myhelper: RS Aggarwal solution class 8 chapter 6 Operations on Algebraic Expressions Exercise 6D

Exercise 6D

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Q1 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 1:

Find each of the following products:
(i) (x + 6)(x + 6)
(ii) (4x + 5y)(4x + 5y)
(iii) (7a + 9b)(7a + 9b)
(iv) $(\frac{2}{3} x + \frac{4}{5} y) (\frac{2}{3} x + \frac{4}{5} y)$
(v) (x² + 7)(x² + 7)
(vi) $(\frac{5}{6} a^{2} + 2) (\frac{5}{6} a^{2} + 2)$

Answer 1:

(i) We have:

$(x + 6) (x + 6) = (x + 6)^{2} = x^{2} + 6^{2} + 2 \times x \times 6 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = x^{2} + 36 + 12 x$

(ii) We have:

$(4 x + 5 y) (4 x + 5 y) = (4 x + 5 y)^{2} = {(4 x)}^{2} + {(5 y)}^{2} + 2 \times 4 x \times 5 y [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = 16 x^{2} + 25 y^{2} + 40 x y$

(iii) We have:
$(7 a + 9 b) (7 a + 9 b) = (7 a + 9 b)^{2} = {(7 a)}^{2} + {(9 b)}^{2} + 2 \times 7 a \times 9 b [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = 49 a^{2} + 81 b^{2} + 126 a b$

(iv) We have:
$(\frac{2}{3} x + \frac{4}{5} y) (\frac{2}{3} x + \frac{4}{5} y) = {(\frac{2}{3} x + \frac{4}{5} y)}^{2} = {(\frac{2}{3} x)}^{2} + {(\frac{4}{5} y)}^{2} + 2 \times \frac{2}{3} x \times \frac{4}{5} y [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = \frac{4}{9} x^{2} + \frac{16}{25} y^{2} + \frac{16}{15} x y$

(v) We have:
$(x^{2} + 7) (x^{2} + 7) = (x^{2} + 7)^{2} = {(x^{2})}^{2} + 7^{2} + 2 \times x^{2} \times 7 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = x^{4} + 49 + 14 x^{2}$

(vi) We have:
$(\frac{5}{6} a^{2} + 2) (\frac{5}{6} a^{2} + 2) = {(\frac{5}{6} a^{2} + 2)}^{2} = {(\frac{5}{6} a^{2})}^{2} + {(2)}^{2} + 2 \times \frac{5}{6} a^{2} \times 2 [using (a + b)^{2} = a^{2} + b^{2} + 2 a b] = \frac{25}{36} a^{4} + 4 + \frac{10}{3} a^{2}$

Q2 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 2:

Find each of the following products:
(i) (x − 4)(x − 4)
(ii) (2x − 3y)(2x − 3y)
(iii) $(\frac{3}{4} x - \frac{5}{6} y) (\frac{3}{4} x - \frac{5}{6} y)$
(iv) $(x - \frac{3}{x}) (x - \frac{3}{x})$
(v) $(\frac{1}{3} x^{2} - 9) (\frac{1}{3} x^{2} - 9)$
(vi) $(\frac{1}{2} y^{2} - \frac{1}{3} y) (\frac{1}{2} y^{2} - \frac{1}{3} y)$

Answer 2:

(i) We have:
$(x - 4) (x - 4) = (x - 4)^{2} = x^{2} - 2 \times x \times 4 + 4^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = x^{2} - 8 x + 16$

(ii) We have:
$(2 x - 3 y) (2 x - 3 y) = (2 x - 3 y)^{2} = {(2 x)}^{2} - 2 \times 2 x \times 3 y + {(3 y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = 4 x^{2} - 12 x y + 9 y^{2}$

(iii) We have:
$(\frac{3}{4} x - \frac{5}{6} y) (\frac{3}{4} x - \frac{5}{6} y) = {(\frac{3}{4} x - \frac{5}{6} y)}^{2} = {(\frac{3}{4} x)}^{2} - 2 \times \frac{3}{4} x \times \frac{5}{6} y + {(\frac{5}{6} y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{9}{16} x^{2} - \frac{15}{12} x y + \frac{25}{36} y^{2}$

(iv) We have:
$(x - \frac{3}{x}) (x - \frac{3}{x}) = {(x - \frac{3}{x})}^{2} = {(x)}^{2} - 2 \times x \times \frac{3}{x} + {(\frac{3}{x})}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = x^{2} - 6 + \frac{9}{x^{2}}$

(v) We have:
$(\frac{1}{3} x^{2} - 9) (\frac{1}{3} x^{2} - 9) = {(\frac{1}{3} x^{2} - 9)}^{2} = {(\frac{1}{3} x^{2})}^{2} - 2 \times \frac{1}{3} x^{2} \times 9 + {(9)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{1}{9} x^{4} - 6 x^{2} + 81$

(vi) We have:
$(\frac{1}{2} y^{2} - \frac{1}{3} y) (\frac{1}{2} y^{2} - \frac{1}{3} y) = {(\frac{1}{2} y^{2} - \frac{1}{3} y)}^{2} = {(\frac{1}{2} y^{2})}^{2} - 2 \times \frac{1}{2} y^{2} \times \frac{1}{3} y + {(\frac{1}{3} y)}^{2} [using (a - b)^{2} = a^{2} - 2 a b + b^{2}] = \frac{1}{4} y^{4} - \frac{1}{3} y^{3} + \frac{1}{9} y^{2}$

Q3 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 3:

Expand:
(i) (8a + 3b)²
(ii) (7x + 2y)²
(iii) (5x + 11)²
(iv) ${(\frac{a}{2} + \frac{2}{a})}^{2}$
(v) ${(\frac{3 x}{4} + \frac{2 y}{9})}^{2}$
(vi) (9x − 10)²
(vii) (x²y − yz²)²
(viii) ${(\frac{x}{y} - \frac{y}{x})}^{2}$
(ix) ${(3 m - \frac{4}{5} n)}^{2}$

Answer 3:

We shall use the identities (a+b)² =a² +b² +2ab and (a-b)² =a² +b² -2ab.

(i) We have:
$(8 a + 3 b)^{2} = {(8 a)}^{2} + 2 \times 8 a \times 3 b + {(3 b)}^{2} = 64 a^{2} + 48 a b + 9 b^{2}$

(ii)We have:
$(7 x + 2 y)^{2} = {(7 x)}^{2} + 2 \times 7 x \times 2 y + {(2 y)}^{2} = 49 x^{2} + 28 x y + 4 y^{2}$

(iii) We have :
$(5 x + 11)^{2} = {(5 x)}^{2} + 2 \times 5 x \times 11 + {(11)}^{2} = 25 x^{2} + 110 x + 121$

(iv) We have:
${(\frac{a}{2} + \frac{2}{a})}^{2} = {(\frac{a}{2})}^{2} + 2 \times \frac{a}{2} \times \frac{2}{a} + {(\frac{2}{a})}^{2} = {\frac{a}{4}}^{2} + 2 + \frac{4}{a^{2}}$

(v) We have:
${(\frac{3 x}{4} + \frac{2 y}{9})}^{2} = {(\frac{3 x}{4})}^{2} + 2 \times \frac{3 x}{4} \times \frac{2 y}{9} + {(\frac{2 y}{9})}^{2} = {\frac{9 x}{16}}^{2} + \frac{1}{3} x y + \frac{4 y^{2}}{81}$

(vi) We have:
$(9 x - 10)^{2} {(9 x)}^{2} - 2 \times 9 x \times 10 + {(10)}^{2} = 81 x^{2} - 180 x + 100$

(vii) We have:
$(x^{2} y - y z^{2})^{2} {(x^{2} y)}^{2} - 2 \times x^{2} y \times y z^{2} + {(y z^{2})}^{2} = x^{4} y^{2} - 2 x^{2} y^{2} z^{2} + y^{2} z^{4}$

(viii) We have:
${(\frac{x}{y} - \frac{y}{x})}^{2} = {(\frac{x}{y})}^{2} - 2 \times \frac{x}{y} \times \frac{y}{x} + {(\frac{y}{x})}^{2} = \frac{x^{2}}{y^{2}} - 2 + \frac{y^{2}}{x^{2}}$

(ix) We have:
${(3 m - \frac{4}{5} n)}^{2} = {(3 m)}^{2} - 2 \times 3 m \times \frac{4}{5} n + {(\frac{4}{5} n)}^{2} = 9 m^{2} - \frac{24 m n}{5} + \frac{16}{25} n^{2}$

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Q4 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 4:

Find each of the following products:
(i) (x + 3)(x − 3)
(ii) (2x + 5)(2x − 5)
(iii) (8 + x)(8 − x)
(iv) (7x + 11y)(7x − 11y)
(v) $(5 x^{2} + \frac{3}{4} y^{2}) (5 x^{2} - \frac{3}{4} y^{2})$
(vi) $(\frac{4 x}{5} - \frac{5 y}{3}) (\frac{4 x}{5} + \frac{5 y}{3})$
(vii) $(x + \frac{1}{x}) (x - \frac{1}{x})$
(viii) $(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y})$

Answer 4:

(i) We have:

$(x + 3) (x - 3) = x^{2} - 9 [using (a + b) (a - b) = a^{2} - b^{2}]$

(ii) We have:

$(2 x + 5) (2 x - 5) = 4 x^{2} - 25 [using (a + b) (a - b) = a^{2} - b^{2}]$

(iii) We have:

$(8 + x) (8 - x) = 64 - x^{2} [using (a + b) (a - b) = a^{2} - b^{2}]$

(iv) We have:

$(7 x + 11 y) (7 x - 11 y) = 49 x^{2} - 121 y^{2} [using (a + b) (a - b) = a^{2} - b^{2}]$

(v) We have:

$(5 x^{2} + \frac{3}{4} y^{2}) (5 x^{2} - \frac{3}{4} y^{2}) = 25 x^{4} - \frac{9}{16} y^{4} [using (a + b) (a - b) = a^{2} - b^{2}]$

(vi) We have:

$(\frac{4 x}{5} - \frac{5 y}{3}) (\frac{4 x}{5} + \frac{5 y}{3}) = \frac{16 x^{2}}{25} - \frac{25 y^{2}}{9} [using (a + b) (a - b) = a^{2} - b^{2})]$

(vii) We have:
$(x + \frac{1}{x}) (x - \frac{1}{x}) = x^{2} - \frac{1}{x^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

(viii) We have:
$(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y}) = \frac{1}{x^{2}} - \frac{1}{y^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

(ix) We have:
$(2 a + \frac{3}{b}) (2 a - \frac{3}{b}) = 4 a^{2} - \frac{9}{b^{2}} [using (a + b) (a - b) = a^{2} - b^{2}]$

Q5 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 5:

Using the formula for squaring a binomial, evaluate the following:
(i) (54)²
(ii) (82)²
(iii) (103)²
(iv) (704)²

Answer 5:

We shall use the identity (a+b)² =a² +b² +2ab.

(i)
${(54)}^{2} = (50 + 4)^{2} = {(50)}^{2} + 2 \times 50 \times 4 + {(4)}^{2} = 2500 + 400 + 16 = 2916$

(ii)
${(82)}^{2} = (80 + 2)^{2} = {(80)}^{2} + 2 \times 80 \times 2 + {(2)}^{2} = 6400 + 320 + 4 = 6724$

(iii)
${(103)}^{2} = (100 + 3)^{2} = {(100)}^{2} + 2 \times 100 \times 3 + {(3)}^{2} = 10000 + 600 + 9 = 10609$

(iv)
${(704)}^{2} = (700 + 4)^{2} = {(700)}^{2} + 2 \times 700 \times 4 + {(4)}^{2} = 490000 + 5600 + 16 = 495616$

Q6 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 6:

Using the formula for squaring a binomial, evaluate the following:
(i) (69)²
(ii) (78)²
(iii) (197)²
(iv) (999)²

Answer 6:

We shall use the identity (a-b)² = a² +b² -2ab.

(i)
${(69)}^{2} = (70 - 1)^{2} = {(70)}^{2} - 2 \times 70 \times 1 + 1 = 4900 - 140 + 1 = 4761$

(ii)
${(78)}^{2} = (80 - 2)^{2} = {(80)}^{2} - 2 \times 80 \times 2 + 4 = 6400 - 320 + 4 = 6084$

(iii)
${(197)}^{2} = (200 - 3)^{2} = {(200)}^{2} - 2 \times 200 \times 3 + 9 = 40000 - 1200 + 9 = 38809$

(iv)
${(999)}^{2} = (1000 - 1)^{2} = {(1000)}^{2} - 2 \times 1000 \times 1 + 1 = 1000000 - 2000 + 1 = 998001$

Q7 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 7:

Find the value of:
(i) (82)² − (18)²
(ii) (128)² − (72)²
(iii) 197 × 203
(iv) $\frac{198 \times 198 - 102 \times 102}{96}$
(v) (14.7 × 15.3)
(vi) (8.63)² − (1.37)²

Answer 7:

We shall use the identity (a-b) (a+b)=a² - b².

(i)
$(82)^{2} - (18)^{2} = (82 - 18) (82 + 18) = (64) (100) = 6400$

(ii)
$(128)^{2} - (72)^{2} = (128 - 72) (128 + 72) = (56) (200) = 11200$

(iii)
$197 \times 203 = (200 - 3) (200 + 3) = {(200)}^{2} - {(3)}^{2} = 40000 - 9 = 39991$

(iv)
$\frac{198 \times 198 - 102 \times 102}{96} = \frac{{(198)}^{2} - {(102)}^{2}}{96} = \frac{(198 - 102) (198 + 102)}{96} = \frac{(96) (300)}{96} = 300$

(v)
$(14.7 \times 15.3) = (15 - 0.3) \times (15 + 0.3) = (15)^{2} - (0.3)^{2} = 225 - 0.09 = 224.91$

(vi)
$(8.63)^{2} - (1.37)^{2} = (8.63 - 1.37) (8.63 + 1.37) = (7.26) (10) = 72.6$

Q8 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 8:

Find the value of the expression (9x² + 24x + 16), when x = 12.

Answer 8:

$(9 x^{2} + 24 x + 16) G i v e n, x = 12 \Rightarrow {(3 x)}^{2} + 2 (3 x) (4) + {(4)}^{2} \Rightarrow {(3 x + 4)}^{2} \Rightarrow {(3 (12) + 4)}^{2} \Rightarrow {(36 + 4)}^{2} \Rightarrow {(40)}^{2} = 1600$

Therefore, the value of the expression (9x² + 24x + 16), when x = 12, is 1600.

Q9 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 9:

Find the value of the expression (64x² + 81y² + 144xy), when x = 11 and $y = \frac{4}{3} .$

Answer 9:

$(64 x^{2} + 81 y^{2} + 144 x y) G i v e n : x = 11 y = \frac{4}{3} \Rightarrow {(8 x)}^{2} + {(9 y)}^{2} + 2 (8 x) (9 y) \Rightarrow {(8 x + 9 y)}^{2} \Rightarrow {(8 (11) + 9 (\frac{4}{3}))}^{2} \Rightarrow {(88 + 12)}^{2} \Rightarrow {(100)}^{2} \Rightarrow 10000$

Therefore, the value of the expression (64x² + 81y² + 144xy), when x = 11 and y = $\frac{4}{3}$ , is 10000.y=43

Q10 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 10:

Find the value of the expression (36x² + 25y² − 60xy), when $x = \frac{2}{3}$ and $y = \frac{1}{5} .$

Answer 10:

$(36 x^{2} + 25 y^{2} - 60 x y) \Rightarrow x = \frac{2}{3}, y = \frac{1}{5} = {(6 x)}^{2} + {(5 y)}^{2} - 2 (6 x) (5 y) = {(6 x - 5 y)}^{2} = {(6 (\frac{2}{3}) - 5 (\frac{1}{5}))}^{2} = {(4 - 1)}^{2} = {(3)}^{2} \Rightarrow 9$

Q11 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 11:

If $(x + \frac{1}{x}) = 4$ , find the values of
(i) $(x^{2} + \frac{1}{x^{2}})$ and
(ii) $(x^{4} + \frac{1}{x^{4}})$ .

Answer 11:

$(i) (x + \frac{1}{x}) = 4 Squaring both the sides : \Rightarrow {(x + \frac{1}{x})}^{2} = {(4)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} + 2 (x) (\frac{1}{x})) = 16 \Rightarrow (x^{2} + \frac{1}{x^{2}}) + 2 = 16 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 16 - 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 14$

Therefore, the value of x²+ $\frac{1}{x^{2}}$ is 14.

$(x^{2} + \frac{1}{x^{2}}) = 14 S q u a r i n g b o t h t h e s i d e s : \Rightarrow (x^{4} + \frac{1}{x^{4}} + 2 (x^{2}) (\frac{1}{x^{2}})) = {(14)}^{2} \Rightarrow (x^{4} + \frac{1}{x^{4}}) + 2 = 196 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 196 - 2 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 194$

Therefore, the value of x⁴+ $\frac{1}{x^{4}}$ is 194.

Q12 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 12:

If $(x - \frac{1}{x}) = 5$ , find the values of
(i) $(x^{2} + \frac{1}{x^{2}})$
(ii) $(x^{4} + \frac{1}{x^{4}}) .$

Answer 12:

$(i) (x - \frac{1}{x}) = 5 \Rightarrow Squaring both the sides : \Rightarrow {(x - \frac{1}{x})}^{2} = {(5)}^{2} \Rightarrow (x^{2} + \frac{1}{x^{2}} - 2 (x) (\frac{1}{x})) = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) - 2 = 25 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 25 + 2 \Rightarrow (x^{2} + \frac{1}{x^{2}}) = 27 Therefore, the value of (x^{2} + \frac{1}{x^{2}}) is 27 .$

$(x^{2} + \frac{1}{x^{2}}) = 27 \Rightarrow Squaring both the sides : \Rightarrow (x^{4} + \frac{1}{x^{4}} + 2 (x^{2}) (\frac{1}{x^{2}})) = {(27)}^{2} \Rightarrow (x^{4} + \frac{1}{x^{4}}) + 2 = 729 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 729 - 2 \Rightarrow (x^{4} + \frac{1}{x^{4}}) = 727$

$Therefore, the value of (x^{4} + \frac{1}{x^{4}}) is 727 .$

Q13 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 13:

Find the continued product:
(i) (x + 1)(x − 1)(x² + 1)
(ii) (x − 3)(x + 3)(x² + 9)
(iii) (3x − 2y)(3x + 2y)(9x² + 4y²)
(iv) (2p + 3)(2p − 3)(4p² + 9)

Answer 13:

$(i) (x + 1) (x - 1) (x^{2} + 1) \Rightarrow (x^{2} - x + x - 1) (x^{2} + 1) \Rightarrow (x^{2} - 1) (x^{2} + 1) \Rightarrow {(x^{2})}^{2} - {(1^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow x^{4} - 1 . Therefore, the product of (x + 1) (x - 1) (x^{2} + 1) is x^{4} - 1 .$

$(i i) (x - 3) (x + 3) (x^{2} + 9) \Rightarrow ({(x)}^{2} - {(3)}^{2}) (x^{2} + 9) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (x^{2} - 9) (x^{2} + 9) \Rightarrow {(x^{2})}^{2} - {(9)}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow x^{4} - 81 Therefore, the product of (x - 3) (x + 3) (x^{2} + 9) is x^{4} - 81 .$

$(iii) (3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) \Rightarrow ({(3 x)}^{2} - {(2 y)}^{2}) (9 x^{2} + 4 y^{2}) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (9 x^{2} - 4 y^{2}) (9 x^{2} + 4 y^{2}) \Rightarrow {(9 x^{2})}^{2} - {(4 y^{2})}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow 81 x^{4} - 16 y^{4} . Therefore, the product of (3 x - 2 y) (3 x + 2 y) (9 x^{2} + 4 y^{2}) is 81 x^{4} - 16 y^{4} .$

$(iv) (2 p + 3) (2 p - 3) (4 p^{2} + 9) \Rightarrow ({(2 p)}^{2} - {(3)}^{2}) (4 p^{2} + 9) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow (4 p^{2} - 9) (4 p^{2} + 9) \Rightarrow {(4 p^{2})}^{2} - {(9)}^{2} [according to the formula a^{2} - b^{2} = (a + b) (a - b)] \Rightarrow 16 p^{4} - 81 . Therefore, the product of (2 p + 3) (2 p - 3) (4 p^{2} + 9) is 16 p^{4} - 81 .$

Q14 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 14:

If x + y = 12 and xy = 14, find the value of (x² + y²).

Answer 14:

$x + y = 12 On squaring both the sides : \Rightarrow {(x + y)}^{2} = {(12)}^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 144 \Rightarrow x^{2} + y^{2} = 144 - 2 x y Given : x y = 14 \Rightarrow x^{2} + y^{2} = 144 - 2 (14) \Rightarrow x^{2} + y^{2} = 144 - 28 \Rightarrow x^{2} + y^{2} = 116 Therefore, the value of x^{2} + y^{2} is 116 .$

Q15 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Question 15:

If x − y = 7 and xy = 9, find the value of (x² + y²).

Answer 15:

$x - y = 7 \Rightarrow On squaring both the sides : \Rightarrow {(x - y)}^{2} = {(7)}^{2} \Rightarrow x^{2} + y^{2} - 2 x y = 49 \Rightarrow x^{2} + y^{2} = 49 + 2 x y Given : x y = 9 \Rightarrow x^{2} + y^{2} = 49 + 2 (9) \Rightarrow x^{2} + y^{2} = 49 + 18 \Rightarrow x^{2} + y^{2} = 67 . Therefore, the value of x^{2} + y^{2} i s 67 .$

RS Aggarwal solution class 8 chapter 6 Operations on Algebraic Expressions Exercise 6D

Exercise 6D

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Q1 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Q6 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Q8 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Q9 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Q10 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Q11 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

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Q12 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

Question 12:

Answer 12:

Q13 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

Question 13:

Answer 13:

Q14 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

Question 14:

Answer 14:

Q15 | Ex-6D | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions

Question 15:

Answer 15:

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