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

## RD Sharma Solutions Class 8 Chapter 6 Exercise 6.6

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

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

here, a = 8a, b = 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:

$(\frac{3a}{2} – \frac{5b}{4})^{2}$ is in the form of $(a – b)^{2} = a^{2} – 2ab + b^{2}$

here, a = $\frac{3a}{2}$, b = $\frac{5b}{4}$

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

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

(x) $(a^{2}b – bc^{2})^{2}$

Sol:

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

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

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

=> $a^{4}b^{2} – 2a^{2}b^{2}c^{2} + b^{2}c^{4}$

Q2 Find the product of the following binomials

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

sol:

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

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

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

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

=> $4x^{2} + 4xy + y^{2}$

(ii) (a + 2b) (a – 2b)

sol:

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

here, a = a, b = 2b

=> $a^{2} – (2b)^{2}$

=> $a^{2} – 4b^{2}$

(iii) $(a^{2} + bc) – (a^{2} – bc)$

sol:

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

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

$(102)^{2}$ can be written as $(100 + 2)^{2}$

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

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 $(100 – 1)^{2}$

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

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 $(1000 + 1)^{2}$

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

=> 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:

Since, $\frac{113 + 87}{2}$ = $\frac{200}{2}$ = 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:

Since, $\frac{95 + 105}{2}$ = $\frac{200}{2}$ = 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:

The numerator is in the form of  (a + b) (a – b) = $a^{2} – b^{2}$

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

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

=> $\frac{(58)^{2} – (42)^{2}}{16}$ = 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) = $a^{2} – b^{2}$

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

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

=> $\frac{(8.63\times 8.63) – (1.37\times 1.37)}{0.726}$  = $\frac{10\times 7.26}{0.726}$

=> $\frac{(8.63\times 8.63) – (1.37\times 1.37)}{0.726}$ = $10\times 10$

=> $\frac{(8.63\times 8.63) – (1.37\times 1.37)}{0.726}$ = 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 $x + \frac{1}{x}$, $x – \frac{1}{x}$

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

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

=> $x – \frac{1}{x} = \pm \sqrt{20}$

=> $x – \frac{1}{x} = \pm 4$

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

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

=> $16x^{2} + 24x + 9$ = $(4(\frac{7}{4}) + 3)^{2}$

=> $16x^{2} + 24x + 9$ = $(7 + 3)^{2}$

=> $16x^{2} + 24x + 9$ = $(10)^{2}$

=> $16x^{2} + 24x + 9$ = 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,

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

= $x^{2} – 2 + \frac{1}{x^{2}}$

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

=> $(x – \frac{1}{x})^{2} = 142 – 2$

=> $(x – \frac{1}{x})^{2} = 140$

=> $x – \frac{1}{x}= \pm \sqrt{40}$

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) and (2x – 3y) and then subtracting them, we get

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

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

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

=> $(14)^{2} – (2)^{2} = 24xy$

=> (14 + 2) (14 – 2) = 24xy

=> $16\times 12 = 24xy$

=> 24xy = 192

=>     xy = $\frac{192}{8}$

=>      xy = 8

hence, 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

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

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

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

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

=> $x^{4} + y^{4} = 841 – 8$

=> $x^{4} + y^{4} = 833$

#### Practise This Question

ColumnAColumnB1.Areaa.67m2.Perimeterb.Rs.6/m23.Cost per m2c.100m2d.Rs.6m2e.100/m2

Which of the following options show the correct match of parameters in column A with possible values in column B?