# RD Sharma Solutions Class 8 Algebraic Expressions And Identities Exercise 6.6

### RD Sharma Class 8 Solutions Chapter 6 Ex 6.6 PDF Free Download

#### Exercise 6.6

Q1 Write the following squares of binomials as trinomials

We know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$ and

$(a – b)^{2} = a^{2} – 2ab + b^{2}$

(i) $(x + 2)^{2}$

Sol:

$(x + 2)^{2}$ is in the form of $(a + b)^{2} = a^{2} + 2ab + b^{2}$

here, a = x , b = 2

⇒ $x^{2} + 2\times x\times 2 + b^{2}$

⇒ $x^{2} + 4x + b^{2}$

(ii) $(8a + 3b)^{2}$

Sol:

$(8a + 3b)^{2}$ is in the form of $(x + y)^{2} = x^{2} + 2xy + y^{2}$

here, x = 8a, y = 3b

⇒ $(8a)^{2} + 2\times (8a)\times (3b) + (3b)^{2}$

⇒ $64^{2} + 48ab + 36b^{2}$

(iii) $(2m + 1)^{2}$

Sol:

$(2m + 1)^{2}$ is in the form of $(a + b)^{2} = a^{2} + 2ab + b^{2}$

here, a =2m, b = 1

⇒ $(2m)^{2} + 2\times (2m)\times (1) + (1)^{2}$

⇒ $4m^{2} + 4m + 1$

(iv) $(9a + \frac{1}{6})^{2}$

Sol:

$(9a + \frac{1}{6})^{2}$ is in the form of $(a + b)^{2} = a^{2} + 2ab + b^{2}$

here, a = 9a, b = $\frac{1}{6}$

⇒ $(9a)^{2} + 2\times (9a)\times (\frac{1}{6}) + (\frac{1}{6})^{2}$

⇒ $81a^{2} + 3a + \frac{1}{36}$

(v) $(x + \frac{x^{2}}{2})^{2}$

Sol:

$(x + \frac{x^{2}}{2})^{2}$ is in the form of $(a + b)^{2} = a^{2} + 2ab + b^{2}$

here, a = x, b = $\frac{x^{2}}{2}$

⇒ $x^{2} + 2\times x\times (\frac{x^{2}}{2}) + (\frac{x^{2}}{2})^{2}$

⇒ $x^{2} + x^{3} + \frac{x^{4}}{4}$

(vi) $(\frac{x}{4} – \frac{y}{3})^{2}$

Sol:

$(\frac{x}{4} – \frac{y}{3})^{2}$ is in the form of $(a – b)^{2} = a^{2} – 2ab + b^{2}$

here, a = $\frac{x}{4}$, b = $\frac{y}{3}$

⇒ $(\frac{x}{4})^{2} – 2\times (\frac{x}{4})\times (\frac{y}{3}) + (\frac{y}{3})^{2}$

⇒ $\frac{x^{2}}{16} – \frac{1}{6}xy + \frac{y^{2}}{9}$

(vii) $(3x – \frac{1}{3x})^{2}$

Sol:

$(3x – \frac{1}{3x})^{2}$ is in the form of $(a – b)^{2} = a^{2} – 2ab + b^{2}$

here, a = 3x, b = $\frac{1}{3x}$

⇒ $(3x)^{2} – 2\times 3x\times (\frac{1}{3x}) + (\frac{1}{3x})^{2}$

⇒ $9x^{2} – 2 + \frac{1}{9x^{2}}$

(viii) $(\frac{x}{y} – \frac{y}{x})^{2}$

Sol:

$(\frac{x}{y} – \frac{y}{x})^{2}$ is in the form of $(a – b)^{2} = a^{2} – 2ab + b^{2}$

here, a = $\frac{x}{y}$, b = $\frac{y}{x}$

⇒ $(\frac{x}{y})^{2} – 2\times (\frac{x}{y})\times (\frac{y}{x}) + (\frac{y}{x})^{2}$

⇒ $\frac{x^{2}}{y^{2}} – 2 + \frac{y^{2}}{x^{2}}$

(ix) $(\frac{3a}{2} – \frac{5b}{4})^{2}$

Sol:

$(\frac{3a}{2} – \frac{5b}{4})^{2}$

Here, x = $\frac{3a}{2}$, y = $\frac{5b}{4}$

⇒ $(\frac{3a}{2})^{2} – 2\times (\frac{3a}{2})\times (\frac{5b}{4}) + (\frac{5b}{4})^{2}$

⇒ $\frac{9a^{2}}{4} – \frac{15ab}{4} + \frac{25b^{2}}{16}$

(x) $(a^{2}b – bc^{2})^{2}$

Sol:

$(a^{2}b – bc^{2})^{2}$ is in the form of $(x – y)^{2} = x^{2} – 2xy + y^{2}$

here, x = $a^{2}b$, y = $bc^{2}$

⇒ $(a^{2}b)^{2} – 2\times (a^{2}b)\times (bc^{2}) + (bc^{2})^{2}$

⇒ $a^{4}b^{2} – 2a^{2}b^{2}c^{2} + b^{2}c^{4}$

Q2. Find the product of the following binomials

(i) (2x + y) (2x + y)

sol:

(2x + y) (2x + y) can be written as $(2x + y)^{2}$

⇒ $(2x + y)^{2}$

⇒ $(2x)^{2} + 2\times (2x)\times (y) + y^{2}$

⇒ $4x^{2} + 4xy + y^{2}$

(ii) (a + 2b) (a – 2b)

sol:

(a + 2b) (a – 2b) can be written as $a^{2} – (2b)^{2}$

⇒ $a^{2} – 4b^{2}$

(iii) $(a^{2} + bc) (a^{2} – bc)$

Sol:

$(a^{2} + bc) (a^{2} – bc)$

= $(a^{2})^{2} – (bc)^{2}$

= $a^{4} – b^{2}c^{2}$

(iv) $(\frac{4x}{5} – \frac{3y}{4}) (\frac{4x}{5} + \frac{3y}{4})$

sol:

we know that (a + b) (a – b) = $a^{2} – b^{2}$

here, a = $\frac{4x}{5}$, b = $\frac{3y}{4}$

⇒ $(\frac{4x}{5})^{2} – (\frac{3y}{4})^{2}$

⇒ $\frac{16x^{2}}{25} – \frac{9y^{2}}{16}$

(v) $(2x + \frac{3}{y}) (2x – \frac{3}{y})$

sol:

we know that (a + b) (a – b) = $a^{2} – b^{2}$

here, a = 2x, b = $\frac{3}{y}$

⇒ $(2x)^{2} – (\frac{3}{y})^{2}$

⇒ $4x^{2} – \frac{9}{y^{2}}$

(vi) $(2a^{3} + b^{3}) (2a^{3} – b^{3})$

sol:

we know that (a + b) (a – b) = $a^{2} – b^{2}$

here, a = $2a^{3}$, b = $b^{3}$

⇒ $(2a^{3})^{2} – (b^{3})^{2}$

⇒ $4a^{6} – b^{6}$

(vii) $(x^{4} + \frac{2}{x^{2}}) (x^{4} – \frac{2}{x^{2}})$

sol:

we know that (a + b) (a – b) = $a^{2} – b^{2}$

here, a= $x^{4}$, b = $\frac{2}{x^{2}}$

⇒ $(x^{4})^{2} – (\frac{2}{x^{2}})^{2}$

⇒ $x^{8} – \frac{4}{x^{4}}$

(viii) $(x^{3} + \frac{1}{x^{3}}) (x^{3} – \frac{1}{x^{3}})$

sol:

we know that (a + b) (a – b) = $a^{2} – b^{2}$

here, a = $x^{3}$, b = $\frac{1}{x^{3}}$

⇒ $(x^{3})^{2} – (\frac{1}{x^{3}})^{2}$

⇒ $x^{6} – \frac{1}{x^{6}}$

Q3 Using the formula for squaring a binomial, evaluate the following

(i) $(102)^{2}$

sol:

102 can be written as (100 + 2)

Using identity, $(a + b)^{2} = a^{2} + 2ab + b^{2}$, we have

$(102)^{2}$ = $(100 + 2)^{2}$

Here, a = 100, b = 2

⇒ $(100 + 2)^{2}$

⇒ $(100)^{2} + 2\times (100)\times 2 + 2^{2}$

⇒ 10000 + 400 + 4

⇒ 10404

(ii) $(99)^{2}$

sol:

$(99)^{2}$ can be written as $(100 – 1)^{2}$

Using identity, $(a – b)^{2} = a^{2} – 2ab + b^{2}$

here, a = 100, b = 1

⇒ $(100 – 1)^{2}$

⇒ $(100)^{2} – 2\times (100)\times 1 + 1^{2}$

⇒ 10000 – 200 + 1

⇒ 9801

(iii) $(1001)^{2}$

sol:

$(1001)^{2}$ can be written as $(1000 + 1)^{2}$

Using identity, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

here, a = 1000, b = 1

⇒ $(1000 + 1)^{2}$

⇒ $(1000)^{2} + 2\times (1000)\times 1 + 1^{2}$

⇒ 1000000 + 2000 + 1

⇒ 1002001

(iv) $(999)^{2}$

sol:

$(999)^{2}$ can be written as $(1000 – 1)^{2}$

we know that, $(a – b)^{2} = a^{2} – 2ab + b^{2}$

here, a = 1000, b = 1

⇒ $(1000 – 1)^{2}$

⇒ $(1000)^{2} – 2\times (1000)\times 1 + 1^{2}$

⇒ 1000000 – 2000 + 1

⇒ 998001

(v) $(703)^{2}$

sol:

$(703)^{2}$ can be written as $(700 + 3)^{2}$

we know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

here, a = 700, b = 3

⇒ $(700 + 3)^{2}$

⇒ $(700)^{2} + 2\times (700)\times 3 + 3^{2}$

⇒ 490000 + 4200 + 9

⇒ 494209

Q4 Simplify the following using the formula: (a + b) (a – b) = $a^{2} – b^{2}$

(i) $(82)^{2} – (18)^{2}$

sol:

$(82)^{2} – (18)^{2}$

here, a = 82, b = 18

⇒ (82 + 18) (82 – 18)

⇒ $100 \times 64$

⇒ 6400

(ii) $(467)^{2} – (33)^{2}$

sol:

$(467)^{2} – (33)^{2}$

here, a = 467, b = 33

⇒ (467 + 33) (467 – 33)

⇒ $500 \times 434$

⇒ 217000

(iii) $(79)^{2} – (69)^{2}$

sol:

$(79)^{2} – (69)^{2}$

here, a = 79, b = 69

⇒ (79 + 69) (79 – 69)

⇒ $148 \times 10$

⇒ 1480

(iv) $197 \times 203$

sol:

Since, $\frac{197 + 203}{2}$ = $\frac{400}{2}$ = 200

$197 \times 203$ can be written as (200 – 3)(200 + 3)

⇒ (200 – 3) (200 + 3)

⇒ $(200)^{2} – (3)^{2}$

⇒ 40000 – 9

⇒ 39991

(v) $113 \times 87$

sol:

Average of 113 and 87 is, $\frac{113 + 87}{2}$ = $\frac{200}{2}$ = 100

$113 \times 87$ can be written as (100 + 13) (100 – 13)

⇒ (100 + 13) (100 – 13)

⇒ $(100)^{2} – (13)^{2}$

⇒ 10000 – 169

⇒ 9831

(vi) $95 \times 105$

sol:

Average of 95 and 105 is, $\frac{95 + 105}{2}$ = $\frac{200}{2}$ = 100

$95 \times 105$ can be written as (100 + 5) (100 – 5)

⇒ (100 + 5) (100 – 5)

⇒ $(100)^{2} – (5)^{2}$

⇒ 10000 – 25

⇒ 9975

(vii) $1.8 \times 2.2$

sol:

Since, $\frac{1.8 + 2.2}{2}$ = $\frac{4}{2}$ = 2

$1.8 \times 2.2$ can be written as (2 + 0.2) (2 – 0.2)

⇒ (2 + 0.2) (2 – 0.2)

⇒ $(2)^{2} – (0.2)^{2}$

⇒ 4 – 0.04

⇒ 3.96

(viii) $9.8 \times 10.2$

sol:

Since, $\frac{9.8 + 10.2}{2}$ = $\frac{20}{2}$ = 10

$9.8 \times 10.2$ can be written as (10 + 0.2) (10 – 0.2)

⇒ (10 + 0.2) (10 – 0.2)

⇒ $(10)^{2} – (0.2)^{2}$

⇒ 100 – 0.04

⇒ 99.96

Q5 Simplify the following using identities

(i) $\frac{(58)^{2} – (42)^{2}}{16}$

sol:

$\frac{(58)^{2} – (42)^{2}}{16}$ = $\frac{(58 + 42) (58 – 42)}{16}$

Using identity: (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $\frac{(58)^{2} – (42)^{2}}{16}$ = $\frac{100\times 16}{16}$

⇒ $\frac{(58)^{2} – (42)^{2}}{16}$ = 100

(ii) $(178\times 178) – (22\times 22)$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $(178\times 178) – (22\times 22)$ = $(178)^{2} – (22)^{2}$

⇒ $(178\times 178) – (22\times 22)$ = (178 + 22) (178 – 22)

⇒ $(178\times 178) – (22\times 22)$ = $200\times 156$

⇒ $(178\times 178) – (22\times 22)$ = 31200

(iii) $\frac{(198\times 198) – (102\times 102)}{96}$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $\frac{(198\times 198) – (102\times 102)}{96}$ = $\frac{(198)^{2} – (102)^{2}}{96}$

⇒ $\frac{(198\times 198) – (102\times 102)}{96}$ = $\frac{(198 + 102) (198 – 102) }{96}$

⇒ $\frac{(198\times 198) – (102\times 102)}{96}$ = $\frac{300\times 96}{96}$

⇒ $\frac{(198\times 198) – (102\times 102)}{96}$ = 300

(iv) $(1.73\times 1.73) – (0.27\times 0.27)$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $(1.73\times 1.73) – (0.27\times 0.27)$ = $(1.73)^{2} – (0.27)^{2}$

⇒ $(1.73\times 1.73) – (0.27\times 0.27)$ = (1.73 + 0.27) (1.73 – 0.27)

⇒ $(1.73\times 1.73) – (0.27\times 0.27)$ = $2\times 1.46$

⇒ $(1.73\times 1.73) – (0.27\times 0.27)$ = 2.92

(v) $\frac{(8.63\times 8.63) – (1.37\times 1.37)}{0.726}$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $\frac{(8.63\times 8.63) – (1.37\times 1.37)}{0.726}$ = $\frac{(8.63)^{2} – (1.37)^{2}}{0.726}$

= $\frac{(8.63+ 1.37) (8.63 – 1.37) }{0.726}$

= $\frac{10\times 7.26}{0.726}$

= $10\times 10$

= 100

Q6 Find the value of x, if:

(i) $4x = (52)^{2} – (48)^{2}$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $4x = (52)^{2} – (48)^{2}$

⇒ 4x = (52 + 48) (52 – 48)

⇒ 4x = $100\times 4$

⇒ 4x = 400

⇒ x = $\frac{400}{4}$

⇒ x = 100

(ii) $14x = (47)^{2} – (33)^{2}$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $14x = (47)^{2} – (33)^{2}$

⇒ 14x = (47 + 33) (47 – 33)

⇒ 14x = $80\times 14$

⇒ 14x = 1120

⇒ x = $\frac{1120}{14}$

⇒ x = 80

(iii) $5x = (50)^{2} – (40)^{2}$

sol:

we know that, (a + b) (a – b) = $a^{2} – b^{2}$

⇒ $5x = (50)^{2} – (40)^{2}$

⇒ 5x = (50 + 40) (50 – 40)

⇒ 5x = $90\times 10$

⇒ 5x = 900

⇒ x = $\frac{900}{5}$

⇒ x = 180

Q7 If $x + \frac{1}{x} = 20$ , find the value of $x^{2} + \frac{1}{x^{2}}$

sol:

Given that,

$x + \frac{1}{x} = 20$

squaring on both sides

⇒ $(x + \frac{1}{x})^{2} = (20)^{2}$

⇒ $(x + \frac{1}{x})^{2} = 400$

we know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

⇒ $x^{2} + 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2} = 400$

⇒ $x^{2} + 2 + (\frac{1}{x})^{2} = 400$

⇒ $x^{2} + \frac{1}{x^{2}} = 400 – 2$

⇒ $x^{2} + \frac{1}{x^{2}} = 398$

hence, $x^{2} + \frac{1}{x^{2}} = 398$

Q8 If $x – \frac{1}{x} = 3$, find the values of $x^{2} + \frac{1}{x^{2}}$, $x^{4} + \frac{1}{x^{4}}$

sol:

Given that,

$x – \frac{1}{x} = 3$

squaring on both sides

⇒ $(x – \frac{1}{x})^{2} = (3)^{2}$

⇒ $(x – \frac{1}{x})^{2} = 9$

we know that, $(a – b)^{2} = a^{2} – 2ab + b^{2}$

⇒ $x^{2} – 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2} = 9$

⇒ $x^{2} – 2 + (\frac{1}{x})^{2} = 9$

⇒ $x^{2} + \frac{1}{x^{2}} = 9 + 2$

⇒ $x^{2} + \frac{1}{x^{2}} = 11$

Again, squaring on both sides

⇒ $(x^{2} + \frac{1}{x^{2}})^{2} = (11)^{2}$

⇒ $(x^{2} + \frac{1}{x^{2}})^{2} = 121$

⇒ $x^{2} + 2\times (x^{2})\times (\frac{1}{x^{2}}) + (\frac{1}{x^{2}})^{2} = 121$

⇒ $x^{4} + 2 + \frac{1}{x^{4}} = 121$

⇒ $x^{4} + \frac{1}{x^{4}} = 121 – 2$

⇒ $x^{4} + \frac{1}{x^{4}} = 119$

hence, $x^{4} + \frac{1}{x^{4}} = 119$

Q9 If $x^{2} + \frac{1}{x^{2}} = 18$, find the values of $x + \frac{1}{x}$, $x – \frac{1}{x}$

sol:

Given that,

$x^{2} + \frac{1}{x^{2}} = 18$, find the values of $x + \frac{1}{x}$, $x – \frac{1}{x}$

Let’s solve first, $x + \frac{1}{x}$

squaring the above equation

$(x + \frac{1}{x})^{2}$ = $x^{2} + 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2}$

= $x^{2} + 2 + \frac{1}{x^{2}}$

⇒ $(x + \frac{1}{x})^{2}$ = $x^{2} + 2 + \frac{1}{x^{2}}$

⇒ $(x + \frac{1}{x})^{2}$ = 18 + 2

⇒ $(x + \frac{1}{x})^{2}$ = 20

⇒ $x + \frac{1}{x} = \pm \sqrt{20}$

consider, $x – \frac{1}{x}$

squaring the above equation

$(x – \frac{1}{x})^{2}$ = $x^{2} – 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2}$

= $x^{2} – 2 + \frac{1}{x^{2}}$

⇒ $(x – \frac{1}{x})^{2}$ = $x^{2} – 2 + \frac{1}{x^{2}}$

⇒ $(x – \frac{1}{x})^{2}$ = 18 – 2

⇒ $(x – \frac{1}{x})^{2}$ = 16

⇒ $x – \frac{1}{x} = \pm \sqrt{16}$

⇒ $x – \frac{1}{x} = \pm 4$

Q10: If x + y = 4 and xy = 2,find the value of $x^{2} + y^{2}$

sol:

Given that,

x + y = 4 and xy = 2

we know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

⇒ $x^{2} + y^{2} = (x + y)^{2} – 2xy$

⇒ $x^{2} + y^{2} = 4^{2} – (2 \times 2)$

⇒ $x^{2} + y^{2} = 16 – 4$

⇒ $x^{2} + y^{2} = 12$

Q11 : If x – y = 7 and xy = 9,find the value of $x^{2} + y^{2}$

sol:

Given that,

x – y = 7 and xy = 9

we know that, $(a – b)^{2} = a^{2} – 2ab + b^{2}$

⇒ $x^{2} – y^{2} = (x – y)^{2} + 2xy$

⇒ $x^{2} – y^{2} = 7^{2} + (2 \times 9)$

⇒ $x^{2} – y^{2} = 49 + 18$

⇒ $x^{2} – y^{2} = 67$

Q12: If 3x + 5y = 11 and xy = 2, find the value of $9x^{2} + 25y^{2}$

sol:

Given that,

3x + 5y = 11 and xy = 2

we know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

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

⇒ $(3x + 5y)^{2} = 9x^{2} + 30xy + 25y^{2}$

⇒ $9x^{2} + 25y^{2} = (3x + 5y)^{2} – 10xy$

⇒ $9x^{2} + 25y^{2} = (11)^{2} – (30\times 2)$

⇒$9x^{2} + 25y^{2} = 121 – 60$

⇒ $9x^{2} + 25y^{2} = 61$

Q13 Find the values of the following expressions

(i) $16x^{2} + 24x + 9$ when x = $\frac{7}{4}$

Sol:

Given, $16x^{2} + 24x + 9$ and x = $\frac{7}{4}$

we know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

$16x^{2} + 24x + 9$ = $(4x + 3)^{2}$

Substituting the value of x i.e. x = $\frac{7}{4}$

= $(4(\frac{7}{4}) + 3)^{2}$

= $(7 + 3)^{2}$

= $(10)^{2}$

= 100

(ii) $64x^{2} + 81y^{2} + 144xy$ when x = 11 and y = $\frac{4}{3}$

sol:

Given, $64x^{2} + 81y^{2} + 144xy$ and x = 11, y = $\frac{4}{3}$

we know that, $(a + b)^{2} = a^{2} + 2ab + b^{2}$

$64x^{2} + 81y^{2} + 144xy$ = $(8x + 9y)^{2}$

⇒ $64x^{2} + 81y^{2} + 144xy$ = $(8(11) + 9(\frac{4}{3}))^{2}$

⇒ $64x^{2} + 81y^{2} + 144xy$ = $(88 + 12)^{2}$

⇒ $64x^{2} + 81y^{2} + 144xy$ = $(100)^{2}$

⇒ $64x^{2} + 81y^{2} + 144xy$ = $10000$

(iii) $81x^{2} + 16y^{2} – 72xy$ when x = $\frac{2}{3}$

and y = $\frac{3}{4}$

sol:

Given that, $81x^{2} + 16y^{2} – 72xy$ and x = $\frac{2}{3}$,

y = $\frac{3}{4}$

we know that, $(a – b)^{2} = a^{2} – 2ab + b^{2}$

$81x^{2} + 16y^{2} – 72xy$ = $(9x – 4y)^{2}$

⇒ $81x^{2} + 16y^{2} – 72xy$ = $(9(\frac{2}{3}) – 4(\frac{3}{4}))^{2}$

⇒ $81x^{2} + 16y^{2} – 72xy$ = $(6 – 3)^{2}$

⇒ $81x^{2} + 16y^{2} – 72xy$ = $3^{2}$

⇒ $81x^{2} + 16y^{2} – 72xy$ = 9

Q14 If $x + \frac{1}{x}$ = 9, find the value of $x^{4} + \frac{1}{x^{4}}$

sol:

Given,

$x + \frac{1}{x}$ = 9

squaring on both sides

$(x + \frac{1}{x})^{2} = 9^{2}$

⇒ $(x + \frac{1}{x})^{2} = 81$

⇒ $x^{2} + 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2} = 81$

⇒ $x^{2} + 2 + \frac{1}{x^{2}} = 81$

⇒ $x^{2} + \frac{1}{x^{2}} = 81 – 2$

⇒ $x^{2} + \frac{1}{x^{2}} = 79$

Again, squaring on both sides

$(x^{2} + \frac{1}{x^{2}})^{2} = (79)^{2}$

⇒ $(x^{2} + \frac{1}{x^{2}})^{2} = 6241$

⇒ $(x^{2})^{2} + 2\times (x^{2})\times (\frac{1}{x^{2}}) + (\frac{1}{x^{2}})^{2}$ = 6241

⇒ $x^{4} + 2 + \frac{1}{x^{4}} = 6241$

⇒ $x^{4} + \frac{1}{x^{4}} = 6241 – 2$

⇒ $x^{4} + \frac{1}{x^{4}} = 6239$

Q15 If $x + \frac{1}{x}$ = 12, find the value of $x – \frac{1}{x}$

sol:

Given that,

$x + \frac{1}{x}$ = 12

squaring on both sides

$(x + \frac{1}{x})^{2} = (12)^{2}$

⇒ $(x + \frac{1}{x})^{2} = 144$

⇒ $x^{2} + 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2} = 144$

⇒ $x^{2} + 2 + \frac{1}{x^{2}} = 144$

⇒ $x^{2} + \frac{1}{x^{2}} = 144 – 2$

⇒ $x^{2} + \frac{1}{x^{2}} = 142$

Here,

$(x – \frac{1}{x})^{2} = x^{2} – 2\times x\times \frac{1}{x} + (\frac{1}{x})^{2}$

= $x^{2} – 2 + \frac{1}{x^{2}}$

⇒ $(x – \frac{1}{x})^{2} = x^{2} – 2 + \frac{1}{x^{2}}$

⇒ $(x – \frac{1}{x})^{2} = 142 – 2$

⇒ $(x – \frac{1}{x})^{2} = 140$

⇒ $x – \frac{1}{x}= \pm \sqrt{140}$

Q16 If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy

Sol:

we know that, $(a + b) (a – b) = a^{2} – b^{2}$

Given, 2x + 3y = 14 and 2x – 3y = 2

squaring 2x + 3y = 14 and 2x – 3y = 2 and then subtracting them, we get

(2x + 3y)^2 = 14^2 and (2x – 3y)^2 = 2^2

Subtract 2nd term form first, we have

(2x + 3y)^2 – (2x – 3y)^2 = 14^2 – 2^2 …….(1)

Solve L.H.S. first

$(2x + 3y)^{2} – (2x – 3y)^{2} = [(2x + 3y) + (2x – 3y)][(2x + 3y) – (2x – 3y)]$

⇒ $(2x + 3y)^{2} – (2x – 3y)^{2} = 4x \times 6y$

⇒ $(2x + 3y)^{2} – (2x – 3y)^{2} = 24xy$

Now, from equation (1)

$(14)^{2} – (2)^{2} = 24xy$

⇒ (14 + 2) (14 – 2) = 24xy

⇒ $16\times 12 = 24xy$

⇒ 24xy = 192

⇒ xy = $\frac{192}{8}$

⇒ xy = 8

Q17 If $x^{2} + y^{2} = 29$ and xy = 2, Find the value of

(i) x + y

sol

Given,

$x^{2} + y^{2} = 29$ and xy = 2

squaring the (x + y)

$(x + y)^{2} = x^{2} + 2\times x\times y + y^{2}$

⇒ $(x + y)^{2} = 29 + (2\times 2)$

⇒ $(x + y)^{2} = 29 + 4$

⇒ $(x + y)^{2} = 33$

⇒ $x + y = \pm \sqrt{33}$

(ii) x – y

sol:

Given,

$x^{2} + y^{2} = 29$ and xy = 2

squaring the (x – y)

$(x – y)^{2} = x^{2} – 2\times x\times y + y^{2}$

⇒ $(x – y)^{2} = 29 – (2\times 2)$

⇒ $(x – y)^{2} = 29 – 4$

⇒ $(x – y)^{2} = 25$

⇒ $x – y = \pm \sqrt{25}$

⇒ $x + y = \pm 5$

(iii) $x^{4} + y^{4}$

Sol:

Given,

$x^{2} + y^{2} = 29$ and xy = 2

$(x^{2} + y^{2})^{2} = x^{4} + 2x^{2}y^{2} + y^{4}$

⇒ $x^{4} + y^{4} = (x^{2} + y^{2})^{2} – 2x^{2}y^{2}$

⇒ $x^{4} + y^{4} = (x^{2} + y^{2})^{2} – 2(xy)^{2}$

⇒ $x^{4} + y^{4} = (29)^{2} – 2(2)^{2}$

⇒ $x^{4} + y^{4} = 841 – 8$

⇒ $x^{4} + y^{4} = 833$