NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities | Class 8th Maths
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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 − 3zy | (ii) 1 + x + x2 | (iii) 4x2y2 − 4x2y2z2 + z2 |
(iv) 3 − pq + qr − rp | (v) \(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\) | (vi) 0.3a − 0.6ab + 0.5b |
Solutions :
Terms | Coefficients | ||
(i) | 5xyz2 − 3zy | There are two terms in this expressions. 5xyz2 and 3zy |
5 and − 3 |
(ii) | 1 + x + x2 | There are two terms in this expressions. 1, x and x2 |
1, 1 and 1 |
(iii) | 4x2y2 − 4x2y2z2 + z2 | There are three terms in this expressions. 4x2y2 , − 4x2y2z2 and z2 |
4, − 4 and 1 |
(vi) | 3 − pq + qr – rp | There are four terms in this expressions. 3, − pq, qr and − rp |
3, − 1, 1 and − 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 − xy |
\(\displaystyle \frac{1}{2}\) , \(\displaystyle \frac{1}{2}\) and − 1 |
(vi) | 0.3a − 0.6ab + 0.5b | There are three terms in this expressions. 0.3, − 0.6ab and 0.5b |
0.3, − 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 − 3y2, 2y − 3y2 + 4y3, 5x − 4y + 3xy, 4z − 15z2, ab + bc + cd + da, pqr, p2q, 2p + 2q
Solutions :
Monomials :- 1000, pqr
Binomials :- x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q
Trinomials :- 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 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 − bc, bc − ca, ca − ab | (ii) a − b + ab, b − c + bc, c − a + ac |
(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2 | (iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl |
Solutions :
(i) ab − bc, bc − ca, ca − ab
on writing similar terms together –
= ab − bc + bc − ca + ca − ab
= ab − ab + bc − bc + ca − ca
= 0
(ii) a − b + ab, b − c + bc, c − a + ac
= a − b + ab + b − c + bc + c − a + ac
on writing similar terms together –
= a − a + b − b + c − c + ab + bc + ac
by adding and substracting same terms –
= ab + bc + ac
(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2
= 2p2q2 − 3pq + 4 + 5 + 7pq − 3p2q2
on writing similar terms together –
= 2p2q2 − 3p2q2 + 7pq − 3pq + 4 + 5
= − p2q2 + 4pq + 9
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
= l2 + m2 + m2 + n2 + n2 + l2 + 2lm + 2mn + 2nl
on writing similar terms together –
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
= 2(l2 + m2 + n2 + lm + mn + nl)
4. (a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11p + 5pq − 2pq2 + 5p2q
Solutions :
(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
= 12a − 9ab + 5b − 3 − (4a − 7ab + 3b + 12)
remove the bracket –
= 12a − 9ab + 5b − 3 − 4a + 7ab − 3b − 12 (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= 12a − 4a + 5b − 3b + 7ab − 9ab − 4 − 12
= 8a + 2b − 2ab − 16
(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
= 5xy − 2yz − 2zx + 10xyz − (3xy + 5yz − 7zx)
remove the bracket –
= 5xy − 2yz − 2zx + 10xyz − 3xy − 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 − 3xy − 2yz − 5yz + 7zx − 2zx + 10xyz
= 2xy − 7yz + 5zx + 10xyz
(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11p + 5pq − 2pq2 + 5p2q
= 18 − 3p − 11p + 5pq − 2pq2 + 5p2q − (4p2q − 3pq + 5pq2 − 8p + 7q − 10)
remove the bracket –
= 18 − 3p − 11q + 5pq − 2pq2 + 5p2q − 4p2q + 3pq − 5pq2 + 8p − 7q + 10) (If there is a negative sign outside the bracket, the symbols of all the sign inside will change.)
on writing similar terms together –
= − 3p + 8p − 11q − 7q + 5pq + 3pq − 2pq2 − 5pq2 + 5p2q − 4p2q
= 5p − 18q + 8pq − 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) − 4p, 7p | (iii) − 4p, 7pq | (iv) 4p3, − 3p |
(v) 4p, 0 |
Solutions :
(i) 4, 7p
= 4 × 7p
= 28p
(ii) − 4p, 7p
= − 4p × 7p
= − 28p2
(iii) − 4p, 7pq
= − 4p × 7pq
= − 28p2q
(iv) 4p3, − 3p
= 4p3 × − 3p
= − 12p4
(v) 4p, 0
= 4p × 0
= 0
2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
Solutions :
(p, q)
length = p
breadth = q
area of rectangle = length × breadth
= p × q
= pq unit square
(10m, 5n)
length = 10m
breadth = 5n
area of rectangle = length × breadth
= 10m × 5n
= 50mn unit square
(20x2, 5y2)
length = 20x2
breadth = 5y2
area of rectangle = length × breadth
= 20x2 × 5y2
= 100x2y2 unit square
(4x, 3x2)
length = 4x
breadth = 3x2
area of rectangle = length × breadth
= 4x × 3x2
= 12x3 unit square
(3mn, 4np)
length = 3mn
breadth = 4np
area of rectangle = length × breadth
= 3mn × 4np
= 12mn2p unit square
3. Complete the table of products.
First monomial → Second monomial ↓ |
2x | − 5y | 3x2 | − 4xy | 7x2y | − 9x2y2 |
2x | 4x2 | |||||
− 5y | − 15x2y | |||||
3x2 | ||||||
− 4xy | ||||||
7x2y | ||||||
− 9x2y2 |
Solutions :
First monomial → Second monomial ↓ |
2x | − 5y | 3x2 | − 4xy | 7x2y | − 9x2y2 |
2x | 4x2 | − 10xy | 6×3 | − 8x2y | 14x3y | − 18x3y2 |
− 5y | − 10xy | 25y2 | − 15x2y | 20xy2 | − 35x2y2 | 45x2y3 |
3x2 | 6x3 | − 15x2y | 9x4 | − 12x3y | 21x4y | − 27x4y2 |
− 4xy | − 8x2y | 20xy2 | − 12x3y | 16x2y2 | − 28x3y2 | 36x3y3 |
7x2y | 14x3y | − 35x2y2 | 21x4y | − 28x3y2 | 49x4y2 | − 63x4y3 |
− 9x2y2 | − 18x3y2 | 45x2y2 | − 27x4y2 | 36x3y3 | − 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 :
(i) 5a, 3a2, 7a4
volume of rectangular box = length × breadth × height
= 5a × 3a2 × 7a4
= 5 × 3 × 7 × a × a2 × a4
= 105a7
(ii) 2p, 4q, 8r
volume of rectangular box = length × breadth × height
= 2p × 4q × 8r
= 2 × 4 × 8 × p × q × r
= 64pqr
(iii) xy, 2x2y, 2xy2
volume of rectangular box = length × breadth × height
= xy × 2x2y × 2xy2
= 1 × 2 × 2 × xy × x2y × xy2
= 4x4y4
(iv) a, 2b, 3c
volume of rectangular box = length × breadth × height
= a × 2b × 3c
= 1 × 2 × 3 × a × b × c
= 6abc
5. Obtain the product of
(i) xy, yz, zx | (ii) a, − a2, a3 | (iii) 2, 4y, 8y2, 16y3 |
(iv) a, 2b, 3c, 6abc | (v) m, − mn, mnp |
Solutions :
(i) xy, yz, zx
= xy × yz × zx
= x2y2z2
(ii) a, − a2, a3
= a × (− a2) × a3
= − a6
(iii) 2, 4y, 8y2, 16y3
= 2 × 4y × 8y2 × 16y3
= 2 × 4 × 8 × 16 × y × y2 × y3
= 1024y3
(iv) a, 2b, 3c, 6abc
= 2 × 3 × a × b × c × abc
= 6a2b2c2
(v) m, − mn, mnp
= m × (− mn) × mnp
= − 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 − b | (iii) a + b, 7a2b2 | (iv) a2 − 9, 4a |
(v) pq + qr + rp, 0 |
Solutions :
(i) 4p, q + r
= 4p × (q + r)
= 4p × q + 4p × r
= 4pq + 4pr
2. Complete the table.
First expression | Second expression | Product | |
(i) | a | b + c + d | ____ |
(ii) | x + y − 5 | 5xy | ____ |
(iii) | p | 6p2 − 7p + 5 | ____ |
(iv) | 4p2q2 | p2 − q2 | ____ |
(v) | a + b + c | abc | ____ |
Solutions :
First expression | Second expression | Product | |
(i) | a | b + c + d | ab + ac + ad |
(ii) | x + y − 5 | 5xy | 5x2y + 5xy2 − 25xy |
(iii) | p | 6p2 − 7p + 5 | 6p3 − 7p2 + 5p |
(iv) | 4p2q2 | p2 − q2 | 4p4q2 − 4p2q4 |
(v) | a + b + c | abc | a2bc + ab2c + abc2 |
(i) a × (b + c + d)
= ab + ac + ad
(ii) (x + y − 5) × 5xy
= 5x2y + 5xy2 − 25xy
(iii) p × (6p2 − 7p + 5)
= 6p3 − 7p2 + 5p
(iv) 4p2q2 × (p2 − q2)
= 4p4q2 − 4p2q4
(v) (a + b + c) × abc
= a2bc + ab2c + abc2
3. Find the product.
(i) (a2) × (2a22) × (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 × x2 × x3 × x4 |
Solutions :
(i) (a2) × (2a22) × (4a26)
= 2 × 4 × a2 × a22 × 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 × x2 × x3 × 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 – 5) + 3 and find its values for (i) x = 3 (ii) \(\displaystyle x=\frac{1}{2}\)
(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1
(iii) a = – 1.
Solutions :
(a) (i) 3x (4x − 5) + 3
= 3x × 4x − 3x × 5 + 3
= 12x2 − 15x + 3 ………………………….. (i)
putting x = 3 equ. in (i)
= 12 × (3)2 − 9 × 5 + 3
= 12 × 9 − 45 + 3
= 108 + 3 − 45
= 111 − 45
= 66
(ii) Putting \(\displaystyle x=\frac{1}{2}\) in equ (i) – 5. (a) Add: p (p − q), q (q − r) and r (r − p)
\(\displaystyle \begin{array}{l}=12\times {{\left( {\frac{1}{2}} \right)}^{2}}-15\times \frac{1}{2}+3\\=12\times \frac{1}{4}-\frac{{15}}{2}+3\\=3-\frac{{15}}{2}+3\\=\frac{{6-15+6}}{2}\\=\frac{{12-15}}{2}\\=\frac{{-3}}{2}\end{array}\)
(b) Add: 2x (z − x − y) and 2y (z − y − x)
(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l )
(d) Subtract: 3a (a + b + c ) − 2b (a − b + c) from 4c (− a + b + c )
Solutions :
(a) Add: p (p − q), q (q − r) and r (r − p)
= p (p − q) + q (q − r) + r (r − p)
= p2 − pq + q2 − qr + r2 − rp
= p2 + q2 + r2 − pq − qr − rp
(b) Add: 2x (z − x − y) and 2y (z − y − x)
= 2x (z − x − y) + 2y (z − y − x)
= 2xz − 2x2 − 2xy + 2yz − 2y2 − 2xy
= − 2x2 − 2y2 − 2xy − 2xy + 2yz + 2xz
= − 2x2 − 2y2 − 4xy + 2yz + 2xz
(c) Subtract: 3l (l − 4m + 5n) from 4l (10n − 3m + 2l )
= 4l (10n − 3m + 2l) − 3l (l − 4m + 5n)
= 4l × 10n − 4l × 3m + 4l + 2l − 3l × l − 3l × (− 4m) − 3l × 5n
= 40ln − 12lm + 8l2 − 3l2 + 12lm − 15ln
= 40ln − 15ln − 12lm + 12lm + 8l2 − 3l2
= 25ln − 0lm + 5l2
= 5l2 + 25ln
(d) Subtract: 3a (a + b + c ) − 2b (a − b + c) from 4c (− a + b + c )
= 4c (− a + b + c) − [3a (a + b + c) − 2b (a − b + c)]
= − 4ac + 4bc + 4c2 − [3a2 + 3ab + 3ac − 2ab + 2a2 – 2bc]
= − 4ac + 4bc + 4c2 − 3a2 − 3ab − 3ac + 2ab − 2b2 + 2bc
= − 4ac − 3ac + 4bc + 2bc − 3ab + 2ab + 4c2 − 3a2 − 2b2
= − 7ac + 6bc − ab + 4c2 − 3a2 − 2b2
= − 3a2 − 2b2 + 4c2 − ab + 6bc − 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 − 3) | (ii) (y − 8) and (3y − 4) |
(iii) (2.5l − 0.5m) and (2.5l + 0.5m) | (iv) (a + 3b) and (x + 5) |
(v) (2pq + 3q2) and (3pq − 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 − 3)
= 2x (4x − 3) + 5 (4x − 3)
= 2x × 4x − 2x × 3 + 5 × 4x − 5 × 3
= 8x2 − 6x + 20x − 15
= 8x2 + 14x − 15
(ii) (y – 8) and (3y − 4)
= y (3y − 4) − 8 (3y − 4)
= y × 3y − y × 4 − 8 × 3y − 8 × (− 4)
= 3y2 − 4y − 24y + 32
= 3y2 − 28y + 32
(iii) (2.5l − 0.5m) and (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)
= 2.5l × 2.5l + 2.5l × 0.5m – 0.5m × 2.5l − 0.5m × 0.5m
= 2.25l2 + 1.25lm − 1.25lm − 0.25m2
= 2.25l2 − 0.25m2
(iv) (a + 3b) and (x + 5)
= a (x + 5) + 3b (x + 5)
= a × x + a × 5 + 3b × x + 3b × 5
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2) and (3pq – 2q2)
= 2pq (3pq − 2q2) + 3q2 (3pq − 2q2)
= 2pq × 3pq − 2q2 × 2pq + 3q2 × 3pq − 3q2 × 2q2
= 2p2q2 − 4pq3 + 9pq3 − 6q4
= 6p2q2 + 5pq3 − 6q4
2. Find the product.
(i) (5 − 2x) (3 + x) | (ii) (x + 7y) (7x − y) |
(iii) (a2 + b) (a + b2) | (iv) (p2 – q2) (2p + q) |
Solutions :
(i) (5 − 2x) (3 + x)
= 5 × (3 + x) − 2x × (3 + x)
= 15 + 5x − 6x − 2x2
= 15 − x − 2x2
(ii) (x + 7y) (7x − y)
= x × (7x − y) + 7y × (7x − y)
= 7x2 − xy + 49xy − 7y2
= 7x2 + 48xy − 7y2
(iii) (a2 + b) (a + b2)
= a2 × (a + b2) + b × (a + b2)
= a3 + a2b2 + ab + b2
(iv) (p2 − q2) (2p + q)
= p2 × (2p + q) − q2 × (2p + q)
= 2p3 + p2q − 2pq2 − q3
3. Simplify.
(i) (x2 − 5) (x + 5) + 25 | (ii) (a2 + 5) (b3 + 3) + 5 |
(iii) (t + s2) (t2 − s) | |
(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd) | |
(v) (x + y) (2x + y) + (x + 2y) (x − y) | (vi) (x + y) (x2 − xy + y2) |
(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y | |
(viii) (a + b + c) (a + b − c) |
Solutions :
(i) (x2 – 5) (x + 5) + 25
= x2 × (x + 5) − 5 × (x + 5) + 25
= x3 + 5x2 − 5x − 25 + 25
= x3 + 5x2 − 5x
(ii) (a2 + 5) (b3 + 3) + 5
= a2 × (b3 + 3) + 5 × (b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20
(iii) (t + s2) (t2 − s)
= t × (t2 − s) + s2 × (t2 − s)
= t3 − st + s2t2 − s3
(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd)
= a × (c − d) + b × (c − d) + a × (c + d) − b × (c + d) + 2ac + 2bd
= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd
= ac + ac + 2ac + bc − bc − bd − bd + 2bd − ad + ad
= 2ac + 2ac − 2bd + 2bd
= 4ac
(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x × (2x + y) + y × (2x + y) + x × (x − y) + 2y × (x − y)
= 2x2 + xy + 2xy + y2 + x2 − xy + 2xy − 2y2
= 2x2 + x2 + xy + 2xy − xy + 2xy + y2 − 2y2
= 3x2 + 5xy − xy − y2
= 3x2 + 4xy − y2
(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y
= 1.5x × (1.5x + 4y + 3) − 4y × (1.5x + 4y + 3) − 4.5x + 12y
= 2.25x2 + 6.0xy + 4.5x – 6.0xy − 16y2 − 12y − 4.5x − 12y
= 2.25x2 + 6.0xy − 6.0xy + 4.5x − 4.5x − 16y2 − 12y − 12y
= 2.25x2 − 16y2
(viii) (a + b + c) (a + b − c)
= a × (a + b − c) + b × (a + b – c) + c × (a + b − c)
= a2 + ab − ac + ab + b2 − bc + ac + bc − c2
= a2 + ab + ab − ac + ac + b2 − bc + bc − c2
= a2 + 2ab + b2 − c2
= a2 + b2 − 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 – 7) (2a – 7) | (iv) |
(v) (1.1m – 0.4) (1.1m + 0.4) | (vi) (a2 + b2) (− a2 + b2) |
(vii) (6x – 7) (6x + 7) | (viii) (– a + c) (– a + c) |
(ix) | (x) (7a – 9b) (7a – 9b) |
Solutions :
(i) (x + 3) (x + 3)
= (x + 3)2
[Use the Identity (a + b)2 = a2 + 2ab + b2 ]
= x2 + 2 × x × 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 × 2y × 5 + 52
= 4y2 + 20y + 25
(iii) (2a – 7) (2a – 7)
= (2a – 7)2
[Use the Identity (a − b)2 = a2 − 2ab + b2 ]
= (2a)2 − 2 × 2a × 7 + 72
= 4a2 − 28a + 49
(v) (1.1m – 0.4) (1.1m + 0.4)
= (1.1m)2 − (0.4)2
[Use the Identity (a)2 − (b)2 = (a + b) (a − b) ]
= (1.1m)2 − (0.4)2
= 1.21m2 − 0.16
(vi) (a2 + b2) (− a2 + b2)
= (b2 + a2) (b2 − a2)
= (b2)2 − (a2)2
[Use the Identity (a)2 − (b)2 = (a + b) (a − b) ]
= b4 − a4
(vii) (6x – 7) (6x + 7)
= (6x)2 − (7)2
[Use the Identity (a)2 − (b)2 = (a + b) (a − b) ]
= 36x2 − 49
(viii) (– a + c) (– a + c)
= (c − a) (c − a)
= (c − a)2
[Use the Identity (a − b)2 = a2 − 2ab + b2 ]
= c2 − 2 × c × a + a2
= c2 − 2ca + a2
(x) (7a − 9b) (7a – 9b)
= (7a – 9b)2
[Use the Identity (a − b)2 = a2 − 2ab + b2 ]
= (7a)2 − 2 × 7a × 9b + (9b)2
= 49a2 − 126ab + 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) |
(iii) (4x – 5) (4x − 1) | (iv) (4x + 5) (4x – 1) |
(v) (2x + 5y) (2x + 3y) | (vi) (2a2 + 9) (2a2 + 5) |
(vii) (xyz – 4) (xyz – 2) |
Solutions :
(i) (x + 3) (x + 7)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(x + 3) (x + 7) = x2 + (3 + 7) x + 3 × 7
= x2 + 10x + 21
(ii) (4x + 5) (4x + 1)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(4x + 5) (4x + 1) = (4x)2 + (5 + 1) 4x + 5 × 1
= 16x2 + 6 × 4x + 5
= 16x2 + 24x + 5
(iii) (4x – 5) (4x – 1) = 16x − 24x + 5
by identity (x − a) (x − b) = x2 − (a + b) x + ab
(4x – 5) (4x – 1) = (4x)2 − (5 + 1) 4x + 5 × 1
= 16x2 − 6 × 4x + 5
= 16x2 − 24x + 5
(iv) (4x + 5) (4x − 1) = 16x2 − 4x + 20x − 5
by identity (x + a) (x − b) = x2 + (a − b) x − ab
(4x + 5) (4x – 1) = (4x)2 + (5 − 1) 4x − 5 × 1
= 16x2 + 4 × 4x − 5
= 16x2 + 16x − 5
(v) (2x + 5y) (2x + 3y)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(2x + 5y) (2x + 3y) = (2x)2 + (5y + 3y) 2x + 5y × 3y
= 4x2 + 8y × 2x + 15y2
= 4x2 + 16xy + 15y2
(vi) (2a2 + 9) (2a2 + 5)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
(2a2 + 9) (2a2 + 5) = (2a2)2 + (9 + 5) 2a2 + 9 × 5
= 4a4 + 14 × 2a2 + 45
= 4a4 + 28a2 + 45
(vii) (xyz – 4) (xyz – 2)
by identity (x − a) (x − b) = x2 − (a + b) x + ab
(xyz – 4) (xyz – 2) = (xyz)2 − (4 + 2) xyz + 4 × 2
= x2y2z2 − 6xyz + 8
3. Find the following squares by using the identities.
(i) (b − 7)2 | (ii) (xy + 3z)2 | (iii) (6x2 − 5y)2 |
(iv) | (v) (0.4p − 0.5q)2 | (iv) (2xy + 5y)2 |
Solutions :
(i) (b − 7)2
by identity (a − b)2 = a2 − 2ab + b2
(b − 7)2 = (b)2 − 2 × b × 7 + (7)2
= b2 − 14b + 49
(ii) (xy + 3z)2
by identity (a + b)2 = a2 + 2ab + b2
(xy + 3z)2 = (xy)2 + 2 × xy × 3z + (3z)2
= x2y2 + 6xyz + 9z2
(iii) (6x2 − 5y)2
by identity (a − b)2 = a2 − 2ab + b2
(6x2 − 5y)2 = (6x2)2 − 2 × 6x2 × 5y + (5y)2
= 36x4 – 60x2y + 25y2
(v) (0.4p − 0.5q)2
by identity (a − b)2 = a2 − 2ab + b2
(0.4p − 0.5q)2 = (0.4p)2 − 2 × 0.4p × 0.5q + (0.5q)2
= 0.16p2 − 0.4pq + 0.25q2
(iv) (2xy + 5y)2
by identity (a + b)2 = a2 + 2ab + b2
(2xy + 5y)2 = (2xy)2 + 2 × 2xy × 5y + (5y)2
= 4x2y2 + 20xy + 25y2
4. Simplify.
(i) (a2 − b2)2 | (ii) (2x + 5)2 − (2x − 5)2 |
(iii) (7m – 8n)2 + (7m + 8n)2 | (iv) (4m + 5n)2 + (5m + 4n)2 |
(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2 | |
(vi) (ab + bc)2 − 2ab2c | (vii) (m2 − n2m)2 + 2m3n2 |
Solutions :
(i) (a2 − b2)2
by identity (a − b)2 = a2 − 2ab + b2
(a2 − b2)2 = (a2)2 − 2 × a2 × b2 + (b2)2
= a4 − 2a2b2 + b4
(ii) (2x + 5)2 − (2x − 5)2
[by identity (a)2 − (b)2 = (a + b) (a − b) ]
(2x + 5)2 − (2x − 5)2 = (2x + 5 + 2x − 5) {2x + 5 − (2x − 5)}
= 4x (2x + 5 − 2x + 5)
= 4x × 10
= 40x
(iii) (7m – 8n)2 + (7m + 8n)2
by identity (a − b)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2
= [(7m)2 − 2 × 7m × 8n + (8n)2] + [(7m)2 + 2 × 7m × 8n + (8n)2]
= 49m2 − 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
= [(4m)2 + 2 × 4m × 5n + (5n)2] + [(5m)2 + 2 × 5m × 4n + (4n)2]
= 16m2 + 40mn + 25n2 + 25m2 + 20mn + 16n2
= 16m2 + 25m2 + 80mn + 25n2 + 16n2
= 41m2 + 80mn + 16n2
(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
by identity (a − b)2 = a2 − 2ab + b2
= [(2.5p)2 − 2 × 2.5p × 1.5q + (1.5q)2] − [(1.5p)2 – 2 × 1.5p × 2.5q + (2.5q)2]
= 6.25p2 − 7.8pq + 2.25q2 − [2.25p2 − 7.8pq + 6.25q2]
= 6.25p2 − 7.8pq + 2.25q2 − 2.25p2 + 7.8pq − 6.25q2
= 6.25p2 − 2.25p2 + 2.25q2 − 6.25q2
= 4p2 − 4q`2
(vi) (ab + bc)2 − 2ab2c
by identity (a + b)2 = a2 + 2ab + b2
= (ab)2 + 2 × ab × bc + (bc)2 − 2ab2c
= a2b2 + 2ab2c + b2c2 − 2ab2c
= a2b2 + b2c2 + 2ab2c − 2ab2c
= a2b2 + b2c2
(vii) (m2 − n2m)2 + 2m3n2
by identity (a − b)2 = a2 − 2ab + b2
= (m2)2 − 2 × m2 × n2m + (n2m)2 + 2m3n2
= m4 − 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2
5. Show that.
(i) (3x + 7)2 − 84x = (3x − 7)2 | (ii) (9p − 5q)2 + 180pq = (9p + 5q)2 |
(iii) | |
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2 | |
(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0 |
Solutions :
(i) (3x + 7)2 − 84x = (3x − 7)2
L.H.S.
(3x + 7)2 − 84x
by identity (a + b)2 = a2 + 2ab + b2
= (3x)2 + 2 × 3x × 7 + 72 − 84x
= 9x2 + 42x + 49 − 84x
= 9x2 − 42x + 49
(because a2 − 2ab + b2 = (a − b)2
therefor 9x2 − 42x + 49 = (3x)2 − 2 × 3x × 7 + (7)2
= (3x − 7)2
L.H.S. = R.H.S.
(ii) (9p − 5q)2 + 180pq = (9p + 5q)2
L.H.S.
(9p − 5q)2 + 180pq
by identity (a − b)2 = a2 − 2ab + b2
= (9p)2 − 2 × 9p × 5q + (5q)2 + 180pq
= 81p2 − 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
(because a2 + 2ab + b2 = (a + b)2
therefor 81p2 + 90pq + 25q2 = (9p)2 − 2 × 9p × 5q + (5q)2
= (9p + 5q)2
L.H.S. = R.H.S.
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
L.H.S.
(4pq + 3q)2 − (4pq − 3q)2
by identity (a + b)2 = a2 + 2ab + b2 and identity (a − b)2 = a2 − 2ab + b2 से –
= (4pq)2 + 2 × 4pq × 3q + (3q)2 − [{4pq)2 − 2 × 4pq × 3q + (3q)2]
= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 − 16p2q2 + 24pq2 − 9q2
= 16p2q2 − 16p2q2 + 9q2 − 9q2 + 24pq2 + 24pq2
= 48pq2
L.H.S. = R.H.S.
(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0
L.H.S.
(a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)
(by identity (a)2 − (b)2 = (a + b) (a − b)
= (a2 − b2) + (b2 − c2) + (c2 − a2)
= a2 − b2 + b2 − c2 + c2 − a2
= a2 − a2 − b2 + b2 − 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 × 303 | (vii) 78 × 82 | (viii) 8.92 |
(ix) 10.5 × 9.5 |
Solutions :
(i) 712
= (70 + 1)2
by identity (a + b)2 = a2 + 2ab + b2
(70 + 1)2 = (70)2 + 2 × 70 × 1 + (1)2
= 4900 + 140 + 1
= 5041
(ii) 992
= (100 − 1)2
by identity (a − b)2 = a2 − 2ab + b2
(100 − 1)2 = (100)2 − 2 × 100 × 1 + (1)2
= 10000 − 200 + 1
= 10001 − 200
= 9801
(iii) 1022
= (100 + 2)2
by identity (a + b)2 = a2 + 2ab + b2
(100 + 2)2 = (100)2 + 2 × 100 × 2 + (2)2
= 10000 + 400 + 4
= 10404
(iv) 9982
= (1000 − 2)2
by identity (a − b)2 = a2 − 2ab + b2
(1000 − 2)2 = (1000)2 − 2 × (1000) × 2 + (2)2
= 1000000 − 4000 + 4
= 1000004 − 4000
= 996004
(vi) 297 × 303
= (300 − 3) (300 + 3)
by identity (a)2 − (b)2 = (a + b) (a − b)
= (300)2 − (3)2
= 90000 − 9
= 89991
(vii) 78 × 82
= (80 − 2) (80 + 2)
by identity (a)2 − (b)2 = (a + b) (a − b)
= (80)2 − (2)2
= 6400 − 4
= 6396
(viii) 8.92
= (10.0 − 1.1)2
by identity (a − b)2 = a2 − 2ab + b2
(10.0 − 1.1)2 = (10)2 − 2 × 10 × 1.1 + (1.1)2
= 100 − 22 + 1.21
= 100.21 − 22
= 79.21
(ix) 10.5 × 9.5
= (10.0 + 0.5) (10.0 – 0.5)
by identity (a)2 − (b)2 = (a + b) (a − b)
= (10)2 − (0.5)2
= 100 − 0.25
= 99.75
7. Using a2 − b2 = (a + b) (a − b), find
(i) 512 − 492 | (ii) (1.02)2 − (0.98)2 |
(iii) 1532 − 1472 | (iv) 12.12 − 7.92 |
Solutions :
(i) 512 − 492
by identity a2 − b2 = (a + b) (a − b)
512 − 492 = (51 + 49) (51 − 49)
= 100 × 2
= 200
(ii) (1.02)2 − (0.98)2
by identity a2 − b2 = (a + b) (a − b)
(1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 − 0.98)
= 2 × 0.04
= 0.08
(iii) 1532 − 1472
by identity a2 − b2 = (a + b) (a − b)
1532 − 1472 = (153 + 147) (153 − 147)
= 300 × 6
= 1800
(iv) 12.12 − 7.92
by identity a2 − b2 = (a + b) (a − b)
12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)
= 20 × 4.2
= 84
8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 × 104 | (ii) 5.1 × 5.2 |
(iii) 103 × 98 | (iv) 9.7 × 9.8 |
Solutions :
(i) 103 × 104
= (100 + 3) (100 + 4)
by identityसर्वसमिका (x + a) (x + b) = x2 + (a + b) x + ab
= (100)2 + (3 + 4) 100 + 3 × 4
= 10000 + 700 + 12
= 10712
(ii) 5.1 × 5.2
= (5.0 + 0.1) (5.0 + 0.2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
= (5.0)2 + (0.1 + 0.2) 5.0 + 0.1 × 0.2
= 25 + 1.5 + 0.02
= 26.52
(iii) 103 × 98
= (100 + 3) (100 – 2)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
= (100)2 + {3 + (− 2} 100 + 3 × (− 2)
= 10000 + 1 × 100 + (− 6)
= 10000 + 100 − 6
= 10100 − 6
= 10094
(iv) 9.7 × 9.8
= (9.0 + 0.7) (9.0 + 0.8)
by identity (x + a) (x + b) = x2 + (a + b) x + ab
= (9)2 + (0.7 + 0.8) 9 + 0.7 × 0.8
= 81 + 1.5 × 9 + 0.56
= 81 + 13.5 + 0.56
= 95.06