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Chapter 1

Rational & Irrational Numbers — Exercise 1(C)

Class - 9 Concise Mathematics Selina

Exercise 1(C)

Question 1(a)

If x = 5−2,x+1x\sqrt{5} - 2, x + \dfrac{1}{x} is equal to :

  1. 252\sqrt{5}

  2. 4

  3. 454\sqrt{5}

  4. -4

Answer

Given,

x = 5−2\sqrt{5} - 2

1x=15−2\dfrac{1}{x} = \dfrac{1}{\sqrt{5} - 2}

Rationalizing,

⇒1x=15−2×5+25+2=5+2(5)2−(2)2=5+25−4=5+21=5+2.⇒x+1x=5−2+5+2=25.\Rightarrow \dfrac{1}{x} = \dfrac{1}{\sqrt{5} - 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] = \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} \\[1em] = \dfrac{\sqrt{5} + 2}{5 - 4} \\[1em] = \dfrac{\sqrt{5} + 2}{1} \\[1em] = \sqrt{5} + 2. \\[1em] \Rightarrow x + \dfrac{1}{x} = \sqrt{5} - 2 + \sqrt{5} + 2 \\[1em] = 2\sqrt{5}.

Hence, Option 1 is the correct option.

Question 1(b)

If x = 1 + 2, then (x+1x)2\sqrt{2}, \text{ then } \Big(x + \dfrac{1}{x}\Big)^2 is :

  1. 222\sqrt{2}

  2. 8

  3. 4

  4. 424\sqrt{2}

Answer

Given,

x = 1+21 + \sqrt{2}

1x=11+2\dfrac{1}{x} = \dfrac{1}{1 + \sqrt{2}}

Rationalizing,

⇒11+2×1−21−2=1−2(1)2−(2)2=1−21−2=1−2−1=−1+2.⇒(x+1x)2=[1+2+(−1+2)]2=[1−1+2+2]2=[22]2=8.\Rightarrow \dfrac{1}{1 + \sqrt{2}} \times \dfrac{1 - \sqrt{2}}{1 - \sqrt{2}} \\[1em] = \dfrac{1 - \sqrt{2}}{(1)^2 - (\sqrt{2})^2} \\[1em] = \dfrac{1 - \sqrt{2}}{1 - 2} \\[1em] = \dfrac{1 - \sqrt{2}}{-1} \\[1em] = -1 + \sqrt{2}. \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = [1 + \sqrt{2} + (-1 + \sqrt{2})]^2 \\[1em] = [1 - 1 + \sqrt{2} + \sqrt{2}]^2 \\[1em] = [2\sqrt{2}]^2 \\[1em] = 8.

Hence, Option 2 is the correct option.

Question 1(c)

227+31243\dfrac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}} is equal to :

  1. 232\sqrt{3}

  2. 323\sqrt{2}

  3. 3

  4. 3+2\sqrt{3} + \sqrt{2}

Answer

Solving,

⇒227+31243⇒2×33+3×2343⇒63+6343⇒12343⇒3.\Rightarrow \dfrac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 \times 3\sqrt{3} + 3 \times 2\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow \dfrac{6\sqrt{3} + 6\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow \dfrac{12\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow 3.

Hence, Option 3 is the correct option.

Question 1(d)

(5−3)2(\sqrt{5} - \sqrt{3})^2 is :

  1. 8+2158 + 2\sqrt{15}

  2. 8+158 + \sqrt{15}

  3. 8−158 - \sqrt{15}

  4. 8−2158 - 2\sqrt{15}

Answer

Solving,

⇒(5−3)2⇒(5)2+(3)2−2×5×3⇒5+3−215⇒8−215.\Rightarrow (\sqrt{5} - \sqrt{3})^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{3})^2 - 2 \times \sqrt{5} \times \sqrt{3} \\[1em] \Rightarrow 5 + 3 - 2\sqrt{15} \\[1em] \Rightarrow 8 - 2\sqrt{15}.

Hence, Option 4 is the correct option.

Question 1(e)

34+7\dfrac{3}{4 + \sqrt{7}} is equal to :

  1. 13(4−7)\dfrac{1}{3}(4 - \sqrt{7})

  2. 3(4−7)3(4 - \sqrt{7})

  3. 13(4+7)\dfrac{1}{3}(4 + \sqrt{7})

  4. 3(4+7)3(4 + \sqrt{7})

Answer

Rationalizing,

⇒34+7×4−74−7⇒3(4−7)(4)2−(7)2⇒3(4−7)16−7⇒3(4−7)9⇒13(4−7).\Rightarrow \dfrac{3}{4 + \sqrt{7}} \times \dfrac{4 - \sqrt{7}}{4 - \sqrt{7}} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{16 - 7} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{9} \\[1em] \Rightarrow \dfrac{1}{3}(4 - \sqrt{7}).

Hence, Option 1 is the correct option.

Question 1(f)

17−5\dfrac{1}{7 - \sqrt{5}} is equal to :

  1. 4(7+5)4(7 + \sqrt{5})

  2. 144(7+5)\dfrac{1}{44}(7 + \sqrt{5})

  3. 144(7−5)\dfrac{1}{44}(7 - \sqrt{5})

  4. 4(7−5)4(7 - \sqrt{5})

Answer

Rationalizing,

⇒17−5×7+57+5⇒7+5(7)2−(5)2⇒7+549−5⇒144(7+5).\Rightarrow \dfrac{1}{7 - \sqrt{5}} \times \dfrac{7 + \sqrt{5}}{7 + \sqrt{5}} \\[1em] \Rightarrow \dfrac{7 + \sqrt{5}}{(7)^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{7 + \sqrt{5}}{49 - 5} \\[1em] \Rightarrow \dfrac{1}{44}(7 + \sqrt{5}).

Hence, Option 2 is the correct option.

Question 1(g)

If x = 2−1, then (x−1x)2\sqrt{2} - 1, \text{ then } \Big(x - \dfrac{1}{x}\Big)^2 is :

  1. 222\sqrt{2}

  2. 8

  3. 4

  4. 2−22 - \sqrt{2}

Answer

Given,

x = 2−1\sqrt{2} - 1

1x=12−1\dfrac{1}{x} = \dfrac{1}{\sqrt{2} - 1}

Rationalizing,

⇒1x=12−1×2+12+1=2+1(2)2−(1)2=2+12−1=2+11=2+1.⇒(x−1x)2=[2−1−(2+1)]2=[2−2−1−1]2=[−2]2=4.\Rightarrow \dfrac{1}{x} = \dfrac{1}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[1em] = \dfrac{\sqrt{2} + 1}{(\sqrt{2})^2 - (1)^2} \\[1em] = \dfrac{\sqrt{2} + 1}{2 - 1} \\[1em] = \dfrac{\sqrt{2} + 1}{1} \\[1em] = \sqrt{2} + 1. \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = [\sqrt{2} - 1 - (\sqrt{2} + 1)]^2 \\[1em] = [\sqrt{2} - \sqrt{2} - 1 - 1]^2 \\[1em] = [-2]^2 \\[1em] = 4.

Hence, Option 3 is the correct option.

Question 1(h)

5−75+7−5+75−7\dfrac{5 - \sqrt{7}}{5 + \sqrt{7}} - \dfrac{5 + \sqrt{7}}{5 - \sqrt{7}} is equal to :

  1. 10710\sqrt{7}

  2. 197\dfrac{1}{9}\sqrt{7}

  3. 1079\dfrac{10\sqrt{7}}{9}

  4. −1079-\dfrac{10\sqrt{7}}{9}

Answer

Given,

⇒5−75+7−5+75−7⇒(5−7)2−(5+7)2(5+7)(5−7)⇒(5)2+(7)2−2×5×7−[(5)2+(7)2+2×5×7]52−(7)2⇒25+7−107−[25+7+107]25−7⇒25−25+7−7−107−10718⇒−20718⇒−1079.\Rightarrow \dfrac{5 - \sqrt{7}}{5 + \sqrt{7}} - \dfrac{5 + \sqrt{7}}{5 - \sqrt{7}} \\[1em] \Rightarrow \dfrac{(5 - \sqrt{7})^2 - (5 + \sqrt{7})^2}{(5 + \sqrt{7})(5 - \sqrt{7})} \\[1em] \Rightarrow \dfrac{(5)^2 + (\sqrt{7})^2 - 2 \times 5 \times \sqrt{7} - [(5)^2 + (\sqrt{7})^2 + 2 \times 5 \times \sqrt{7}]}{5^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \dfrac{25 + 7 - 10\sqrt{7} - [25 + 7 + 10\sqrt{7}]}{25 - 7} \\[1em] \Rightarrow \dfrac{25 - 25 + 7 - 7 - 10\sqrt{7} - 10\sqrt{7}}{18} \\[1em] \Rightarrow \dfrac{-20\sqrt{7}}{18} \\[1em] \Rightarrow \dfrac{-10\sqrt{7}}{9}.

Hence, Option 4 is the correct option.

Question 2

State, with reason, which of the following are surds and which are not :

(i) 180\sqrt{180}

(ii) 274\sqrt[4]{27}

(iii) 1285\sqrt[5]{128}

(iv) 643\sqrt[3]{64}

(v) 253.403\sqrt[3]{25}.\sqrt[3]{40}

(vi) −1253\sqrt[3]{-125}

(vii) π\sqrt{π}

(viii) 3+2\sqrt{3 + \sqrt{2}}

Answer

(i) Given,

180=2×2×3×3×5=2×3×5=65\sqrt{180} = \sqrt{2 \times 2 \times 3 \times 3 \times 5} = 2 \times 3 \times \sqrt{5} = 6\sqrt{5}, which is an irrational number.

Since, 180 is a rational number and 180\sqrt{180} is an irrational number.

Hence, 180\sqrt{180} is a surd.

(ii) Given,

274=334=(3)34\sqrt[4]{27} = \sqrt[4]{3^3} = (3)^{\dfrac{3}{4}}, which is irrational.

Since, 27 is a rational number and 274\sqrt[4]{27} is an irrational number.

Hence, 274\sqrt[4]{27} is a surd.

(iii) Given,

1285=275=25×225=(25)15×(22)15=2×(2)25\sqrt[5]{128} = \sqrt[5]{2^7} = \sqrt[5]{2^5 \times 2^2} \\[1em] = (2^5)^{\frac{1}{5}} \times (2^2)^{\frac{1}{5}} \\[1em] = 2 \times (2)^{\frac{2}{5}}

which is irrational.

Since, 128 is a rational number and 1285\sqrt[5]{128} is an irrational number.

Hence, 1285\sqrt[5]{128} is a surd.

(iv) Given,

643=433=43×13\sqrt[3]{64} = \sqrt[3]{4^3} = 4^{3 \times \dfrac{1}{3}} = 4, which is rational.

Since, 64 is rational and 643\sqrt[3]{64} is also rational.

Hence, 643\sqrt[3]{64} is not a surd.

(v) Given,

253.403=10003=1033=103×13\sqrt[3]{25}.\sqrt[3]{40} = \sqrt[3]{1000} = \sqrt[3]{10^3} = 10^{3 \times \dfrac{1}{3}} = 10, which is rational.

Since, 1000 is rational and 10 is also rational.

Hence, 253.403\sqrt[3]{25}.\sqrt[3]{40} is not a surd.

(vi) Given,

−1253=(−5)33=(−5)3×13=−5\sqrt[3]{-125} = \sqrt[3]{(-5)^3} = (-5)^{3 \times \dfrac{1}{3}} = -5, which is rational,

Since, -125 is rational and -5 is also rational.

Hence, −1253\sqrt[3]{-125} is not a surd.

(vii) Given,

π\sqrt{π}

Since, π is irrational and π\sqrt{π} is also irrational.

Hence, π\sqrt{π} is not a surd.

(viii) Given,

3+2\sqrt{3 + \sqrt{2}}

Since, 3+23 + \sqrt{2} is irrational.

Hence, 3+2\sqrt{3 + \sqrt{2}} is not a surd.

Question 3

Write the lowest rationalizing factor of :

(i) 525\sqrt{2}

(ii) 24\sqrt{24}

(iii) 5−3\sqrt{5} - 3

(iv) 7−77 - \sqrt{7}

(v) 18−50\sqrt{18} - \sqrt{50}

(vi) 5−2\sqrt{5} - \sqrt{2}

(vii) 13+3\sqrt{13} + 3

Answer

(i) Given,

⇒52\Rightarrow 5\sqrt{2}

Rationalizing,

⇒52×2⇒10.\Rightarrow 5\sqrt{2} \times \sqrt{2} \\[1em] \Rightarrow 10.

Hence, lowest rationalizing factor of 52=25\sqrt{2} = \sqrt{2}.

(ii) Given,

⇒24⇒26.\Rightarrow \sqrt{24} \\[1em] \Rightarrow 2\sqrt{6}.

Rationalizing,

⇒26×6⇒12.\Rightarrow 2\sqrt{6} \times \sqrt{6} \\[1em] \Rightarrow 12.

Hence, lowest rationalizing factor of 24=6\sqrt{24} = \sqrt{6}.

(iii) Given,

⇒5−3\Rightarrow \sqrt{5} - 3

Rationalizing,

⇒(5−3)×(5+3)⇒(5)2−32⇒5−9⇒−4.\Rightarrow (\sqrt{5} - 3) \times (\sqrt{5} + 3) \\[1em] \Rightarrow (\sqrt{5})^2 - 3^2 \\[1em] \Rightarrow 5 - 9 \\[1em] \Rightarrow -4.

Hence, lowest rationalizing factor of 5−3=5+3\sqrt{5} - 3 = \sqrt{5} + 3.

(iv) Given,

⇒7−7\Rightarrow 7 - \sqrt{7}

Rationalizing,

⇒(7−7)×(7+7)⇒(7)2−(7)2⇒49−7⇒42.\Rightarrow (7 - \sqrt{7}) \times (7 + \sqrt{7}) \\[1em] \Rightarrow (7)^2 - (\sqrt{7})^2 \\[1em] \Rightarrow 49 - 7 \\[1em] \Rightarrow 42.

Hence, lowest rationalizing factor of 7−7=7+77 - \sqrt{7} = 7 + \sqrt{7}.

(v) Given,

⇒18−50⇒32−52\Rightarrow \sqrt{18} - \sqrt{50}\\[1em] \Rightarrow 3\sqrt{2} - 5\sqrt{2}

Rationalizing,

⇒(32−52)×(2)⇒(6−10)⇒−4\Rightarrow (3\sqrt{2} - 5\sqrt{2}) \times (\sqrt{2}) \\[1em] \Rightarrow (6 - 10) \\[1em] \Rightarrow -4

Hence, lowest rationalizing factor of 18−50=2\sqrt{18} - \sqrt{50} = \sqrt{2}.

(vi) Given,

⇒5−2\Rightarrow \sqrt{5} - \sqrt{2}

Rationalizing,

⇒(5−2)×(5+2)⇒(5)2−(2)2⇒5−2⇒3.\Rightarrow (\sqrt{5} - \sqrt{2}) \times (\sqrt{5} + \sqrt{2}) \\[1em] \Rightarrow (\sqrt{5})^2 - (\sqrt{2})^2 \\[1em] \Rightarrow 5 - 2 \\[1em] \Rightarrow 3.

Hence, lowest rationalizing factor of 5−2=5+2\sqrt{5} - \sqrt{2} = \sqrt{5} + \sqrt{2}.

(vii) Given,

⇒13+3\Rightarrow \sqrt{13} + 3

Rationalizing,

⇒(13+3)(13−3)⇒(13)2−32⇒13−9⇒4.\Rightarrow (\sqrt{13} + 3)(\sqrt{13} - 3) \\[1em] \Rightarrow (\sqrt{13})^2 - 3^2 \\[1em] \Rightarrow 13 - 9 \\[1em] \Rightarrow 4.

Hence, lowest rationalizing factor of 13+3=13−3\sqrt{13} + 3 = \sqrt{13} - 3.

Question 4

Rationalize the denominators of:

(i) 235\dfrac{2\sqrt{3}}{\sqrt{5}}

(ii) 6−56+5\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}

Answer

(i) Given,

235\dfrac{2\sqrt{3}}{\sqrt{5}}

Let us rationalize the denominator,

⇒23×55×5⇒215(5)2⇒2155\Rightarrow \dfrac{2\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} \\[1em] \Rightarrow \dfrac{2\sqrt{15}}{(\sqrt{5})^2}\\[1em] \Rightarrow \dfrac{2\sqrt{15}}{5}

Hence,235=2155\dfrac{2\sqrt{3}}{\sqrt{5}} = \dfrac{2\sqrt{15}}{5}

(ii) Given, 6−56+5\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}

Let us rationalize the denominator,

⇒(6−5)×(6−5)(6+5)×(6−5)=(6−5)2(6)2−(5)2=(6)2+(5)2−2×6×56−5=6+5−2×301=11−230\Rightarrow \dfrac{(\sqrt{6} - \sqrt{5}) \times (\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})\times (\sqrt{6} - \sqrt{5})}\\[1em] = \dfrac{(\sqrt{6} - \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2}\\[1em] = \dfrac{(\sqrt{6})^2 + (\sqrt{5})^2 - 2 \times \sqrt{6} \times \sqrt{5}}{6 - 5}\\[1em] = \dfrac{6 + 5 - 2 \times \sqrt{30}}{1}\\[1em] = 11 - 2\sqrt{30}

Hence, 6−56+5=11−230\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} = 11 - 2\sqrt{30}.

Question 5(i)

Find the values of 'a' and 'b':

2+32−3=a+b3\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}

Answer

Given,

Equation : 2+32−3=a+b3\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}

Rationalizing L.H.S. of the above equation :

⇒2+32−3×2+32+3⇒(2+3)222−(3)2⇒22+(3)2+2×2×322−(3)2⇒4+3+434−3⇒7+43.\Rightarrow \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{4 + 3 + 4\sqrt{3}}{4 - 3} \\[1em] \Rightarrow 7 + 4\sqrt{3}.

Comparing 7+43 with a+b37 + 4\sqrt{3} \text{ with } a + b\sqrt{3}, we get :

a = 7 and b = 4.

Hence, a = 7 and b = 4.

Question 5(ii)

Find the values of 'a' and 'b':

7−27+2=a7+b\dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b

Answer

Given,

Equation : 7−27+2=a7+b\dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b

Rationalizing L.H.S. of the above equation :

⇒7−27+2×7−27−2⇒(7−2)2(7)2−22⇒(7)2+22−2×7×27−4⇒7+4−473⇒11−473⇒−473+113.\Rightarrow \dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} \times \dfrac{\sqrt{7} - 2}{\sqrt{7} - 2} \\[1em] \Rightarrow \dfrac{(\sqrt{7} - 2)^2}{(\sqrt{7})^2 - 2^2} \\[1em] \Rightarrow \dfrac{(\sqrt{7})^2 + 2^2 - 2 \times \sqrt{7} \times 2}{7 - 4} \\[1em] \Rightarrow \dfrac{7 + 4 - 4\sqrt{7}}{3} \\[1em] \Rightarrow \dfrac{11 - 4\sqrt{7}}{3} \\[1em] \Rightarrow -\dfrac{4\sqrt{7}}{3} + \dfrac{11}{3}.

Comparing −473+113 with a7+b-\dfrac{4\sqrt{7}}{3} + \dfrac{11}{3} \text{ with } a\sqrt{7} + b, we get :

a = −43 and b=113-\dfrac{4}{3}\text{ and } b = \dfrac{11}{3}.

Hence, a = −43 and b=113-\dfrac{4}{3}\text{ and } b = \dfrac{11}{3}.

Question 5(iii)

Find the values of 'a' and 'b':

33−2=a3−b2\dfrac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}

Answer

Given,

Equation : 33−2=a3−b2\dfrac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}

Rationalizing L.H.S. of the above equation :

⇒33−2×3+23+2⇒3(3+2)(3)2−(2)2⇒33+323−2⇒33+321⇒33+32.\Rightarrow \dfrac{3}{\sqrt{3} - \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{3(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3\sqrt{3} + 3\sqrt{2}}{3 - 2} \\[1em] \Rightarrow \dfrac{3\sqrt{3} + 3\sqrt{2}}{1} \\[1em] \Rightarrow 3\sqrt{3} + 3\sqrt{2}.

Comparing 33+32 with a3−b23\sqrt{3} + 3\sqrt{2}\text{ with } a\sqrt{3} - b\sqrt{2}, we get :

a = 3 and b = -3.

Hence, a = 3 and b = -3.

Question 6(i)

Simplify :

2223+1+1723−1\dfrac{22}{2\sqrt{3} + 1} + \dfrac{17}{2\sqrt{3} - 1}

Answer

(i) Solving,

⇒2223+1+1723−1⇒22(23−1)+17(23+1)(23+1)(23−1)⇒443−22+343+17(23)2−12⇒783−512−1⇒783−511.\Rightarrow \dfrac{22}{2\sqrt{3} + 1} + \dfrac{17}{2\sqrt{3} - 1} \\[1em] \Rightarrow \dfrac{22(2\sqrt{3} - 1) + 17(2\sqrt{3} + 1)}{(2\sqrt{3} + 1)(2\sqrt{3} - 1)} \\[1em] \Rightarrow \dfrac{44\sqrt{3} - 22 + 34\sqrt{3} + 17}{(2\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \dfrac{78\sqrt{3} - 5}{12 - 1} \\[1em] \Rightarrow \dfrac{78\sqrt{3} - 5}{11}.

Hence, solution = 783−511.\dfrac{78\sqrt{3} - 5}{11}.

Question 6(ii)

Simplify :

26−2−36+2\dfrac{\sqrt{2}}{\sqrt{6} - \sqrt{2}} - \dfrac{\sqrt{3}}{\sqrt{6} + \sqrt{2}}

Answer

Solving,

⇒26−2−36+2⇒2(6+2)−3(6−2)(6−2)(6+2)⇒12+2−18+6(6)2−(2)2⇒23+2−32+66−2⇒23+2−32+64.\Rightarrow \dfrac{\sqrt{2}}{\sqrt{6} - \sqrt{2}} - \dfrac{\sqrt{3}}{\sqrt{6} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{\sqrt{2}(\sqrt{6} + \sqrt{2}) - \sqrt{3}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} \\[1em] \Rightarrow \dfrac{\sqrt{12} + 2 - \sqrt{18} + \sqrt{6}}{(\sqrt{6})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{6 - 2} \\[1em] \Rightarrow \dfrac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4}.

Hence, solution = 23+2−32+64\dfrac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4}.

Question 7

If x = 5−25+2 and y=5+25−2\dfrac{\sqrt{5} - 2}{\sqrt{5} + 2} \text{ and } y = \dfrac{\sqrt{5} + 2}{\sqrt{5} - 2}; find :

(i) x2

(ii) y2

(iii) xy

(iv) x2 + y2 + xy

Answer

(i) Substituting value of x, we get :

⇒x2=(5−25+2)2=(52+22−2×5×2(5)2+22+2×5×2)=5+4−455+4+45=9−459+45.\Rightarrow x^2 = \Big(\dfrac{\sqrt{5} - 2}{\sqrt{5} + 2}\Big)^2 \\[1em] = \Big(\dfrac{\sqrt{5}^2 + 2^2 - 2 \times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 + 2 \times \sqrt{5} \times 2}\Big) \\[1em] = \dfrac{5 + 4 - 4\sqrt{5}}{5 + 4 + 4\sqrt{5}} \\[1em] = \dfrac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}}.

Rationalizing,

=9−459+45×9−459−45=(9−45)292−(45)2=92+(45)2−2×9×4581−80=81+80−7251=161−725.= \dfrac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}} \times \dfrac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} \\[1em] = \dfrac{(9 - 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \dfrac{9^2 + (4\sqrt{5})^2 - 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \dfrac{81 + 80 - 72\sqrt{5}}{1} \\[1em] = 161 - 72\sqrt{5}.

Hence, x2 = 161−725.161 - 72\sqrt{5}.

(ii) Substituting value of y, we get :

⇒y2=(5+25−2)2=(5)2+22+2×5×2(5)2+22−2×5×2=5+4+455+4−45=9+459−45\Rightarrow y^2 = \Big(\dfrac{\sqrt{5} + 2}{\sqrt{5} - 2}\Big)^2 \\[1em] = \dfrac{(\sqrt{5})^2 + 2^2 + 2\times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 - 2\times \sqrt{5} \times 2} \\[1em] = \dfrac{5 + 4 + 4\sqrt{5}}{5 + 4 - 4\sqrt{5}} \\[1em] = \dfrac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}}

Rationalizing,

=9+459−45×9+459+45=(9+45)292−(45)2=92+(45)2+2×9×4581−80=81+80+7251=161+725.= \dfrac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}} \times \dfrac{9 + 4\sqrt{5}}{9 + 4\sqrt{5}} \\[1em] = \dfrac{(9 + 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \dfrac{9^2 + (4\sqrt{5})^2 + 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \dfrac{81 + 80 + 72\sqrt{5}}{1} \\[1em] = 161 + 72\sqrt{5}.

Hence, y2 = 161+725.161 + 72\sqrt{5}.

(iii) Substituting values of x and y, we get :

⇒xy=5−25+2×5+25−2=1.\Rightarrow xy = \dfrac{\sqrt{5} - 2}{\sqrt{5} + 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} - 2} \\[1em] = 1.

Hence, xy = 1.

(iv) Substituting value of x2, y2 and xy, we get :

⇒x2+y2+xy=161−725+161+725+1=323.\Rightarrow x^2 + y^2 + xy = 161 - 72\sqrt{5} + 161 + 72\sqrt{5} + 1 \\[1em] = 323.

Hence, x2 + y2 + xy = 323.

Question 8

If m = 13−22 and n=13+22\dfrac{1}{3 - 2\sqrt{2}} \text{ and } n = \dfrac{1}{3 + 2\sqrt{2}}, find :

(i) m2

(ii) n2

(iii) mn

Answer

(i) Substituting value of m, we get :

⇒m2=(13−22)2=12(3−22)2=132+(22)2−2×3×22=19+8−122=117−122\Rightarrow m^2 = \Big(\dfrac{1}{3 - 2\sqrt{2}}\Big)^2 \\[1em] = \dfrac{1^2}{(3 - 2\sqrt{2})^2} \\[1em] = \dfrac{1}{3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2}} \\[1em] = \dfrac{1}{9 + 8 - 12\sqrt{2}} \\[1em] = \dfrac{1}{17 - 12\sqrt{2}}

Rationalizing,

=117−122×17+12217+122=17+122172−(122)2=17+122289−288=17+1221=17+122.= \dfrac{1}{17 - 12\sqrt{2}} \times \dfrac{17 + 12\sqrt{2}}{17 + 12\sqrt{2}} \\[1em] = \dfrac{17 + 12\sqrt{2}}{17^2 - (12\sqrt{2})^2} \\[1em] = \dfrac{17 + 12\sqrt{2}}{289 - 288} \\[1em] = \dfrac{17 + 12\sqrt{2}}{1} \\[1em] = 17 + 12\sqrt{2}.

Hence, m2 = 17+12217 + 12\sqrt{2}.

(ii) Substituting value of n, we get :

⇒n2=(13+22)2=12(3+22)2=132+(22)2+2×3×22=19+8+122=117+122\Rightarrow n^2 = \Big(\dfrac{1}{3 + 2\sqrt{2}}\Big)^2 \\[1em] = \dfrac{1^2}{(3 + 2\sqrt{2})^2} \\[1em] = \dfrac{1}{3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2}} \\[1em] = \dfrac{1}{9 + 8 + 12\sqrt{2}} \\[1em] = \dfrac{1}{17 + 12\sqrt{2}}

Rationalizing,

=117+122×17−12217−122=17−122172−(122)2=17−122289−288=17−1221=17−122.= \dfrac{1}{17 + 12\sqrt{2}} \times \dfrac{17 - 12\sqrt{2}}{17 - 12\sqrt{2}} \\[1em] = \dfrac{17 - 12\sqrt{2}}{17^2 - (12\sqrt{2})^2} \\[1em] = \dfrac{17 - 12\sqrt{2}}{289 - 288} \\[1em] = \dfrac{17 - 12\sqrt{2}}{1} \\[1em] = 17 - 12\sqrt{2}.

Hence, n2 = 17−12217 - 12\sqrt{2}.

(iii) Substituting value of m and n, we get :

⇒mn=13−22×13+22=132+3×22−22×3−(22)2=19+62−62−8=19−8=11=1.\Rightarrow mn = \dfrac{1}{3 - 2\sqrt{2}} \times \dfrac{1}{3 + 2\sqrt{2}} \\[1em] = \dfrac{1}{3^2 + 3 \times 2\sqrt{2} - 2\sqrt{2} \times 3 - (2\sqrt{2})^2} \\[1em] = \dfrac{1}{9 + 6\sqrt{2} - 6\sqrt{2} - 8} \\[1em] = \dfrac{1}{9 - 8} \\[1em] = \dfrac{1}{1} \\[1em] = 1.

Hence, mn = 1.

Question 9

If x = 23+222\sqrt{3} + 2\sqrt{2}, find :

(i) 1x\dfrac{1}{x}

(ii) x+1xx + \dfrac{1}{x}

(iii) (x+1x)2\Big(x + \dfrac{1}{x}\Big)^2

Answer

(i) Substituting value of x, we get :

⇒1x=123+22\Rightarrow \dfrac{1}{x} = \dfrac{1}{2\sqrt{3} + 2\sqrt{2}}

Rationalizing,

=123+22×23−2223−22=23−22(23)2−(22)2=23−2212−8=2(3−2)4=3−22.= \dfrac{1}{2\sqrt{3} + 2\sqrt{2}} \times \dfrac{2\sqrt{3} - 2\sqrt{2}}{2\sqrt{3} - 2\sqrt{2}} \\[1em] = \dfrac{2\sqrt{3} - 2\sqrt{2}}{(2\sqrt{3})^2 - (2\sqrt{2})^2} \\[1em] = \dfrac{2\sqrt{3} - 2\sqrt{2}}{12 - 8} \\[1em] = \dfrac{2(\sqrt{3} - \sqrt{2})}{4} \\[1em] = \dfrac{\sqrt{3} - \sqrt{2}}{2}.

Hence, 1x=3−22\dfrac{1}{x} = \dfrac{\sqrt{3} - \sqrt{2}}{2}.

(ii) Substituting values of x and 1x\dfrac{1}{x} we get :

⇒x+1x=23+22+3−22=43+42+3−22=53+322.\Rightarrow x + \dfrac{1}{x} = 2\sqrt{3} + 2\sqrt{2} + \dfrac{\sqrt{3} - \sqrt{2}}{2} \\[1em] = \dfrac{4\sqrt{3} + 4\sqrt{2} + \sqrt{3} -\sqrt{2}}{2} \\[1em] = \dfrac{5\sqrt{3} + 3\sqrt{2}}{2}.

Hence, x+1x=(53+322)x + \dfrac{1}{x} = \Big(\dfrac{5\sqrt{3} + 3\sqrt{2}}{2}\Big).

(iii) Substituting value of (x+1x)\Big(x + \dfrac{1}{x}\Big), we get :

⇒(x+1x)2=(53+322)2=(53)2+(32)2+2×53×3222=75+18+3064=93+3064.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = \Big(\dfrac{5\sqrt{3} + 3\sqrt{2}}{2}\Big)^2 \\[1em] = \dfrac{(5\sqrt{3})^2 + (3\sqrt{2})^2 + 2 \times 5\sqrt{3} \times 3\sqrt{2}}{2^2} \\[1em] = \dfrac{75 + 18 + 30\sqrt{6}}{4} \\[1em] = \dfrac{93 + 30\sqrt{6}}{4}.

Hence, (x+1x)2=93+3064\Big(x + \dfrac{1}{x}\Big)^2 = \dfrac{93 + 30\sqrt{6}}{4}.

Question 10

If x = 1−21 - \sqrt{2}, find the value of (x−1x)3\Big(x - \dfrac{1}{x}\Big)^3.

Answer

Given,

x = 1 - 2\sqrt{2}

∴1x=11−2\therefore \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}}

Rationalizing,

⇒1x=11−2×1+21+2⇒1+212−(2)2⇒1+21−2⇒1+2−1⇒−(1+2)⇒−1−2.\Rightarrow \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}} \times \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1em] \Rightarrow \dfrac{1 + \sqrt{2}}{1^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - 2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{2}}{-1} \\[1em] \Rightarrow -(1 + \sqrt{2}) \\[1em] \Rightarrow -1 - \sqrt{2}.

Substituting values we get :

⇒(x−1x)3=[1−2−(−1−2)]3=[1+1−2+2]3=[2]3=8.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = [1 - \sqrt{2} - (-1 - \sqrt{2})]^3 \\[1em] = [1 + 1 - \sqrt{2} + \sqrt{2}]^3 \\[1em] = [2]^3 \\[1em] = 8.

Hence, (x−1x)3\Big(x - \dfrac{1}{x}\Big)^3 = 8.

Question 11

If x = 5 - 262\sqrt{6}, find : x2+1x2x^2 + \dfrac{1}{x^2}

Answer

Given,

x = 5 - 262\sqrt{6}

∴1x=15−26\therefore \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}}

Rationalizing,

⇒1x=15−26×5+265+26=5+2652−(26)2=5+2625−24=5+261=5+26.\Rightarrow \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} \times \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1em] = \dfrac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] = \dfrac{5 + 2\sqrt{6}}{25 - 24} \\[1em] = \dfrac{5 + 2\sqrt{6}}{1} \\[1em] = 5 + 2\sqrt{6}.

By formula,

x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2

Substituting values we get :

⇒x2+1x2=(5−26+5+26)2−2=102−2=100−2=98.\Rightarrow x^2 + \dfrac{1}{x^2} = (5 - 2\sqrt{6} + 5 + 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 98.

Question 12

If 2=1.4 and 3\sqrt{2} = 1.4 \text{ and } \sqrt{3} = 1.7, find the value of each of the following, correct to one decimal place :

(i) 13−2\dfrac{1}{\sqrt{3} - \sqrt{2}}

(ii) 13+22\dfrac{1}{3 + 2\sqrt{2}}

(iii) 2−33\dfrac{2 - \sqrt{3}}{\sqrt{3}}

Answer

(i) Given,

⇒13−2\Rightarrow \dfrac{1}{\sqrt{3} - \sqrt{2}}

Rationalizing,

⇒13−2×3+23+2⇒3+2(3)2−(2)2⇒3+23−2⇒3+21⇒3+2⇒1.7+1.4⇒3.1\Rightarrow \dfrac{1}{\sqrt{3} - \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{\sqrt{3} + \sqrt{2}}{3 - 2} \\[1em] \Rightarrow \dfrac{\sqrt{3} + \sqrt{2}}{1} \\[1em] \Rightarrow \sqrt{3} + \sqrt{2} \\[1em] \Rightarrow 1.7 + 1.4 \\[1em] \Rightarrow 3.1

Hence, 13−2\dfrac{1}{\sqrt{3} - \sqrt{2}} = 3.1

(ii) Given,

⇒13+22\Rightarrow \dfrac{1}{3 + 2\sqrt{2}}

Rationalizing,

⇒13+22×3−223−22⇒3−2232−(22)2⇒3−229−8⇒3−221⇒3−22⇒3−2×1.4⇒3−2.8⇒0.2\Rightarrow \dfrac{1}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{1} \\[1em] \Rightarrow 3 - 2\sqrt{2} \\[1em] \Rightarrow 3 - 2 \times 1.4 \\[1em] \Rightarrow 3 - 2.8 \\[1em] \Rightarrow 0.2

Hence, 13+22\dfrac{1}{3 + 2\sqrt{2}} = 0.2

(iii) Given,

⇒2−33\Rightarrow \dfrac{2 - \sqrt{3}}{\sqrt{3}}

Rationalizing,

⇒2−33×33⇒23−33⇒2×1.7−33⇒3.4−33⇒0.43⇒0.1\Rightarrow \dfrac{2 - \sqrt{3}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{2\sqrt{3} - 3}{3} \\[1em] \Rightarrow \dfrac{2 \times 1.7 - 3}{3} \\[1em] \Rightarrow \dfrac{3.4 - 3}{3} \\[1em] \Rightarrow \dfrac{0.4}{3} \\[1em] \Rightarrow 0.1

Hence, 2−33\dfrac{2 - \sqrt{3}}{\sqrt{3}} = 0.1

Question 13

Evaluate :

4−54+5+4+54−5\dfrac{4 - \sqrt{5}}{4 + \sqrt{5}} + \dfrac{4 + \sqrt{5}}{4 - \sqrt{5}}

Answer

Solving,

⇒4−54+5+4+54−5⇒(4−5)2+(4+5)2(4+5)(4−5)⇒42+(5)2−2×4×5+42+(5)2+2×4×542−(5)2⇒16+5−85+16+5+8516−5⇒4211⇒3911.\Rightarrow \dfrac{4 - \sqrt{5}}{4 + \sqrt{5}} + \dfrac{4 + \sqrt{5}}{4 - \sqrt{5}} \\[1em] \Rightarrow \dfrac{(4 - \sqrt{5})^2 + (4 + \sqrt{5})^2}{(4 + \sqrt{5})(4 - \sqrt{5})} \\[1em] \Rightarrow \dfrac{4^2 + (\sqrt{5})^2 - 2 \times 4 \times \sqrt{5} + 4^2 + (\sqrt{5})^2 + 2 \times 4 \times \sqrt{5}}{4^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{16 + 5 - 8\sqrt{5} + 16 + 5 + 8\sqrt{5}}{16 - 5} \\[1em] \Rightarrow \dfrac{42}{11} \\[1em] \Rightarrow 3\dfrac{9}{11}.

Hence, 4−54+5+4+54−5=3911\dfrac{4 - \sqrt{5}}{4 + \sqrt{5}} + \dfrac{4 + \sqrt{5}}{4 - \sqrt{5}} = 3\dfrac{9}{11}.

Question 14

If 2+52−5=x and 2−52+5=y\dfrac{2 + \sqrt{5}}{2 - \sqrt{5}} = x \text{ and } \dfrac{2 - \sqrt{5}}{2 + \sqrt{5}} = y; find the value of x2 - y2.

Answer

Given,

x = 2+52−5\dfrac{2 + \sqrt{5}}{2 - \sqrt{5}}

Rationalizing,

⇒x=2+52−5×2+52+5=(2+5)222−(5)2=22+(5)2+2×2×54−5=4+5+45−1=−(9+45).⇒x2=[−(9+45)]2=92+(45)2+2×9×45=81+80+725=161+725.\Rightarrow x = \dfrac{2 + \sqrt{5}}{2 - \sqrt{5}} \times \dfrac{2 + \sqrt{5}}{2 + \sqrt{5}} \\[1em] = \dfrac{(2 + \sqrt{5})^2}{2^2 - (\sqrt{5})^2} \\[1em] = \dfrac{2^2 + (\sqrt{5})^2 + 2\times 2 \times \sqrt{5}}{4 - 5} \\[1em] = \dfrac{4 + 5 + 4\sqrt{5}}{-1} \\[1em] = -(9 + 4\sqrt{5}). \\[1em] \Rightarrow x^2 = [-(9 + 4\sqrt{5})]^2 \\[1em] = 9^2 + (4\sqrt{5})^2 + 2\times 9 \times 4\sqrt{5} \\[1em] = 81 + 80 + 72\sqrt{5} \\[1em] = 161 + 72\sqrt{5}.

Given,

y = 2−52+5\dfrac{2 - \sqrt{5}}{2 + \sqrt{5}}

Rationalizing,

⇒y=2−52+5×2−52−5=(2−5)222−(5)2=22+(5)2−2×2×54−5=4+5−45−1=−(9−45).⇒y2=[−(9−45)]2=92+(45)2−2×9×45=81+80−725=161−725.\Rightarrow y = \dfrac{2 - \sqrt{5}}{2 + \sqrt{5}} \times \dfrac{2 - \sqrt{5}}{2 - \sqrt{5}} \\[1em] = \dfrac{(2 - \sqrt{5})^2}{2^2 - (\sqrt{5})^2} \\[1em] = \dfrac{2^2 + (\sqrt{5})^2 - 2\times 2 \times \sqrt{5}}{4 - 5} \\[1em] = \dfrac{4 + 5 - 4\sqrt{5}}{-1} \\[1em] = -(9 - 4\sqrt{5}). \\[1em] \Rightarrow y^2 = [-(9 - 4\sqrt{5})]^2 \\[1em] = 9^2 + (4\sqrt{5})^2 - 2\times 9 \times 4\sqrt{5} \\[1em] = 81 + 80 - 72\sqrt{5} \\[1em] = 161 - 72\sqrt{5}.

Substituting values of x2 and y2, we get :

⇒x2−y2=(161+725)−(161−725)=161−161+725+725=1445.\Rightarrow x^2 - y^2 = (161 + 72\sqrt{5}) - (161 - 72\sqrt{5}) \\[1em] = 161 - 161 + 72\sqrt{5} + 72\sqrt{5} \\[1em] = 144\sqrt{5}.

Hence, x2 - y2 = 1445144\sqrt{5}.

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