NCERT solutions for class 8 Maths Chapter 9 Algebraic Expressions

Solution for Exercise 9.1

1.Identify the terms, their coefficients for each of the following expressions :
(i) \(5xyz^2 – 3zy\)

(ii) \(1 + x + x^2\)

(iii) \(4x^2y^2 – 4x^2y^2z^2 + z^2\)

(iv) \(3 – pq + qr -rp\)

(v) \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\)

(vi)\( 0.3a – 0.6ab + 0.5b\)


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Answer :

(i) We have the expression \(5xyz^2 – 3zy\), the terms are \(5xyz^2\) and \(-3zy\).
Coefficient of \(xyz^2\) in the term \(5xyz^2\) is 5.

Coefficient of \(zy\) in the term \(– 3yz\) is \(– 3\).

(ii)We have the expression \(1 + x + x^2\), the terms are 1, x and \(x^2\).

Coefficient of the term 1 is 1.

Coefficient of x in the term x is 1.

Coefficient of \(x^2\) in the term \(x^2\) is 1.

(iii) We have the expression \(4xy – 4xyz + z\) , the terms are \(4x^2y^2, – 4x^2y^2z^2\) and \(z^2\).

Coefficient of \(x^2y^2\) in the term \(4x^2y^2\) is 4.

Coefficient of \(x^2y^2z^2\) in the term \(– 4x^2y^2z^2\) is – 4.

Coefficient of \(z^2\) in the term \(z^2\) is 1.

(iv) We have the expression \(3 – pq + qr – rp\), the terms are 3, – pq, qr and – rp.

Coefficient of the term 3 is 3.

Coefficient of pq in the term – pq is -1.

Coefficient of qr in the term qr is 1.

Coefficient of rp in the term – rp is -1.

(v) We have the expression \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\), the terms are \(\frac { x }{ 2 } ,\frac { 7 }{ 2 }\) and \(– xy\).

Coefficient of x in the term \(\frac { x }{ 2 }\) is \(\frac { 1 }{ 2 }\)

Coefficient of y in the term \(\frac { y }{ 2 }\) is \(\frac { 1 }{ 2 }\)

Coefficient of xy in the term \(– xy\) is -1.

(vi) In the expression \(0.3a -0.6 ab + 0.5 b\), the terms are \(0.3a, – 006ab \)and \(0.5b\).

Coefficient of a in the term 0.3a is 0.3.

Coefficient of ab in the term – 0.6ab is – 0.6.

Coefficient of b in the term 0.5b is 0.5.

2.Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? \(x + y, 1000, x + x^2 + x^3 + x^4, 7 + y + 5x, 2y – 3y^2, 2y – 3y^2 + 4y^3\),\( 5x – 4y + 3xy, 4z – 15z^2, ab + bc+cd + da, pqr, p^2q + pq^2, 2p + 2q.\)


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Answer :

The calssification of the given expressions are:

Monomials : 1000, pqr

Binomials : \(x + y, 2y – 3y^2, 4z -15z^2, p^2q + pq^2, 2p + 2q.\)

Trinomials : \(7 + y + 5x, 2y – 3y^2 + 4y^3, 5x – 4y + 3x.\)

And the polynomials which don't have any category:

\(x + x^2 + x^3 + x^4, ab + be + cd + da.\)

3.Add the following :
(i) \(ab – be, be – ca, ea – ab\)

(ii)\( a – b + ab, b-c + be, c – a + ac\)

(iii) \(2p^2q^2 – 3pq + 4, 5 + 7pq – 3p^2q^2\)

(iv)\( l^2 + m^2, m^2 + n^2, n^2 + l^2, 2lm + 2mn + 2nl\)


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Answer :

(i) We have the expressions as:\( ab – bc, bc – ca, ca – ab\)
Adding them :

\((ab – bc) + (bc – ca) + (ca – ab) \)

\(\Rightarrow ab – bc + bc – ca + ca – ab \)

\(\Rightarrow (ab – ab) + (bc – bc) + (ca – ca)\quad\) [Placing the like terms together]

\(\Rightarrow 0 + 0 + 0\)

\(\Rightarrow 0\)

(ii) We have the expressions as:\(a – b + ab, b – c + bc, c – a + ac\)

Adding these we have:

\((a – b + ab) + (b – c + bc) + (c – a + ac) \)

\(\Rightarrow a – b + ab + b – c + bc + c – a + ac\)

\(\Rightarrow (a – a) + (b – b) + (c – c) + ab + bc + ac\quad \)[Placing all the like terms together]

\(\Rightarrow 0 + 0 + 0 + ab + bc + ac\)

\(\Rightarrow ab + bc + ac\)

(iii)We have the expressions as:\(2p^2q^2 – 3pq + 4, 5 + 7pq – 3p^2q^2\)

Adding these we get: \(2p^2q^2 – 3pq + 4 + 5 + 7pq – 3p^2q^2\)

\(\Rightarrow (2p^2q^2– 3p^2q^2)+(– 3pq+ 7pq)+ (4 + 5)\quad\)[Placing like terms together]

\(\Rightarrow -p^2q^2 + 4pq+ 9\)

(iv)We have the expressions as:\( l^2 + m^2, m^2 + n^2, n^2 + l^2, 2lm + 2mn + 2nl\)
Adding these we get: \((l^2 + m^2)+( m^2 + n^2)+( n^2 + l^2)+ (2lm + 2mn + 2nl)\)

\(\Rightarrow (l^2+l^2)+ (m^2+m^2)+ (n^2+n^2)+(2lm + 2mn + 2nl)\quad\)[Placing like terms together]

\(\Rightarrow (2l^2)+ (2m^2)+ (2n^2)+(2lm + 2mn + 2nl\)

\(\Rightarrow 2(l^2+ m^2+n^2+lm + mn + nl)\)

4.(a) Subtract \(4a – 7ab + 3b + 12\) from \(12a – 9ab + 5b – 3\)
(b) Subtract \(3xy + 5yz – 7zx\) from \(5xy – 2yz – 2zx + 10xyz\)

(c) Subtract \(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10 \) from \(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q\)


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Answer :

(a)We have the expressions to subtract:\(4a – 7ab + 3b + 12\) from \(12a – 9ab + 5b – 3\)
Subtracting these we get: \((12a – 9ab + 5b – 3)-(4a – 7ab + 3b + 12 )\)

\((12a – 9ab + 5b – 3)+(-4a +7ab - 3b - 12 )\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (12a-4a)+ (– 9ab+7ab)+ (+ 5b- 3b )+( – 3 - 12 )\quad\)[Placing like terms together]

\(\Rightarrow (8a)+ (-2ab)+ (2b)+(-15)\)

\(\Rightarrow (8a-2ab+ 2b-15)\)

(b)We have the expressions to subtract:\(3xy + 5yz – 7zx\) from \(5xy – 2yz – 2zx + 10xyz\)
Subtracting these we get: \((5xy – 2yz – 2zx + 10xyz)-(3xy + 5yz – 7zx)\)

\((5xy – 2yz – 2zx + 10xyz)+(-3xy - 5yz + 7zx )\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (5xy-3xy )+ (– 2yz - 5yz)+ (– 2zx+ 7zx )+( + 10xyz )\quad\)[Placing like terms together]

\(\Rightarrow (2xy)+ (-7yz)+ (5zx)+(10xyz)\)

\(\Rightarrow (2xy-7yz+5zx+10xyz)\)

(c)We have the expressions to subtract:\(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10 \) from \(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q\)
Subtracting these we get: \((18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q )-(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10)\)

\(\Rightarrow(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q)+(-4p^2q +3pq-5pq^2 +8p-7q+ 10)\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (+18+10)+(-3p+8p)+(-11q-7q)+(5pq+3pq)+(-2pq^2-5pq^2)+(+ 5p^2q-4p^2q) \quad\)[Placing like terms together]

\(\Rightarrow (+28)+ (+5p)+(-18q)+(8pq)+(-7pq^2)+(+p^2q)\)

\(\Rightarrow (+28+5p-18q+8pq-7pq^2+p^2q)\)

4.Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) \(5a, 3a^2, 7a^4\)

(ii) \(2p, 4q, 8r\)

(iii) \(xy, 2x^2y, 2xy^2\)

(iv) \(a, 2b, 3c\)


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Answer :

(i)Given that, length = 5a, breadth = \(3a^2\), height = \(7a^4\)

∴Volume of the box =\( l \times b \times h = 5a \times 3a^2 \times 7a^4 = 105 a^7 \;\)cu. units

(ii)Given that, length = 2p, breadth = 4q, height = 8r

∴Volume of the box =\( l \times b \times h = 2p \times 4q \times 8r = 64pqr\;\) cu. units

(iii)Given that, length = xy, breadth =\( 2x^2y\), height = 2xy^2\)

Volume of the box =\( l \times b \times h = xy \times 2x^2y \times 2xy^2 = (1 \times 2 \times 2) \times xy \times x^2y \times xy^2 = 4x^4y^4 \;\)cu. units

(iv)Given that, length = a, breadth = 2b, height = 3c

∴Volume of the box = \(l \times b \times h = a \times 2b \times 3c = (1 \times 2 \times 3)abc = 6 abc\;\) cu. units

Solution for 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


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Answer :

(i)Multiplying the given expressions: \( 4 x\times 7p = (4 \times 7) \times p = 28p\)

(ii) Multiplying the given expressions:\(– 4 p \times 7p = (- 4 \times 7) \times (p\times p)\)

\(= -28p^{(1 + 1)} = – 28p^2\)

(iii)Multiplying the given expressions: \(-4p\times 7pq = (- 4 \times 7) \times (p \times p \times q)\)

\(= -28 p^{(1+1)}q = – 28p^2q\)

(iv)Multiplying the given expressions: \(4p3 \times – 3p = (4 \times – 3) \times (p^3 \times p)\)

\(= – 12p^{(3+1)} = =-12p^4\)

(v) Multiplying the given expressions:\(4p \times 0 = (4 \times 0) \times p = 0 \times p = 0\)

2.Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
\((p, q); (10m, 5n); (20x^2, 5y^2); (4x, 3x^2); (3mn, 4np)\)


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Answer :

(i) Given that: Length = p units and breadth = q units

Area of the rectangle = \(length \times breadth = p \times q = pq\) sq units

(ii)Given that: Length = 10 m units, breadth = 5n units

∴Area of the rectangle =\(length \times breadth = 10 m \times 5 n = (10 \times 5) \times m \times n = 50 mn\; \)sq units

(iii) Given that: Length = 20x2 units, breadth = 5y2 units

∴Area of the rectangle = length × breadth = 20x2 × 5y2 = (20 × 5) × x2 × y2 = 100x2y2 sq units

(iv)Given that: Length = 4x units, breadth = 3x2 units

∴Area of the rectangle = \(length \times breadth = 4x \times 3x^2 = (4 \times 3) \times x \times x^2 = 12x^3\;\) sq units

(v) Given that: Length = 3mn units, breadth = 4np units

∴Area of the rectangle = \(length \times breadth = 3mn \times 4np = (3 \times 4) \times mn \times np = 12mn^2p\;\) sq units

3.Complete the table of products :


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Answer :

Completed table is as under :

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


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Answer :

(i) Product=\(xy \times yz \times zx = x^2y^2z^2\)

(ii) Product=\(a \times (-a^2) \times a^3 = (-a)^6\)

(iii)Product=\(2 \times 4y \times 8y^2 \times 16y^3 = (2 \times 4 \times 8 \times 16) \times y \times y^2 \times y^3 = 1024y^6\)

(iv)Product=\( a \times 2b \times 3c \times 6abc = (1 \times 2 \times 3 \times6) \times a \times b \times c \times abc = 36 a^2b^2c^2\)

(v)Product=\( m \times (-mn) \times mnp = [1 \times (-1) \times 1 ]m \times mn \times mnp = -m^3n^2p\)

Solution for 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) \(a + b, 7a^2b^2\)

(iv) \(a^2 – 9, 4a\)

(v) \(pq + qr + rp, 0\)


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Answer :

Multiplication of the expressions are as follows:
(i) \(4p \times (q + r) = (4p \times q) + (4p \times r) = 4pq + 4pr\)

(ii) \(ab, a – b = ab \times (a – b) = (ab \times a) – (ab \times b) = a^2b – ab^2\)

(iii) \((a + b) \times 7a^2b^2 = (a \times 7a^2b^2) + (b \times 7a^2b^2) = 7a^3b^2 + 7a^2b^3\)

(iv) \((a2 – 9) \times 4a = (a^2 \times 4a) – (9 \times 4a) = 4a^3 – 36a\)

(v) \((pq + qr + rp) \times 0 = 0\)

2. Complete the table.


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Answer :

Products of the given expressions in the table are done as follows:
.(i) \(a \times (b + c + d) = (a \times b) + (a \times c) + (a \times d) = ab + ac + ad\)

(ii) \((x + y – 5) (5xy) = (x \times 5xy) + (y \times 5xy) – (5 \times 5xy) = 5x^2y + 5xy^2 – 25xy\)

(iii) \(p \times (6p^2 – 7p + 5) = (p \times 6p^2) – (p \times 7p) + (p \times 5) = 6p^3 – 7p^2 + 5p\)

(iv) \(4p^2q^2 \times (p^2 – q^2) = 4p^2q^2 \times p^2 – 4p^2q^2 \times q^2 = 4p^4q^2 – 4p^2q^4\)

(v) \((a + b + c) \times (abc) = (a \times abc) + (b \times abc) + (c \times abc) = a^2bc + ab^2c + abc^2\)

Therefore, the completed table is as follows:


3.Find the products
(i)\((a^2)\times(2a^{22})\times(4a^{26})\)

(ii)\((\frac{2}3xy)\times(-\frac{9}{10}x^2y^2)\)

(iii)\((-\frac{10}3pq^3)\times(\frac{6}{5}p^3q)\)

(iv)\(x\times x^2\times x^3\times x^4\)


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Answer :

(i)\((a^2)\times(2a^{22})\times(4a^{26})\)

\(=(1\times 2\times 4)\times a^{2+22+26}=8a^{50}\)

(ii)\((\frac{2}3xy)\times(-\frac{9}{10}x^2y^2)\)

\(=(\frac{2}3\times (-\frac{9}{10}))\times x^{(1+2})y^{(1+2)}\)

(iii)\((-\frac{10}3pq^3)\times(\frac{6}{5}p^3q)\)

\(=(-\frac{10}3\times\frac{6}{5})\times p^{(1+3)}q^{(1+3)}\)

\(=-4p^4q^4\)

(iv)\(x\times x^2\times x^3\times x^4\)

\(= x^{(1+2+3+4)}=x^{10\)

4.(a) Simplify: \(3x(4x – 5) + 3\) and find its values for (i) x = 3 (ii) x = \(\frac { 1 }{ 2 }\).
(b) Simplify: \(a(a^2 + a + 1) + 5\) and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1


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Answer :

(a) Given the expression: \(3x(4x – 5) + 3 = 4x \times 3x – 5 \times 3x + 3 = 12x^2 – 15x + 3\)
(i) So for x = 3, we have

\(12 \times (3)2 – 15 \times 3 + 3 = 12 \times 9 – 45 + 3 = 108 – 42 = 66\)

(ii)For x=\(\frac{1}2\), we have:

\(\Rightarrow 12(\frac{1}2)^2 - 15(\frac{1}2)+3\)

\(\Rightarrow 12\times(\frac{1}4) - \frac{15}2+3\)

\(\Rightarrow 3 - \frac{15}2+3\)

\(\Rightarrow \frac{6-15+6}2=\frac{12-15}2=-\frac{3}2\)

(b) We have \(a(a^2 + a + 1) + 5\)

\(= (a^2 \times a) + (a \times a) + (1 \times a) + 5\)

\(= a^3 + a^2 + a + 5\)

(i) For a = 0, we have:

\(= (0)^3 + (0)^2 + (0) + 5 = 5\)

(ii) For a = 1, we have:

\(= (1)^3 + (1)^2 + (1) + 5 = 1 + 1 + 1 + 5 = 8\)

(iii) For a = -1, we have:

\(= (-1)^3 + (-1)^2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4\)

5.(a) \(Add: p(p – q), q(q – r) and r(r – p)\)
(b) \(Add: 2x(z – x – y) and 2y(z – y – x)\)

(c) \(Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)\)

(d) \(Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)\)


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Answer :

(a) \(p(p – q) + q(q – r) + r(r – p)\)

=\( (p \times p) – (p \times q) + (q \times q) – (q \times r) + (r \times r) – (r \times p)\)[Opening the brackets]

= \(p^2 – pq + q^2 – qr + r^2 – rp\)

=\( p^2 + q^2 + r^2 – pq – qr – rp\)

(b) \(2x(z – x – y) + 2y(z – y – x)\)

= \((2x \times z) – (2x \times x) – (2x \times y) + (2y \times z) – (2y \times y) – (2y \times x)\)[Opening the brackets]

= \(2xz – 2x^2 – 2xy + 2yz – 2y^2 – 2xy\)

\(= -2x^2 – 2y^2 + 2xz + 2yz – 4xy\)

\(= -2x^2 – 2y^2 – 4xy + 2yz + 2xz\)

(c) \(4l(10n – 3m + 2l) – 3l(l – 4m + 5n)\)

\(= (4l \times 10n) – (4l \times 3m) + (4l \times 2l) – (3l \times l) – (3l \times -4m) – (3l ×\times 5n)\)[Opening the brackets]

\(= 40ln – 12lm + 8l^2 – 3l^2 + 12lm – 15ln\)

\(= (40ln – 15ln) + (-12lm + 12lm) + (8l^2 – 3l^2)\)

\(= 25ln + 0 + 5l^2\)

\(= 25ln + 5l^2\)

\(= 5l^2 + 25ln\)

(d)\( [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]\)

\(= (-4ac + 4bc + 4c^2) – (3a^2 + 3ab + 3ac – 2ab + 2b^2 – 2bc)\)[Opening the brackets]

\(= -4ac + 4bc + 4c^2 – 3a^2 – 3ab – 3ac + 2ab – 2b^2 + 2bc\)

\(= -3a^2 – 2b^2 + 4c^2 – ab + 6bc – 7ac\)

Solution for 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) \((2pq + 3q^2)\) and \((3pq – 2q^2)\)

(vi) \((\frac { 3 }{ 4 }a^2 + 3b^2)\) and \(4(a^2 – \frac { 2 }{ 3 } b^2)\)


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Answer :

(i)Multiplying the given expressions as:\( (2x + 5) \times (4x – 3)\)

\(= 2x \times (4x – 3) + 5 \times (4x – 3)\)

\(= (2x \times 4x) – (3 \times 2x) + (5 \times 4x) – (5 \times 3)\)

\(= 8x^2 – 6x + 20x – 15\)

\(= 8x^2 + 14x – 15\)

(ii)Multiplying the given expressions as: \((y – 8) \times (3y – 4)\)

\(= y \times (3y – 4) – 8 \times (3y – 4)\)

\(= (y \times 3y) – (y \times 4) – (8 \times 3y) + (-8 \times -4)\)

\(= 3y^2 – 4y – 24y + 32\)

\(= 3y^2 – 28y + 32\)

(iii)Multiplying the given expressions as: \((2.5l – 0.5m) × (2.5l + 0.5m)\)

\(= (2.5l \times 2.5l) + (2.5l \times 0.5m) – (0.5m \times 2.5l) – (0.5m \times 0.5m)\)

\(= 6.25l^2 + 1.25ml – 1.25ml – 0.25m^2\)

\(= 6.25l^2 + 0 – 0.25m^2\)

\(= 6.25l^2 – 0.25m^2\)

(iv)Multiplying the given expressions as: \((a + 3b) \times (x + 5)\)

\(= a \times (x + 5) + 36 \times (x + 5)\)

\(= (a \times x) + (a \times 5) + (36 \times x) + (36 \times 5)\)

\(= ax + 5a + 3bx + 15b\)

(v) Multiplying the given expressions as:\((2pq + 3q^2) \times (3pq – 2q^2)\)

\(= 2pq \times (3pq – 2q^2) + 3q^2 (3pq – 2q^2)\)

\(= (2pq \times 3pq) – (2pq \times 2q^2) + (3q^2 × 3pq) – (3q^2 \times 2q^2)\)

\(= 6p^2q^2 – 4pq^3 + 9pq^3 – 6q^4\)

\(= 6p^2q^2 + 5pq^3 – 6q^4\)

(vi)Multiplying the given expressions as:\((\frac { 3 }{ 4 }a^2 + 3b^2) \times 4(a^2 – \frac { 2 }{ 3 } b^2)\)

\(=(\frac{3}4a^2+3b^2)\times(4a^2-\frac{8}3b^2)\)

\(=\frac{3}4a^2\times (4a^2-\frac{8}3b^2)+3b^2\times(4a^2-\frac{8}3b^2)\)

\(=(\frac{3}4a^2\times 4b^2) - (\frac{3}4a^2\times \frac{8}3b^2) + (3b^2\times 4a^2)-(3b^2\times\frac{8}3b^2)\)

\(=3a^4-2a^2b^2+12a^2b^2-8b^4\)

\(=3a^4+10a^2b^2-8b^4\)

2.Find the product:
(i) \((5 – 2x) (3 + x)\)

(ii)\( (x + 7y) (7x – y)\)

(iii) \((a^2 + b) (a + b^2)\)

(iv)\( (p^2 – q^2)(2p + q)\)


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Answer :

(i)Product of the expression:\( (5 – 2x) (3 + x)\)

\(= 5(3 + x) – 2x(3 + x)\)

\(= (5 \times 3) + (5 \times x) – (2x \times 3) – (2x \times x)\)

\(= 15 + 5x – 6x – 2x^2\)

(ii)Product of the expression: \((x + 7y) (7x – y)\)

\(= x(7x – y) + 7y(7x – y)\)

\(= (x \times 7x) – (x \times y) + (7y \times 7x) – (7y \times y)\)

\(= 7x^2 – xy + 49xy – 7y^2\)

\(= 7x^2 + 48xy – 7y^2\)

(iii)Product of the expression: \((a^2 + b) (a + b^2)\)

\(= a^2 (a + b^2) + b(a + b^2)\)

\(= (a^2 \times a) + (a^2 \times b^2) + (b \times a) + (b \times b^2)\)

\(= a^3 + a^2b^2 + ab + b^3\)

(iv)Product of the expression:\( (p^2 – q^2)(2p + q)\)

\(= p^2(2p + q) – q^2(2p + q)\)

\(= (p^2 \times 2p) + (p^2 \times q) – (q^2 \times 2p) – (q^2 \times q)\)

\(= 2p^3 + p^2q – 2pq^2 – q^3\)

3.Simplify: (i) \((x^2 – 5) (x + 5) + 25\)

(ii) \((a^2 + 5)(b^3 + 3) + 5\)

(iii) \((t + s^2) (t^2 – s)\)

(iv)\( (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)\)

(v)\( (x + y) (2x + y) + (x + 2y) (x – y)\)

(vi) \((x + y)(x^2 – xy + y^2)\)

(vii)\( (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y\)

(viii)\( (a + b + c) (a + b – c)\)


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Answer :

(i)Simplifying the expression, we have: \((x^2 – 5) (x + 5) + 25\)

\(= x^2 \times (x + 5) + 5\times (x + 5) + 25\)

\(= x^3 + 5x^2 – 5x – 25 + 25\)

\(= x^3 + 5x^2 – 5x + 0\)

\(= x^3 + 5x^2 – 5x\)

(ii)Simplifying the expression, we have: \((a^2 + 5)(b^3 + 3) + 5\)

\(= a^2\times (b^3 + 3) + 5\times (b^3 + 3) + 5\)

\(= a^2b^3 + 3a^2 + 5b^3 + 15 + 5\)

\(= a^2b^3 + 3a^2 + 5b^3 + 20\)

(iii)Simplifying the expression, we have:\( (t + s^2) (t^2 – s)\)

\(= t\times (t^2 – s) + s^2 \times (t^2 – s)\)

\(= t^3 – st + s^2t^2 – s^3\)

\(= t^3 + s^2t^2 – st – s^3\)

(iv) Simplifying the expression, we have:\((a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)\)

\(= a\times(c – d) + b\times(c – d) + a\times(c + d) – b\times(c + d) + 2ac + 2bd\)

\(= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd\)

\(= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd\)

\(= 4ac + 0 + 0 + 0\)

\(= 4ac\)

(v) Simplifying the expression, we have:\((x + y) (2x + y) + (x + 2y) (x – y)\)

\(= x\times(2x + y) + y\times(2x + y) + x\times(x – y) + 2y\times(x – y)\)

\(= 2x^2 + xy + 2xy + y^2 + x^2 – xy + 2xy – 2y^2\)

\(= 2x^2 + x^2 + xy + 2xy – xy + 2xy + y^2 – 2y^2\)

\(= 3x^2 + 4xy – y^2\)

(vi)Simplifying the expression, we have:\( (x + y)(x^2 – xy + y^2)\)

\(= x\times(x^2 – xy + y^2) + y\times (x^2 – xy + y^2)\)

\(= x^3 – x^2y + x^2y + xy^2 – xy^2 + y^3\)

\(= x^3 – 0 + 0 + y^3\)

\(= x^3 + y^3\)

(vii)Simplifying the expression, we have:\( (1.5x – 4y)(1.5x + 4y + 3) – 4.5x.+ 12y\)

\(= 1.5x \times (1.5x + 4y + 3) – 4y\times(1.5x + 4y + 3) – 4.5x + 12y\)

\(= 2.25x^2 + 6xy + 4.5x – 6xy – 16y^2 – 12y – 4.5x + 12y\)

\(= 2.25x^2 + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y^2\)

\(= 2.25x^2 + 0 + 0 + 0 – 16y62\)

\(= 2.25x^2 – 16y^2\)

(viii)Simplifying the expression, we have: \((a + b + c) (a + b – c)\)

\(= a(a + b – c) + b(a + b – c) + c(a + b – c)\)

\(= a^2 + ab – ac + ab + b^2 – bc + ac + bc – c^2\)

\(= a^2 + ab + ab – bc + bc – ac + ac + b^2 – c^2\)

\(= a^2 + 2ab + b^2 – c^2 + 0 + 0\)

\(= a^2 + 2ab + b^2 – c^2\)

Solution for 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)\( (3a – \frac { 1 }{ 2 }) (3a – \frac { 1 }{ 2 })\)

(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)\( (\frac { x }{ 2 } + \frac { 3y }{ 4 }) (\frac { x }{ 2 } + \frac { 3y }{ 4 })\)

(x) \((7a – 9b) (7a – 9b)\)


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Answer :

(i)Given: \((x + 3) (x + 3)\)
\(=(x+3)^2\)

\(=(x)^2+2\times x \times 3+(3)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=x^2+6x+9\)

(ii)Given: \((2y + 5) (2y + 5)\)
\(=(2y + 5)^2\)

\(=(2y)^2+2\times 2y \times 5+(5)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=4y^2+20y+25\)

(iii)Given: \((2a - 7) (2a - 7)\)
\(=(2a - 7)^2\)

\(=(2a)^2-2\times (2a) \times 7+(7)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=4a^2-28a+49\)

(iv)Given: \((3a - \frac{1}2) (3a - \frac{1}2)\)
\(=(3a - \frac{1}2)^2\)

\(=(3a)^2-2\times (3a) \times (\frac{1}2)+(\frac{1}2)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=9a^2-3a+\frac{1}4\)

(v)Given: \((1.1m - 0.4) (1.1m + 0.4)\)
\(=((1.1m)^2 - (0.4)^2\quad \)[∵\((a-b)(a+b)=a^2-b^2\)]

\(=1.21a^2-0.16\)

(vi)Given: \((a^2 + b^2) (-a^2 + b^2)\)
\(=(b^2 + a^2)(b^2 -a^2) \)

\(=(b^2)^2-(a^2)^2\quad\)[∵\((a-b)(a+b)=a^2-b^2\)]

\(=(b^4 - a^4)\)

(vii)Given: \((6x - 7) (6x + 7)\)
\(=(6x)^2 - (7)^2\quad \)[∵\((a-b)(a+b)=a^2-b^2\)]

\(=36x^2-49\)

(viii)Given: \((-a+c) (-a+c)\)
\(=[(-a)+c]^2\)

\(=(-a)^2+2\times (-a) \times c+(c)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=a^2-2ac+c^2\)

(ix)Given:\( (\frac { x }{ 2 } + \frac { 3y }{ 4 }) (\frac { x }{ 2 } + \frac { 3y }{ 4 })\)

\(=(\frac { x }{ 2 } + \frac { 3y }{ 4 })^2\)

\(=(\frac { x }{ 2 })^2+2\times (\frac { x }{ 2 }) \times (\frac { 3y }{ 4 })+(\frac { 3y }{ 4 })^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=\frac { x^2 }{ 4 }+\frac { 3 }{ 4 }xy+\frac { 9y^2 }{ 16 }\)

(x)Given: \((7a - 9b) (7a - 9b)\)
\(=(7a - 9b)^2\)

\(=(7a)^2-2\times (7a) \times (9b)+(9b)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=49a^2-126ab+81b^2\)

2.Use the identity \((x + a)(x + b) = x^2 + (a + b)x + ab\) 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) \((xyz – 4) (xyz – 2)\)


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Answer :

(i)We have: \((x + 3) (x + 7)\)

\(=x^2 + (3 + 7)x + 3\times 7\)

\(= x^2 +10x + 21\)

(ii)We have: \((4x + 5) (4x + 1)\)

\(=(4x)^2 + (5 + 1)(4x) + 5\times 1\)

\(= 16x^2 +24x + 5\)

(iii)We have: \((4x - 5) (4x - 1)\)

\(=(4x)^2 - (5 + 1)(4x) + (-5)\times(-1)\)

\(= 16x^2 -24x + 5\)

(iv)We have: \((4x + 5) (4x - 1)\)

\(=(4x)^2 + (5 - 1)(4x) + (5)\times(-1)\)

\(= 16x^2 +16x - 5\)

(v)We have: \((2x + 5y) (2x + 3y)\)

\(=(2x)^2 + (5y +3y)(2x) + (5y)\times (3y)\)

\(=(2x)^2 + (8y)(2x) + (5y)\times (3y)\)

\(= 4x^2 +16xy + 15y^2\)

(vi)We have: \((2a^2 + 9) (2a^2 + 5)\)

\(=(2a^2)^2 + (9+5)(2a^2) + (5)\times (9)\)

\(=(2a^2)^2 + (14)(2a^2) + 45\)

\(= 4a^4 +28a^2 + 45\)

(vii)We have: \((xyz - 4) (xyz - 2)\)

\(=(xyz)^2 - (4+2)(xyz) + (-4)\times (-2)\)

\(=(xyz)^2 - (6)(xyz) + 8\)

\(= x^2y^2z^2 -6xyz + 8\)

3.Find the following squares by using the identities.
(i) \((b – 7)^2\)

(ii) \((xy + 3z)^2\)

(iii) \((6x^2 – 5y)^2\)

(iv) \((\frac { 2 }{ 3 } m + \frac { 3 }{ 2 } n)^2\)

(v) \((0.4p – 0.5q)^2\)

(vi) \((2xy + 5y)^2\)


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Answer :

(i)Given: \((b – 7)^2\)

\(=(b)^2-2(b)(7)+(7)^2\quad\)[using \((a-b)^2=a^2-2ab+b^2\)]

\(=(b)^2-14b+49\)

(ii)Given: \((xy+3z)^2\)

\(=(xy)^2+2(xy)(3z)+(3z)^2\quad\)[using \((a+b)^2=a^2+2ab+b^2\)]

\(=x^2y^2+6xyz+9y^2\)

(iii)Given:\((6x^2 – 5y)^2\)

\(=(6x^2)^2-2(6x^2)(5y)+(5y)^2\quad\)[using \((a-b)^2=a^2-2ab+b^2\)]

\(=36x^4-60x^2y+25y^2\)

(iv)Given: \((\frac { 2 }{ 3 } m + \frac { 3 }{ 2 } n)^2\)

\(=(\frac { 2 }{ 3 } m)^2+2(\frac { 2 }{ 3 } m)(\frac { 3 }{ 2 } n)+(\frac { 3 }{ 2 } n)^2\quad\)[using \((a+b)^2=a^2+2ab+b^2\)]

\(=\frac { 4 }{ 9 } m^2+2mn+\frac { 9 }{ 4 } n^2\)

(v)Given: \((0.4p – 0.5q)^2\)

\(=(0.4p)^2-2(0.4p)(0.5q)+(0.5q)^2\quad\)[using \((a-b)^2=a^2-2ab+b^2\)]

\(=0.16p^2-0.4pq+0.25q^2\)

(vi)Given: \((2xy+5y)^2\)

\(=(2xy)^2+2(2xy)(5y)+(5y)^2\quad\)[using \((a+b)^2=a^2+2ab+b^2\)]

\(=4x^2y^2+20xy^2+25y^2\)

4.Simplify:
(i)\( (a^2 – b^2)^2\)

(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) \((ab + bc)^2 – 2ab^2c\)

(vii) \((m^2 – n^2m)^2 + 2m^3n^2\)


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Answer :

(i)Given:\((a^2-b^2)^2\)
\(=(a^2)^2-2a^2b^2+(b^2)^2\quad\)[use \((a-b)^2=a^2-2ab+b^2\)]

\(=a^4-2a^2b^2+b^4\)

(ii)Given:\((2x + 5)^2 – (2x – 5)^2\)

\(=[(2x)^2+2(2x)(5)+(5)^2]-[(2x)^2-2(2x)(5)+(5)^2]\)

\(=[4x^2+20x+25]-[4x^2-20x+25]\)

\(=4x^2+20x+25-4x^2+20x-25]\)

\(=(4x^2-4x^2)+(+20x+20x)+(+25-25)\quad \)[Placing same terms together]

\(=40x\)

(iii)Given:\((7m – 8n)^2 + (7m + 8n)^2\)

\(=[(7m)^2-2(7m)(8n)+(8n)^2] + [(7m)^2+2(7m)(8n)+(8n)^2]\)

\(=[49m^2-112mn+64n^2]+[49m^2+112mn+64n^2]\)

\(=(49m^2+49m^2)+(-112mn+112mn)+(64n^2+64n^2)\quad \)[Placing same terms together]

\(=98m^2+128n^2\)

(iv)Given:\((4m + 5n)^2 + (5m + 4n)^2\)

\(=[(4m)^2+2(4m)(5n)+(5n)^2] + [(5m)^2+2(5m)(4n)+(4n)^2]\)

\(=[16m^2+40mn+25n^2]+[25m^2+40mn+16n^2]\)

\(=(16m^2+25m^2)+(40mn+40mn)+(25n^2+16n^2)\quad\)[Placing same terms together]

\(=41m^2+80mn+41n^2\)

(v)Given: \((2.5p – 1.5q)^2 – (1.5p – 2.5q)^2\)

\(=[(2.5p)^2-2(2.5p)(1.5q)+(1.5q)^2] - [(1.5p)^2-2(1.5p)(2.5q)+(2.5q)^2]\)

\(=[6.25p^2-7.5pq+2.25q^2]-[2.25p^2-7.5pq+6.25q^2]\)

\(=6.25p^2-7.5pq+2.25q^2-2.25p^2+7.5pq-6.25q^2\)

\(=(6.25p^2-2.25p^2)+(-7.5pq+7.5pq)+(2.25q^2-6.25q^2)\quad\)[Placing same terms together]

\(=4p^2-4q^2\)

(vi)Given:\((ab + bc)^2 – 2ab^2c\)

\(=(ab)^2+2(ab)(bc)+(bc)^2-2ab^2c\)

\(=a^2b^2+2ab^2c+b^2c^2-2ab^2c\)

\(=a^2b^2+b^2c^2\)

(vii)Given:\((m^2 – n^2m)^2 + 2m^3n^2\)
\(=(m^2)^2-2(m^2)(n^2m)+(n^2m)^2+2m^3n^2\)

\(=m^4-2m^3n^2+n^4m^2+2m^3n^2\)

\(=m^4+n^4m^2\)

5. Show that:
(i)\( (3x + 7)^2 – 84x = (3x – 7)^2\)

(ii) \((9p – 5q)^2 + 180pq = (9p + 5q)^2\)

(iii) \((\frac { 4 }{ 3 } m – \frac { 3 }{ 4 } n)^2 + 2mn = \frac { 16 }{ 9 } m^2 + \frac { 9 }{ 16 } n^2\)

(iv) \((4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2\)

(v) \((a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0\)


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Answer :

(i)Given that: \( (3x + 7)^2 – 84x = (3x – 7)^2\)

To show LHS=RHS

LHS:\( (3x + 7)^2 – 84x\)

\(\Rightarrow (3x)^2+2(3x)(7)+7^2-84x\quad\)[using \((a+b)^2=a^2+2ab+b^2\)]

\(\Rightarrow 9x^2+42x+49-84x\)]

\(\Rightarrow 9x^2+42x-84x+49\)]

\(\Rightarrow 9x^2-42x+49\)

\(\Rightarrow (3x)^2-2(3x)(7)+7^2\)

\(\Rightarrow (3x – 7)^2=RHS\)[∵ \((a-b)^2=a^2-2ab+b^2\)]

Hence LHS=RHS

Hence proved
(ii)Given that: \((9p – 5q)^2 + 180pq = (9p + 5q)^2\)

To show LHS=RHS

LHS:\((9p – 5q)^2 + 180pq\)

\(\Rightarrow (9p)^2-2(9p)(5q)+(5q)^2+ 180pq\quad\)[using \((a-b)^2=a^2-2ab+b^2\)]

\(\Rightarrow 81p^2-90pq+25q^2+180pq\)]

\(\Rightarrow 81x^2-90pq+180pq+25q^2\)]

\(\Rightarrow 81x^2+90pq+25q^2\)

\(\Rightarrow (9p)^2+2(9p)(5q)+(5q)^2\)

\(\Rightarrow (9p + 5q)^2=RHS\)[using \((a+b)^2=a^2+2ab+b^2\)]

Hence LHS=RHS

Hence proved
(iii)Given: \((\frac { 4 }{ 3 } m – \frac { 3 }{ 4 } n)^2 + 2mn = \frac { 16 }{ 9 } m^2 + \frac { 9 }{ 16 } n^2\)

To show LHS=RHS

LHS: \((\frac { 4 }{ 3 } m – \frac { 3 }{ 4 } n)^2 + 2mn \)

\(\Rightarrow (\frac { 4 }{ 3 } m)^2-2(\frac { 4 }{ 3 } m)(\frac { 3 }{ 4 } n)+(\frac { 3 }{ 4 } n)^2+ 2mn \quad \)[using \((a-b)^2=a^2-2ab+b^2\)]

\(\Rightarrow \frac { 16 }{ 9 } m^2-2mn+\frac { 9 }{ 16 }n^2+2mn\)]

\(\Rightarrow \frac { 16 }{ 9 } m^2-2mn+2mn+\frac { 9 }{ 16 }n^2\)]

\(\Rightarrow \frac { 16 }{ 9 } m^2+\frac { 9 }{ 16 }n^2\)=RHS

Hence LHS=RHS

Hence proved
(iv)Given: \((4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2\)

To show LHS=RHS

LHS:\((4pq + 3q)^2 – (4pq – 3q)^2 \)

\(=[(4pq)^2+2(4qp)(3q)+(3q)^2] - [(4pq)^2-2(4pq)(3q)+(3q)^2]\)

\(=[16p^2q^2+24pq^2+9q^2]-[16p^2q^2-24pq^2+9q^2]\)

\(=16p^2q^2+24pq^2+9q^2-16p^2q^2+24pq^2-9q^2\)

\(=(16p^2q^2-16p^2q^2)+(+24pq^2+24pq^2)+(9q^2-9q^2)\quad\)[Placing same terms together]

\(=48pq^2=RHS\)

Hence, LHS=RHS

Hence proved

(v)Given:\((a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0\)

To show LHS=RHS

LHS:\((a^2-b^2)+(b^2-c^2)+(c^2-a^2)\quad\)[∵ \((x+y)(x-y)=x^2-y^2\)]

\(=a^2-b^2+b^2-c^2+c^2-a^2\)

\(=0=RHS\)

Hence, LHS=RHS

Hence proved

6.Using identities, evaluate:
(i) \(71^2\)

(ii) \(99^2\)

(iii) \(102^2\)

(iv) \(998^2\)

(v) \(5.2^2\)

(vi) \(297 \times 303\)

(vii) \(78 \times 82\)

(viii) \(8.9^2\)

(ix) \(1.05 \times 9.5\)


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Answer :

(i)To find: \(71^2\)

\(=(70+1)^2\)

\(=(70)^2+2(70)(1)+1^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=4900+140+1\)

\(=5041\)

Thus, we have \(71^2=5041\)

(ii)To find: \(99^2\)

\(=(100-1)^2\)

\(=(100)^2-2(100)(1)+1^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=10000-200+1\)

\(=9801\)

Thus, we have \(99^2=9801\)

(iii)To find: \(102^2\)

\(=(100+2)^2\)

\(=(100)^2+2(100)(2)+2^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=10000+400+4\)

\(=10404\)

Thus, we have \(102^2=10404\)

(iv)To find: \(998^2\)

\(=(1000-2)^2\)

\(=(1000)^2-2(1000)(2)+2^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=1000000-4000+4\)

\(=996004\)

Thus, we have \(998^2=996004\)

(v)To find: \(5.2^2\)

\(=(5+0.2)^2\)

\(=(5)^2+2(5)(0.2+(0.2)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=25+2+0.04\)

\(=27.04\)

Thus, we have \((5.2)^2=27.04\)

(vi)To find:\(297\times 303\)

\((300-3)(300+3)=(300)^2-(3)^2\quad\)[using \((a-b)(a+b)=a^2-b^2\)]>

\(=90000-9\)

\(=89991\)

Hence we have,\(297\times303=89991\)

(vii)To find:\(78\times 82\)

\((80-2)(80+2)=(80)^2-(2)^2\quad\)[using \((a-b)(a+b)=a^2-b^2\)]>

\(=6400-4\)

\(=6396\)

Hence we have,\(78\times82=6396\)

(viii)To find: \(8.9^2\)

\(=(9-0.1)^2\)

\(=(9)^2-2(9)(0.1)+(0.1)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=81-1.8+0.01\)

\(=79.21\)

Thus, we have \((8.9)^2=79.21\)

(ix)To find:\(1.05\times 9.5\)

\((1+0.05)(10-0.5)=1(10-0.5)+0.05(10-0.5)\)>

\(=10-0.5+0.05\times10-0.05\times0.5\)

\(=10-0.5+0.5-0.025\)

\(=10.5-0.525\)

\(=9.975\)

Hence we have,\(1.05\times9.5=9.975\)

7. Using \(a^2 – b^2 = (a + b) (a – b)\), find
(i) \(512 – 492\)

(ii) \((1.02)^2 – (0.98)^2\)

(iii) \(153^2 – 147^2\)

(iv) \(12.1^2 – 7.9^2\)


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Answer :

(i) \(512 – 492 = (51 + 49) (51 – 49) \)

\(= 100 \times 2 = 200\)

(ii)\( (1.02)^2 – (0.98)^2 = (1.02 + 0.98) (1.02 – 0.98) \)<\br>
\(= 2.00 \times 0.04 = 0.08\)

(iii) \(153^2 – 147^2 = (153 + 147) (153 – 147)\)

\(= 300 \times 6 = 1800\)

(iv) \(12.1^2 – 7.9^2 = (12.1 + 7.9) (12.1 – 7.9)\)

\( = 20.0 \times 4.2 = 84\)

8. Using \((x + a) (x + b) = x^2 + (a + b)x + ab\), find)

(i) \(103 × 104\)

(ii) \(5.1 × 5.2\)

(iii) \(103 × 98\)

(iv) \(9.7 × 9.8\)

(i) \(103 \times 104 = (100 + 3)(100 + 4)\)

\( = (100)^2 + (3 + 4) (100) + 3 \times 4 = 10000 + 700 + 12\)

\( = 10712\)

(ii) \(5.1 \times 5.2 = (5 + 0.1) (5 + 0.2)\)

\( = (5)^2 + (0.1 + 0.2) (5) + 0.1 \times 0.2 \)

\(= 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52\)

(iii) \(103 \times 98 = (100 + 3) (100 – 2)\)

\( = (100)^2 + (3 – 2) (100) + 3 \times (-2) = 10000 + 100 – 6 \)

\(= 10100 – 6 = 10094\)

(iv) \(9.7 \times 9.8 = (10 – 0.3) (10 – 0.2) \)

\(= (10)^2 – (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 \)

\(= 95 + 0.06 = 95.06\)



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