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

Answer: 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

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

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

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

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

⇒ \(x^{4} + y^{4} = 841 – 8\)

⇒ \(x^{4} + y^{4} = 833\)

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