
(c) Subtract 4p2q тИТ 3pq + 5pq2 тИТ 8p + 7q тИТ 10 from 18 тИТ 3p тИТ 11p + 5pq тИТ 2pq2 + 5p2q
Solutions :
(a) Subtract 4a тИТ 7ab + 3b + 12 from 12a тИТ 9ab + 5b тИТ 3
= 12a тИТ 9ab + 5b тИТ 3 тИТ (4a тИТ 7ab + 3b + 12)
remove the bracket –
= 12a тИТ 9ab + 5b тИТ 3 тИТ 4a + 7ab тИТ 3b тИТ 12┬а (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= 12a тИТ 4a + 5b тИТ 3b + 7ab тИТ 9ab тИТ 4 тИТ 12
= 8a + 2b тИТ 2ab тИТ 16
(b) Subtract 3xy + 5yz тИТ 7zx from 5xy тИТ 2yz тИТ 2zx + 10xyz
= 5xy тИТ 2yz тИТ 2zx + 10xyz тИТ (3xy + 5yz тИТ 7zx)
remove the bracket –
= 5xy тИТ 2yz тИТ 2zx + 10xyz тИТ 3xy тИТ 5yz + 7zx (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= 5xy тИТ 3xy тИТ 2yz тИТ 5yz + 7zx тИТ 2zx + 10xyz
= 2xy тИТ 7yz + 5zx + 10xyz
(c) Subtract 4p2q тИТ 3pq + 5pq2 тИТ 8p + 7q тИТ 10 from 18 тИТ 3p тИТ 11p + 5pq тИТ 2pq2 + 5p2q
= 18 тИТ 3p тИТ 11p + 5pq тИТ 2pq2 + 5p2q тИТ (4p2q тИТ 3pq + 5pq2 тИТ 8p + 7q тИТ 10)
remove the bracket –
= 18 тИТ 3p тИТ 11q + 5pq тИТ 2pq2 + 5p2q тИТ 4p2q + 3pq тИТ 5pq2 + 8p тИТ 7q + 10)┬а (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= тИТ 3p + 8p тИТ 11q тИТ 7q + 5pq + 3pq тИТ 2pq2 тИТ 5pq2 + 5p2q тИТ 4p2q
= 5p тИТ 18q + 8pq тИТ 7pq2 + p2q
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.2
Exercise 9.2
1. Find the product of the following pairs of monomials.
(i) 4, 7p | (ii) тИТ 4p, 7p | (iii) тИТ 4p, 7pq | (iv) 4p3, тИТ 3p |
(v) 4p, 0 | ┬а | ┬а | ┬а |
Solutions :┬а
(i) 4, 7p
= 4 ├Ч 7p
= 28p
(ii)┬а тИТ 4p, 7p
= тИТ 4p ├Ч 7p
= тИТ 28p2
(iii) тИТ 4p, 7pq
= тИТ 4p ├Ч 7pq
= тИТ 28p2q
(iv) 4p3, тИТ 3p
= 4p3 ├Ч тИТ 3p
= тИТ 12p4
(v) 4p, 0
= 4p ├Ч 0
= 0
2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.┬а┬а
(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
Solutions :
(p, q)
length = p
breadth = q
area of rectangle = length ├Ч breadth
= p ├Ч q
= pq unit square
(10m, 5n)
length = 10m
breadth = 5n
area of rectangle = length ├Ч breadth
= 10m ├Ч 5n
= 50mn unit square
(20x2, 5y2)
length = 20x2
breadth = 5y2
area of rectangle = length ├Ч breadth
= 20x2 ├Ч 5y2
= 100x2y2 unit square
(4x, 3x2)
length = 4x
breadth = 3x2
area of rectangle = length ├Ч breadth
= 4x ├Ч 3x2
= 12x3 unit square
(3mn, 4np)
length = 3mn
breadth = 4np
area of rectangle = length ├Ч breadth
= 3mn ├Ч 4np
= 12mn2p unit square
3. Complete the table of products.
First monomial тЖТ Second monomial тЖУ |
2x | тИТ 5y┬а | 3x2 | тИТ 4xy | 7x2y | тИТ 9x2y2 |
2x | 4x2 | ┬а | ┬а | ┬а | ┬а | ┬а |
тИТ 5y | ┬а | ┬а | тИТ 15x2y | ┬а | ┬а | ┬а |
3x2 | ┬а | ┬а | ┬а | ┬а | ┬а | ┬а |
тИТ 4xy | ┬а | ┬а | ┬а | ┬а | ┬а | ┬а |
7x2y | ┬а | ┬а | ┬а | ┬а | ┬а | ┬а |
тИТ 9x2y2 | ┬а | ┬а | ┬а | ┬а | ┬а | ┬а |
Solutions :┬а
First monomial тЖТ Second monomial тЖУ |
2x | тИТ 5y┬а | 3x2 | тИТ 4xy | 7x2y | тИТ 9x2y2 |
2x | 4x2 | тИТ 10xy | 6×3 | тИТ 8x2y | 14x3y | тИТ 18x3y2 |
тИТ 5y | тИТ 10xy | 25y2 | тИТ 15x2y | 20xy2 | тИТ 35x2y2 | 45x2y3 |
3x2 | 6x3 | тИТ 15x2y | 9x4 | тИТ 12x3y | 21x4y | тИТ 27x4y2 |
тИТ 4xy | тИТ 8x2y | 20xy2 | тИТ 12x3y | 16x2y2 | тИТ 28x3y2 | 36x3y3 |
7x2y | 14x3y | тИТ 35x2y2 | 21x4y | тИТ 28x3y2 | 49x4y2 | тИТ 63x4y3 |
тИТ 9x2y2 | тИТ 18x3y2 | 45x2y2 | тИТ 27x4y2 | 36x3y3 | тИТ 63x4y3 | 81x4y4 |
4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a2, 7a4 | (ii) 2p, 4q, 8r | (iii) xy, 2x2y, 2xy2 | (iv) a, 2b, 3c |
Solutions :┬а
(i) 5a, 3a2, 7a4
volume of rectangular box = length ├Ч breadth ├Ч height
= 5a ├Ч 3a2 ├Ч 7a4
= 5 ├Ч 3 ├Ч 7 ├Ч a ├Ч a2 ├Ч a4
= 105a7
(ii) 2p, 4q, 8r
volume of rectangular box = length ├Ч breadth ├Ч height
= 2p ├Ч 4q ├Ч 8r
= 2 ├Ч 4 ├Ч 8 ├Ч p ├Ч q ├Ч r
= 64pqr
(iii) xy, 2x2y, 2xy2
volume of rectangular box = length ├Ч breadth ├Ч height
= xy ├Ч 2x2y ├Ч 2xy2
= 1 ├Ч 2 ├Ч 2 ├Ч xy ├Ч x2y ├Ч xy2
= 4x4y4
(iv) a, 2b, 3c
volume of rectangular box = length ├Ч breadth ├Ч height
= a ├Ч 2b ├Ч 3c
= 1 ├Ч 2 ├Ч 3 ├Ч a ├Ч b ├Ч c
= 6abc
5. Obtain the product of
(i) xy, yz, zx | (ii) a, тИТ a2, a3 | (iii) 2, 4y, 8y2, 16y3 |
(iv) a, 2b, 3c, 6abc | (v) m, тИТ mn, mnp | ┬а |
Solutions :
(i) xy, yz, zx
= xy ├Ч yz ├Ч zx┬а
= x2y2z2
(ii) a, тИТ a2, a3┬а
= a ├Ч (тИТ a2) ├Ч a3┬а
= тИТ a6
(iii) 2, 4y, 8y2, 16y3
= 2 ├Ч 4y ├Ч 8y2┬а├Ч 16y3
= 2 ├Ч 4 ├Ч 8 ├Ч 16 ├Ч y ├Ч y2 ├Ч y3
= 1024y3
(iv) a, 2b, 3c, 6abc
= 2 ├Ч 3 ├Ч a ├Ч b ├Ч c ├Ч abc
= 6a2b2c2
(v) m, тИТ mn, mnp
= m ├Ч (тИТ mn) ├Ч mnp
= тИТ m2n2p
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.3
Exercise 9.3
1. Carry out the multiplication of the expressions in each of the following pairs
(i) 4p, q + r | (ii) ab, a тИТ b | (iii) a + b, 7a2b2 | (iv) a2 тИТ 9, 4a |
(v) pq + qr + rp, 0 | ┬а | ┬а | ┬а |
Solutions :
(i) 4p, q + r┬а
= 4p ├Ч (q + r)
= 4p ├Ч q + 4p ├Ч r
= 4pq + 4pr
2. Complete the table.
┬а | First expression | Second expression | Product |
(i) | a | b + c + d┬а | ____ |
(ii) | x + y тИТ 5 | 5xy | ____ |
(iii) | p | 6p2 тИТ 7p + 5 | ____ |
(iv) | 4p2q2 | p2 тИТ q2 | ____ |
(v) | a + b + c | abc | ____ |
Solutions :
┬а | First expression | Second expression | Product |
(i) | a | b + c + d┬а | ab + ac + ad |
(ii) | x + y тИТ 5 | 5xy | 5x2y + 5xy2 тИТ 25xy |
(iii) | p | 6p2 тИТ 7p + 5 | 6p3 тИТ 7p2 + 5p |
(iv) | 4p2q2 | p2 тИТ q2 | 4p4q2 тИТ 4p2q4 |
(v) | a + b + c | abc | a2bc + ab2c + abc2 |
(i) a ├Ч (b + c + d)
= ab + ac + ad
(ii) (x + y тИТ 5) ├Ч 5xy
= 5x2y + 5xy2 тИТ 25xy
(iii) p ├Ч (6p2 тИТ 7p + 5)
= 6p3 тИТ 7p2 + 5p
(iv) 4p2q2 ├Ч (p2 тИТ q2)
= 4p4q2 тИТ 4p2q4
(v) (a + b + c) ├Ч abc
= a2bc + ab2c + abc2
3. Find the product.
(i) (a2) ├Ч (2a22) ├Ч (4a26) | (ii) \(\displaystyle \left( {\frac{2}{3}xy} \right)\times \left( {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right)\) |
(iii) \(\displaystyle \left( {-\frac{{10}}{3}p{{q}^{3}}} \right)\times \left( {\frac{6}{5}{{p}^{3}}q} \right)\) | (iv) x ├Ч x2 ├Ч x3 ├Ч x4 |
Solutions :
(i) (a2) ├Ч (2a22) ├Ч (4a26)
= 2 ├Ч 4 ├Ч a2 ├Ч a22 ├Ч a26
= 8a50
(ii) \(\displaystyle \left( {\frac{2}{3}xy} \right)\times \left( {\frac{{-9}}{{10}}{{x}^{2}}{{y}^{2}}} \right)\)
\(\displaystyle \begin{array}{l}=\frac{2}{3}\times \left( {\frac{{-9}}{{10}}} \right)\left( {xy\times {{x}^{2}}{{y}^{2}}} \right)\\=\frac{{-18}}{{30}}{{x}^{3}}{{y}^{3}}\\=\frac{{-3}}{5}{{x}^{3}}{{y}^{3}}\end{array}\)
(iii) \(\displaystyle \left( {-\frac{{10}}{3}p{{q}^{3}}} \right)\times \left( {\frac{6}{5}{{p}^{3}}q} \right)\) (iv) x ├Ч x2 ├Ч x3 ├Ч x4
\(\displaystyle \begin{array}{l}=\left( {-\frac{{10}}{3}} \right)\times \frac{6}{5}\times p{{q}^{3}}\times {{p}^{3}}q\\=-\frac{{60}}{{15}}{{p}^{4}}{{q}^{4}}\\=-4{{p}^{4}}{{q}^{4}}\end{array}\)
= x10
4. (a) Simplify 3x (4x тАУ 5) + 3 and find its values for (i) x = 3 (ii) \(\displaystyle x=\frac{1}{2}\)┬а
(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1
(iii) a = тАУ 1.
Solutions :
(a) (i) 3x (4x тИТ 5) + 3
= 3x ├Ч 4x тИТ 3x ├Ч 5 + 3
= 12x2 тИТ 15x + 3 ………………………….. (i)
putting x = 3 equ. in (i)
= 12 ├Ч (3)2 тИТ 9 ├Ч 5 + 3
= 12 ├Ч 9 тИТ 45 + 3
= 108 + 3 тИТ 45
= 111 тИТ 45
= 66
(ii) Putting \(\displaystyle x=\frac{1}{2}\) in equ (i) – 5. (a) Add: p (p тИТ q), q (q тИТ r) and r (r тИТ p)┬а
\(\displaystyle \begin{array}{l}=12\times {{\left( {\frac{1}{2}} \right)}^{2}}-15\times \frac{1}{2}+3\\=12\times \frac{1}{4}-\frac{{15}}{2}+3\\=3-\frac{{15}}{2}+3\\=\frac{{6-15+6}}{2}\\=\frac{{12-15}}{2}\\=\frac{{-3}}{2}\end{array}\)
(b) Add: 2x (z тИТ x тИТ y) and 2y (z тИТ y тИТ x)┬а
(c) Subtract: 3l (l тИТ 4m + 5n) from 4l (10n тИТ 3m + 2l )┬а
(d) Subtract: 3a (a + b + c ) тИТ 2b (a тИТ b + c) from 4c (тИТ a + b + c )┬а
Solutions :┬а
(a) Add: p (p тИТ q), q (q тИТ r) and r (r тИТ p)┬а
= p (p тИТ q) + q (q тИТ r) + r (r тИТ p)
= p2 тИТ pq┬а+ q2 тИТ qr┬а+ r2 тИТ rp
= p2 + q2 + r2 тИТ pq тИТ qr тИТ rp
(b) Add: 2x (z тИТ x тИТ y) and 2y (z тИТ y тИТ x)┬а
= 2x (z тИТ x тИТ y) + 2y (z тИТ y тИТ x)
= 2xz┬атИТ 2x2 тИТ 2xy┬а+ 2yz┬атИТ 2y2 тИТ 2xy
= тИТ 2x2 тИТ 2y2 тИТ 2xy┬атИТ 2xy┬а+ 2yz┬а+ 2xz
= тИТ 2x2 тИТ 2y2 тИТ 4xy + 2yz┬а+ 2xz
(c) Subtract: 3l (l тИТ 4m + 5n) from 4l (10n тИТ 3m + 2l )┬а
= 4l (10n тИТ 3m + 2l) тИТ 3l (l тИТ 4m + 5n)
= 4l ├Ч 10n тИТ 4l ├Ч 3m + 4l + 2l тИТ 3l ├Ч l тИТ 3l ├Ч (тИТ 4m) тИТ 3l ├Ч 5n
= 40ln┬атИТ 12lm┬а+ 8l2 тИТ 3l2 + 12lm┬атИТ 15ln
= 40ln┬атИТ 15ln┬атИТ 12lm┬а+ 12lm┬а+ 8l2 тИТ 3l2
= 25ln┬атИТ 0lm + 5l2
= 5l2 + 25ln
(d) Subtract: 3a (a + b + c ) тИТ 2b (a тИТ b + c) from 4c (тИТ a + b + c )┬а
= 4c (тИТ a + b┬а+ c) тИТ [3a (a + b┬а+ c) тИТ 2b (a тИТ b┬а+ c)]
= тИТ 4ac + 4bc + 4c2 тИТ [3a2 + 3ab┬а+ 3ac тИТ 2ab┬а+ 2a2 – 2bc]
= тИТ 4ac + 4bc + 4c2 тИТ 3a2 тИТ 3ab┬атИТ 3ac + 2ab тИТ 2b2 + 2bc
= тИТ 4ac тИТ 3ac + 4bc + 2bc тИТ 3ab┬а+ 2ab┬а+ 4c2 тИТ 3a2 тИТ 2b2
= тИТ 7ac + 6bc тИТ ab┬а+ 4c2 тИТ 3a2 тИТ 2b2
=┬а тИТ 3a2 тИТ 2b2 + 4c2 тИТ ab + 6bc тИТ 7ac
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.4
Exercise 9.4
1. Multiply the binomials.
(i) (2x + 5) and (4x тИТ 3) | (ii) (y тИТ 8) and (3y тИТ 4) |
(iii) (2.5l тИТ 0.5m) and (2.5l + 0.5m)┬а | (iv) (a + 3b) and (x + 5) |
(v) (2pq + 3q2) and (3pq тИТ 2q2) | (vi) \(\displaystyle \left( {\frac{3}{4}{{a}^{2}}+3{{b}^{2}}} \right)\) and \(\displaystyle \left( {{{a}^{2}}-\frac{2}{3}{{b}^{2}}} \right)\) |
Solutions :
(i) (2x + 5) and (4x тИТ 3)
= 2x (4x тИТ 3) + 5 (4x тИТ 3)
= 2x ├Ч 4x тИТ 2x ├Ч 3 + 5 ├Ч 4x тИТ 5 ├Ч 3
= 8x2 тИТ 6x + 20x тИТ 15
= 8x2 + 14x тИТ 15
(ii) (y – 8) and (3y тИТ 4)
= y (3y тИТ 4) тИТ 8 (3y тИТ 4)
= y ├Ч 3y тИТ y ├Ч 4 тИТ 8 ├Ч 3y тИТ 8 ├Ч (тИТ 4)
= 3y2 тИТ 4y тИТ 24y + 32
= 3y2 тИТ 28y + 32
(iii) (2.5l тИТ 0.5m) and (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) тИТ 0.5m (2.5l + 0.5m)
= 2.5l ├Ч 2.5l + 2.5l ├Ч 0.5m – 0.5m ├Ч 2.5l тИТ 0.5m ├Ч 0.5m
= 2.25l2 + 1.25lm тИТ 1.25lm тИТ 0.25m2
= 2.25l2 тИТ 0.25m2
(iv) (a + 3b) and (x + 5)
= a (x + 5) + 3b (x + 5)
= a ├Ч x + a ├Ч 5 + 3b ├Ч x + 3b ├Ч 5
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2) and (3pq тАУ 2q2)
= 2pq (3pq тИТ 2q2) + 3q2 (3pq тИТ 2q2)
= 2pq ├Ч 3pq тИТ 2q2 ├Ч 2pq + 3q2 ├Ч 3pq тИТ 3q2 ├Ч 2q2
= 2p2q2 тИТ 4pq3 + 9pq3 тИТ 6q4
= 6p2q2 + 5pq3 тИТ 6q4
2. Find the product.
(i) (5 тИТ 2x) (3 + x) | (ii) (x + 7y) (7x тИТ y) |
(iii) (a2 + b) (a + b2) | (iv) (p2 тАУ q2) (2p + q) |
Solutions :
(i) (5 тИТ 2x) (3 + x)
= 5 ├Ч (3 + x) тИТ 2x ├Ч (3 + x)
= 15 + 5x тИТ 6x тИТ 2x2
= 15 тИТ x тИТ 2x2
(ii) (x + 7y) (7x тИТ y)
= x ├Ч (7x тИТ y) + 7y ├Ч (7x тИТ y)
= 7x2 тИТ xy┬а+ 49xy тИТ 7y2
= 7x2 + 48xy тИТ 7y2
(iii) (a2 + b) (a + b2)
= a2 ├Ч (a + b2) + b ├Ч (a + b2)
= a3 + a2b2 + ab + b2
(iv) (p2 тИТ q2) (2p + q)
= p2 ├Ч (2p + q) тИТ q2 ├Ч (2p + q)
= 2p3 + p2q тИТ 2pq2 тИТ q3
3. Simplify.
(i) (x2 тИТ 5) (x + 5) + 25 | (ii) (a2 + 5) (b3 + 3) + 5 |
(iii) (t + s2) (t2 тИТ s) | ┬а |
(iv) (a + b) (c тИТ d) + (a тИТ b) (c + d) + 2 (ac + bd)┬а | |
(v) (x + y) (2x + y) + (x + 2y) (x тИТ y) | (vi) (x + y) (x2 тИТ xy + y2) |
(vii) (1.5x тИТ 4y) (1.5x + 4y + 3) тИТ 4.5x + 12y | ┬а |
(viii) (a + b + c) (a + b тИТ c) |
Solutions :
(i) (x2 тАУ 5) (x + 5) + 25
= x2 ├Ч (x + 5) тИТ 5 ├Ч (x + 5) + 25
= x3 + 5x2 тИТ 5x тИТ 25 + 25
= x3 + 5x2 тИТ 5x
(ii) (a2 + 5) (b3 + 3) + 5
= a2 ├Ч (b3 + 3) + 5 ├Ч (b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20
(iii) (t + s2) (t2 тИТ s)
= t ├Ч (t2 тИТ s) + s2 ├Ч (t2 тИТ s)
= t3 тИТ st + s2t2 тИТ s3
(iv) (a + b) (c тИТ d) + (a тИТ b) (c + d) + 2 (ac + bd)
= a ├Ч (c тИТ d) + b ├Ч (c тИТ d) + a ├Ч (c + d) тИТ b ├Ч (c + d) + 2ac + 2bd
= ac тИТ ad + bc тИТ bd + ac + ad тИТ bc тИТ bd + 2ac + 2bd
= ac + ac + 2ac + bc тИТ bc тИТ bd тИТ bd + 2bd тИТ ad + ad┬а
= 2ac + 2ac тИТ 2bd + 2bd
= 4ac
(v) (x + y) (2x + y) + (x + 2y) (x тАУ y)
= x ├Ч (2x + y) + y ├Ч (2x + y) + x ├Ч (x тИТ y) + 2y ├Ч (x тИТ y)
= 2x2 + xy + 2xy + y2 + x2 тИТ xy + 2xy тИТ 2y2
= 2x2 + x2 + xy + 2xy тИТ xy + 2xy + y2 тИТ 2y2
= 3x2 + 5xy тИТ xy тИТ y2
= 3x2 + 4xy тИТ y2
(vii) (1.5x тИТ 4y) (1.5x + 4y + 3) тИТ 4.5x + 12y
= 1.5x ├Ч (1.5x + 4y + 3) тИТ 4y ├Ч (1.5x + 4y + 3) тИТ 4.5x + 12y
= 2.25x2 + 6.0xy + 4.5x – 6.0xy тИТ 16y2 тИТ 12y тИТ 4.5x тИТ 12y
= 2.25x2 + 6.0xy тИТ 6.0xy + 4.5x тИТ 4.5x тИТ 16y2 тИТ 12y тИТ 12y
= 2.25x2 тИТ 16y2
(viii) (a + b + c) (a + b тИТ c)
= a ├Ч (a + b тИТ c) + b ├Ч (a + b тАУ c)┬а + c ├Ч (a + b тИТ c)
= a2 + ab тИТ ac + ab + b2 тИТ bc + ac + bc тИТ c2
= a2 + ab + ab тИТ ac + ac + b2 тИТ bc + bc тИТ c2
= a2 + 2ab + b2 тИТ c2
= a2 + b2 тИТ c2 + 2ab
NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities
class 8th maths
Ex 9.5
Exercise 9.5
1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3) | (ii) (2y + 5 ) (2y + 5) |
(iii) (2a тАУ 7) (2a тАУ 7) | (iv)┬а |
(v) (1.1m тАУ 0.4) (1.1m + 0.4)┬а | (vi) (a2 + b2) (тИТ a2 + b2) |
(vii) (6x тАУ 7) (6x + 7)┬а | (viii) (тАУ a + c) (тАУ a + c) |
(ix)┬а | (x) (7a тАУ 9b) (7a тАУ 9b) |
Solutions :
(i) (x + 3) (x + 3)
= (x + 3)2
[Use the Identity (a + b)2 = a2 + 2ab + b2 ]
= x2 + 2 ├Ч x ├Ч 3 + 32
= x2 + 6x + 9
(ii) (2y + 5 ) (2y + 5)
= (2y + 5)2
[Use the Identity (a + b)2 = a2 + 2ab + b2 ]
= (2y)2 + 2 ├Ч 2y ├Ч 5 + 52
= 4y2 + 20y + 25
(iii) (2a тАУ 7) (2a тАУ 7)
= (2a тАУ 7)2
[Use the Identity (a тИТ b)2 = a2 тИТ 2ab + b2 ]
= (2a)2 тИТ 2 ├Ч 2a ├Ч 7 + 72
= 4a2 тИТ 28a + 49
(v) (1.1m тАУ 0.4) (1.1m + 0.4)
= (1.1m)2 тИТ (0.4)2
[Use the Identity (a)2 тИТ (b)2 = (a + b) (a тИТ b) ]
= (1.1m)2 тИТ (0.4)2
= 1.21m2 тИТ 0.16
(vi) (a2 + b2) (тИТ a2 + b2)
= (b2 + a2) (b2 тИТ a2)
= (b2)2 тИТ (a2)2
[Use the Identity (a)2 тИТ (b)2 = (a + b) (a тИТ b) ]
= b4 тИТ a4
(vii) (6x тАУ 7) (6x + 7)
= (6x)2 тИТ (7)2
[Use the Identity (a)2 тИТ (b)2 = (a + b) (a тИТ b) ]
= 36x2 тИТ 49
(viii) (тАУ a + c) (тАУ a + c)
= (c тИТ a) (c тИТ a)
= (c тИТ a)2
[Use the Identity (a тИТ b)2 = a2 тИТ 2ab + b2 ]
= c2 тИТ 2 ├Ч c ├Ч a + a2
= c2 тИТ 2ca + a2
(x) (7a тИТ 9b) (7a тАУ 9b)
= (7a тАУ 9b)2
[Use the Identity (a тИТ b)2 = a2 тИТ 2ab + b2 ]
= (7a)2 тИТ 2 ├Ч 7a ├Ч 9b + (9b)2
= 49a2 тИТ 126ab┬а+ 81b2
2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7) | (ii) (4x + 5) (4x + 1)┬а |
(iii) (4x тАУ 5) (4x тИТ 1) | (iv) (4x + 5) (4x тАУ 1) |
(v) (2x + 5y) (2x + 3y)┬а | (vi) (2a2 + 9) (2a2 + 5)┬а |
(vii) (xyz тАУ 4) (xyz тАУ 2) | ┬а |
Solutions :
(i) (x + 3) (x + 7)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
(x + 3) (x + 7) = x2 + (3 + 7) x + 3 ├Ч 7
= x2 + 10x + 21
(ii) (4x + 5) (4x + 1)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
(4x + 5) (4x + 1) = (4x)2 + (5 + 1) 4x + 5 ├Ч 1
= 16x2 + 6 ├Ч 4x + 5
= 16x2 + 24x + 5
(iii) (4x тАУ 5) (4x тАУ 1) = 16x тИТ 24x + 5
by identity (x тИТ a) (x тИТ b) = x2 тИТ (a + b) x + ab┬а
(4x тАУ 5) (4x тАУ 1) = (4x)2 тИТ (5 + 1) 4x + 5 ├Ч 1
= 16x2 тИТ 6 ├Ч 4x + 5
= 16x2 тИТ 24x + 5
(iv) (4x + 5) (4x тИТ 1) = 16x2┬а тИТ 4x + 20x тИТ 5
by identity (x + a) (x тИТ b) = x2 + (a тИТ b) x тИТ ab┬а
(4x + 5) (4x тАУ 1) = (4x)2 + (5 тИТ 1) 4x тИТ 5 ├Ч 1
= 16x2 + 4 ├Ч 4x тИТ 5
= 16x2 + 16x тИТ 5┬а
(v) (2x + 5y) (2x + 3y)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
(2x + 5y) (2x + 3y) = (2x)2 + (5y + 3y) 2x + 5y ├Ч 3y
= 4x2 + 8y ├Ч 2x + 15y2
= 4x2 + 16xy + 15y2
(vi) (2a2 + 9) (2a2 + 5)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
(2a2 + 9) (2a2 + 5) = (2a2)2 + (9 + 5) 2a2 + 9 ├Ч 5
= 4a4 + 14 ├Ч 2a2 + 45
= 4a4 + 28a2 + 45
(vii) (xyz тАУ 4) (xyz┬атАУ 2)
by identity (x тИТ a) (x тИТ b) = x2 тИТ (a + b) x + ab┬а
(xyz тАУ 4) (xyz тАУ 2) = (xyz)2 тИТ (4 + 2) xyz┬а+ 4 ├Ч 2
= x2y2z2 тИТ 6xyz + 8
3. Find the following squares by using the identities.
(i) (b тИТ 7)2 | (ii) (xy + 3z)2 | (iii) (6x2 тИТ 5y)2 |
(iv)┬а | (v) (0.4p тИТ 0.5q)2 | (iv) (2xy + 5y)2 |
Solutions :
(i) (b тИТ 7)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(b тИТ 7)2 = (b)2 тИТ 2 ├Ч b ├Ч 7 + (7)2
= b2 тИТ 14b + 49
(ii) (xy + 3z)2
by identity (a + b)2 = a2 + 2ab + b2┬а
(xy + 3z)2 = (xy)2 + 2 ├Ч xy ├Ч 3z + (3z)2
= x2y2 + 6xyz┬а+ 9z2
(iii) (6x2 тИТ 5y)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2
(6x2 тИТ 5y)2 = (6x2)2 тИТ 2 ├Ч 6x2 ├Ч 5y + (5y)2
= 36x4 – 60x2y + 25y2
(v) (0.4p тИТ 0.5q)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(0.4p тИТ 0.5q)2 = (0.4p)2 тИТ 2 ├Ч 0.4p ├Ч 0.5q + (0.5q)2
= 0.16p2 тИТ 0.4pq┬а+ 0.25q2
(iv) (2xy + 5y)2
by identity (a + b)2 = a2 + 2ab + b2┬а
(2xy + 5y)2 = (2xy)2 + 2 ├Ч 2xy ├Ч 5y + (5y)2
= 4x2y2 + 20xy + 25y2
4. Simplify.
(i) (a2 тИТ b2)2 | (ii) (2x + 5)2 тИТ (2x тИТ 5)2 |
(iii) (7m – 8n)2 + (7m + 8n)2 | (iv) (4m + 5n)2 + (5m + 4n)2 |
(v) (2.5p тИТ 1.5q)2 тИТ (1.5p тИТ 2.5q)2 | |
(vi) (ab + bc)2 тИТ 2ab2c | (vii) (m2 тИТ n2m)2 + 2m3n2 |
Solutions :
(i) (a2 тИТ b2)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(a2 тИТ b2)2 = (a2)2 тИТ 2 ├Ч a2 ├Ч b2 + (b2)2
= a4 тИТ 2a2b2 + b4
(ii) (2x + 5)2 тИТ (2x тИТ 5)2
[by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b) ]
(2x + 5)2 тИТ (2x тИТ 5)2 = (2x + 5 + 2x тИТ 5) {2x + 5 тИТ (2x тИТ 5)}
= 4x (2x + 5 тИТ 2x + 5)
= 4x ├Ч 10
= 40x
(iii) (7m – 8n)2 + (7m + 8n)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2 and (a + b)2 = a2 + 2ab + b2┬а
= [(7m)2 тИТ 2 ├Ч 7m ├Ч 8n + (8n)2] + [(7m)2 + 2 ├Ч 7m ├Ч 8n + (8n)2]
= 49m2 тИТ 112mn + 64n2 + 49m2 + 112mn + 64n2
= 49m2 + 64n2 + 49m2 + 64n2
= 98m2 + 128n2
(iv) (4m + 5n)2 + (5m + 4n)2
by identity (a + b)2 = a2 + 2ab + b2┬а
= [(4m)2 + 2 ├Ч 4m ├Ч 5n + (5n)2] + [(5m)2 + 2 ├Ч 5m ├Ч 4n + (4n)2]
= 16m2 + 40mn┬а+ 25n2 + 25m2 + 20mn┬а+ 16n2
= 16m2 + 25m2 + 80mn + 25n2 + 16n2
= 41m2 + 80mn┬а+ 16n2
(v) (2.5p тИТ 1.5q)2 тИТ (1.5p тИТ 2.5q)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
= [(2.5p)2 тИТ 2 ├Ч 2.5p ├Ч 1.5q + (1.5q)2] тИТ [(1.5p)2 – 2 ├Ч 1.5p ├Ч 2.5q + (2.5q)2]
= 6.25p2 тИТ 7.8pq┬а+ 2.25q2 тИТ [2.25p2 тИТ 7.8pq┬а+ 6.25q2]
= 6.25p2 тИТ 7.8pq + 2.25q2 тИТ 2.25p2 + 7.8pq┬атИТ 6.25q2
= 6.25p2 тИТ 2.25p2 + 2.25q2 тИТ 6.25q2
= 4p2 тИТ 4q`2
(vi) (ab + bc)2 тИТ 2ab2c
by identity (a + b)2 = a2 + 2ab + b2
= (ab)2 + 2 ├Ч ab ├Ч bc + (bc)2 тИТ 2ab2c
= a2b2 + 2ab2c┬а+ b2c2 тИТ 2ab2c
= a2b2 + b2c2 + 2ab2c┬атИТ 2ab2c
= a2b2┬а+ b2c2
(vii) (m2 тИТ n2m)2 + 2m3n2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2 ┬а
= (m2)2 тИТ 2 ├Ч m2 ├Ч n2m┬а+ (n2m)2 + 2m3n2
= m4 тИТ 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2
5. Show that.
(i) (3x + 7)2 тИТ 84x = (3x тИТ 7)2 | (ii) (9p тИТ 5q)2 + 180pq┬а= (9p + 5q)2 |
(iii) | ┬а |
(iv) (4pq┬а+ 3q)2 тИТ (4pq┬атИТ 3q)2 = 48pq2 | ┬а |
(v) (a тИТ b) (a + b) + (b тИТ c) (b + c) + (c тИТ a) (c + a) = 0 | ┬а |
Solutions :
(i) (3x + 7)2 тИТ 84x = (3x тИТ 7)2
L.H.S.
(3x + 7)2 тИТ 84x
by identity (a + b)2 = a2 + 2ab + b2 ┬а
= (3x)2 + 2 ├Ч 3x ├Ч 7 + 72 тИТ 84x
= 9x2 + 42x + 49 тИТ 84x
= 9x2 тИТ 42x + 49
(because a2 тИТ 2ab + b2 = (a тИТ b)2┬а
therefor 9x2 тИТ 42x + 49 = (3x)2 тИТ 2 ├Ч 3x ├Ч 7 + (7)2
= (3x тИТ 7)2
L.H.S. = R.H.S.
(ii) (9p тИТ 5q)2 + 180pq┬а= (9p + 5q)2
L.H.S.
(9p тИТ 5q)2 + 180pq
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
= (9p)2 тИТ 2 ├Ч 9p ├Ч 5q + (5q)2 + 180pq
= 81p2 тИТ 90pq┬а+ 25q2 + 180pq
= 81p2 + 90pq + 25q2
(because a2 + 2ab + b2 = (a + b)2┬а
therefor 81p2 + 90pq + 25q2 = (9p)2 тИТ 2 ├Ч 9p ├Ч 5q + (5q)2
= (9p + 5q)2
L.H.S. = R.H.S.
(iv) (4pq + 3q)2 тИТ (4pq тИТ 3q)2 = 48pq2
L.H.S.
(4pq┬а+ 3q)2 тИТ (4pq┬атИТ 3q)2
by identity (a + b)2 = a2 + 2ab + b2 and┬аidentity (a тИТ b)2 = a2 тИТ 2ab + b2 рд╕реЗ –┬а
= (4pq)2 + 2 ├Ч 4pq ├Ч 3q + (3q)2 тИТ [{4pq)2 тИТ 2 ├Ч 4pq ├Ч 3q + (3q)2]
= 16p2q2 + 24pq2 + 9q2 тИТ [16p2q2 тИТ 24pq2 + 9q2]
= 16p2q2┬а+ 24pq2 + 9q2 тИТ 16p2q2 + 24pq2 тИТ 9q2
= 16p2q2 тИТ 16p2q2┬а+ 9q2 тИТ 9q2 + 24pq2 + 24pq2
= 48pq2
L.H.S. = R.H.S.
(v) (a тИТ b) (a + b) + (b тИТ c) (b + c) + (c тИТ a) (c + a) = 0
L.H.S.
(a тИТ b) (a + b) + (b тИТ c) (b + c) + (c тИТ a) (c + a)
(by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (a2 тИТ b2) + (b2 тИТ c2) + (c2 тИТ a2)
= a2 тИТ b2 + b2 тИТ c2 + c2 тИТ a2
= a2 тИТ a2 тИТ b2 + b2 тИТ c2 + c2
= 0
L.H.S. = R.H.S.
6. Using identities, evaluate.
(i) 712 | (ii) 992 | (iii) 1022 | (iv) 9982 |
(v) 5.22 | (vi) 297 ├Ч 303 | (vii) 78 ├Ч 82 | (viii) 8.92 |
(ix) 10.5 ├Ч 9.5 | ┬а | ┬а | ┬а |
Solutions :
(i) 712
= (70 + 1)2
by identity (a + b)2 = a2 + 2ab + b2 ┬а
(70 + 1)2 = (70)2 + 2 ├Ч 70 ├Ч 1 + (1)2
= 4900 + 140 + 1
= 5041
(ii) 992
= (100 тИТ 1)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(100 тИТ 1)2 = (100)2 тИТ 2 ├Ч 100 ├Ч 1 + (1)2
= 10000 тИТ 200 + 1
= 10001 тИТ 200
= 9801
(iii) 1022
= (100 + 2)2
by identity (a + b)2 = a2 + 2ab + b2┬а
(100 + 2)2 = (100)2 + 2 ├Ч 100 ├Ч 2 + (2)2
= 10000 + 400 + 4
= 10404
(iv) 9982
= (1000 тИТ 2)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(1000 тИТ 2)2 = (1000)2 тИТ 2 ├Ч (1000) ├Ч 2 + (2)2
= 1000000 тИТ 4000 + 4
= 1000004 тИТ 4000
= 996004
(vi) 297 ├Ч 303
= (300 тИТ 3) (300 + 3)
by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (300)2 тИТ (3)2
= 90000 тИТ 9
= 89991
(vii) 78 ├Ч 82
= (80 тИТ 2) (80 + 2)
by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (80)2 тИТ (2)2
= 6400 тИТ 4
= 6396
(viii) 8.92
= (10.0 тИТ 1.1)2
by identity (a тИТ b)2 = a2 тИТ 2ab + b2┬а
(10.0 тИТ 1.1)2 = (10)2 тИТ 2 ├Ч 10 ├Ч 1.1 + (1.1)2
= 100 тИТ 22 + 1.21
= 100.21 тИТ 22
= 79.21
(ix) 10.5 ├Ч 9.5
= (10.0 + 0.5) (10.0 – 0.5)
by identity (a)2 тИТ (b)2 = (a + b) (a тИТ b)┬а
= (10)2 тИТ (0.5)2
= 100 тИТ 0.25
= 99.75
7. Using a2 тИТ b2 = (a + b) (a тИТ b), find┬а
(i) 512 тИТ 492 | (ii) (1.02)2 тИТ (0.98)2 |
(iii) 1532 тИТ 1472 | (iv) 12.12 тИТ 7.92 |
Solutions :
(i) 512 тИТ 492
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
512 тИТ 492 = (51 + 49) (51 тИТ 49)
= 100 ├Ч 2
= 200
(ii) (1.02)2 тИТ (0.98)2
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
(1.02)2 тИТ (0.98)2 = (1.02 + 0.98) (1.02 тИТ 0.98)
= 2 ├Ч┬а0.04
= 0.08
(iii) 1532 тИТ 1472
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
1532 тИТ 1472 = (153 + 147) (153 тИТ 147)
= 300 ├Ч 6
= 1800
(iv) 12.12 тИТ 7.92
by identity a2 тИТ b2 = (a + b) (a тИТ b)┬а
12.12 тИТ 7.92 = (12.1 + 7.9) (12.1 тИТ 7.9)
= 20 ├Ч 4.2
= 84
8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 ├Ч 104 | (ii) 5.1 ├Ч 5.2 |
(iii) 103 ├Ч 98 | (iv) 9.7 ├Ч 9.8 |
Solutions :
(i) 103 ├Ч 104
= (100 + 3) (100 + 4)
by identityрд╕рд░реНрд╡рд╕рдорд┐рдХрд╛ (x + a) (x + b) = x2 + (a + b) x + ab┬а
= (100)2 + (3 + 4) 100 + 3 ├Ч 4
= 10000 + 700 + 12
= 10712
(ii) 5.1 ├Ч 5.2
= (5.0 + 0.1) (5.0 + 0.2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
= (5.0)2 + (0.1 + 0.2) 5.0 + 0.1 ├Ч 0.2
= 25 + 1.5 + 0.02
= 26.52
(iii) 103 ├Ч 98
= (100 + 3) (100 – 2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab┬а
= (100)2 + {3 + (тИТ 2} 100 + 3 ├Ч (тИТ 2)
= 10000 + 1 ├Ч 100 + (тИТ 6)
= 10000 + 100 тИТ 6
= 10100 тИТ 6
= 10094
(iv) 9.7 ├Ч 9.8
= (9.0 + 0.7) (9.0 + 0.8)
by identity┬а(x + a) (x + b) = x2 + (a + b) x + ab┬а
= (9)2 + (0.7 + 0.8) 9 + 0.7 ├Ч 0.8
= 81 + 1.5 ├Ч 9 + 0.56
= 81 + 13.5 + 0.56
= 95.06