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Chapter 1

Rational and Irrational Numbers — Exercise 1.5

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 1.5

Question 1

Rationalise the denominator of the following :

(i)345(ii)573(iii)34−7(iv)1732+1(v)1641−5(vi)17−6(vii)15+2(viii)2+32−3\begin{matrix} \text{(i)} & \dfrac{3}{4\sqrt{5}} \\[1.5em] \text{(ii)} & \dfrac{5\sqrt{7}}{\sqrt{3}} \\[1.5em] \text{(iii)} & \dfrac{3}{4 - \sqrt{7}} \\[1.5em] \text{(iv)} & \dfrac{17}{3\sqrt{2} + 1} \\[1.5em] \text{(v)} & \dfrac{16}{\sqrt{41}-5} \\[1.5em] \text{(vi)} & \dfrac{1}{\sqrt{7} - \sqrt{6}} \\[1.5em] \text{(vii)} & \dfrac{1}{\sqrt{5} + \sqrt{2}} \\[1.5em] \text{(viii)} & \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1.5em] \end{matrix}

Answer

(i)\text{(i)}

345\dfrac{3}{4\sqrt{5}}

Let us rationalise the denominator,

Then,

345=3×545×5⇒354×5⇒3520\dfrac{3}{4\sqrt{5}} = \dfrac{3×\sqrt{5}}{4\sqrt{5} × \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{3\sqrt{5}}{4 × 5} \\[1.5em] \bold{\Rightarrow\dfrac{3\sqrt{5}}{20}}

(ii)\text{(ii)}

573\dfrac{5\sqrt{7}}{\sqrt{3}}

Let us rationalise the denominator,

Then,

573=57×33×3⇒5213\dfrac{5\sqrt{7}}{\sqrt{3}} = \dfrac{5\sqrt{7}×\sqrt{3}}{\sqrt{3} × \sqrt{3}} \\[1.5em] \bold{\Rightarrow\dfrac{5\sqrt{21}}{3}}

(iii)\text{(iii)}

34−7\dfrac{3}{4 - \sqrt{7}}

Let us rationalise the denominator,

Then,

34−7=34−7×4+74+7⇒3(4+7)(4)2−(7)2⇒3(4+7)(16)−(7)⇒3(4+7)9⇒(4+7)3\dfrac{3}{4 - \sqrt{7}} = \dfrac{3}{4 - \sqrt{7}} × \dfrac{4 + \sqrt{7}}{4 + \sqrt{7}} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{(16) - (7)} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{9} \\[1.5em] \bold{\Rightarrow\dfrac{(4 + \sqrt{7})}{3}} \\[1.5em]

(iv)\text{(iv)}

1732+1\dfrac{17}{3\sqrt{2}+1}

Let us rationalise the denominator,

Then,

1732+1=1732+1×32−132−1⇒17(32−1)(32)2−1⇒17(32−1)17⇒(32−1)\dfrac{17}{3\sqrt{2} + 1} = \dfrac{17}{3\sqrt{2} + 1}×\dfrac{3\sqrt{2} - 1}{3\sqrt{2} - 1} \\[1.5em] \Rightarrow\dfrac{17(3\sqrt{2} - 1)}{(3\sqrt{2})^2 - 1} \\[1.5em] \Rightarrow\dfrac{17(3\sqrt{2} - 1)}{17} \\[1.5em] \bold{\Rightarrow(3\sqrt{2} - 1)} \\[1.5em]

(v)\text{(v)}

1641−5\dfrac{16}{\sqrt{41}-5}

Let us rationalise the denominator,

Then,

1641−5=1641−5×41+541+5⇒16(41+5)(41)2−52⇒16(41+5)41−25⇒16(41+5)1641+5\dfrac{16}{\sqrt{41} - 5} = \dfrac{16}{\sqrt{41} - 5}×\dfrac{\sqrt{41} + 5}{\sqrt{41} + 5} \\[1.5em] \Rightarrow\dfrac{16({\sqrt{41} + 5})}{(\sqrt{41})^2 - 5^2} \\[1.5em] \Rightarrow\dfrac{16({\sqrt{41} + 5})}{41 - 25} \\[1.5em] {\Rightarrow\dfrac{16({\sqrt{41} + 5})}{16}} \\[1.5em] \bold{\sqrt{41} + 5}

(vi)\text{(vi)}

17−6\dfrac{1}{\sqrt{7} - \sqrt{6}}

Let us rationalise the denominator,

Then,

17−6=17−6×7+67+6⇒(7+6)(7)2−(6)2⇒7+67−6⇒7+6\dfrac{1}{\sqrt{7} - \sqrt{6}} = \dfrac{1}{\sqrt{7} - \sqrt{6}} × \dfrac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} \\[1.5em] \Rightarrow\dfrac{({\sqrt{7} + \sqrt{6}})}{(\sqrt{7})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\dfrac{\sqrt{7} + \sqrt{6}}{7 - 6 } \\[1.5em] \bold{{\Rightarrow\sqrt{7} + \sqrt{6}}} \\[1.5em]

(vii)\text{(vii)}

15+2\dfrac{1}{\sqrt{5}+\sqrt{2}}

Let us rationalise the denominator,

Then,

15+2=15+2×5−25−2⇒5−2(5)2−(2)2⇒5−23\dfrac{1}{\sqrt{5} + \sqrt{2}} = \dfrac{1}{\sqrt{5} +\sqrt{2}} × \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{2}}}{(\sqrt{5})^2 - (\sqrt{2})^2} \\[1.5em] \bold{\Rightarrow\dfrac{{\sqrt{5} - \sqrt{2}}}{3}} \\[1.5em]

(viii)\text{(viii)}

2+32−3\dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}

Let us rationalise the denominator,

Then,

2+32−3=2+32−3×2+32+3=(2+3)2(2)2−(3)2=(2)2+(3)2+2×2×3(2)2−(3)2=2+3+2232−3=−(5+26)\dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} × \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} \\[1.5em] = \dfrac{{(\sqrt{2} + \sqrt{3})^2}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{{(\sqrt{2})^2 + (\sqrt{3})^2} + 2 × \sqrt{2} × \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{2 + 3 + 2\sqrt{2}\sqrt{3}}{2 - 3} \\[1.5em] = \bold{-(5 + 2\sqrt{6})} \\[1.5em]

Question 2

Simplify each of the following by rationalising the denominator:

(i)7+357−35(ii)3−223+22(iii)5−3147+214\begin{matrix} \text{(i)} & \dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} \\[1.5em] \text{(ii)} & \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \\[1.5em] \text{(iii)} & \dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} \\[1.5em] \end{matrix}

Answer

(i)\text(i)

7+357−35\dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}}

Let us rationalise the denominator,

Then,

7+357−35=7+357−35×7+357+35⇒(7+35)272−(35)2⇒72+(35)2+2×7×3549−45⇒49+45+42549−45⇒94+42549−45⇒2×(47+215)4⇒(47+215)2\dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} = \dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} × \dfrac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(7 + 3\sqrt{5})^2}{7^2 - (3\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{7^2 + (3\sqrt{5})^2 + 2 × 7 × 3\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\dfrac{49 + 45 + 42\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\dfrac{94 + 42\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\dfrac{2 × (47 + 21\sqrt{5})}{4} \\[1.5em] \bold{\Rightarrow\dfrac{(47 + 21\sqrt{5})}{2}} \\[1.5em]

(ii)\text(ii)

3−223+22\dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}

3−223+22=3−223+22×3−223−22⇒(3−22)232−(22)2⇒32+(22)2−2×3×229−8⇒9+8−1221⇒(17−122)1⇒17−122\dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} × \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(3 - 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2} \\[1.5em] \Rightarrow\dfrac{3^2 + (2\sqrt{2})^2 - 2 × 3 × 2\sqrt{2}}{9-8} \\[1.5em] \Rightarrow\dfrac{9 + 8 - 12\sqrt{2}}{1} \\[1.5em] \Rightarrow\dfrac{(17 - 12\sqrt{2})}{1} \\[1.5em] \bold{\Rightarrow{17 - 12\sqrt{2}}} \\[1.5em]

(iii)\text(iii)

5−3147+214\dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}}

5−3147+214=5−3147+214×7−2147−214⇒(5−314)(7−214)72−(214)2⇒35−1014−2114+(6×14)49−56⇒119−3114−7⇒−119+31147\dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} = \dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} × \dfrac{7 - 2\sqrt{14}}{7 - 2\sqrt{14}} \\[1.5em] \Rightarrow\dfrac{(5 - 3\sqrt{14})(7 - 2\sqrt{14})}{7^2 - (2\sqrt{14})^2} \\[1.5em] \Rightarrow\dfrac{35 - 10\sqrt{14} - 21\sqrt{14} + (6 × 14)}{49 - 56} \\[1.5em] \Rightarrow\dfrac{119 - 31\sqrt{14}}{-7} \\[1.5em] \bold{\Rightarrow\dfrac{-119+31\sqrt{14}}{7}} \\[1.5em]

Question 3

Simplify : 7310+3−256+5−3215+32\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}

Answer

7310+3−256+5−3215+32\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} ....(i)\qquad \text{....(i)}

Simplifying each term individually,

7310+3\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}}

Let us rationalize its denominator,

7310+3=7310+3×10−310−3⇒(73)(10−3)(10)2−(3)2⇒73×10−73×3(10)2−(3)2⇒73×10−73×3(10)2−(3)2⇒730−(7×3)10−3⇒(730−(7×3)7)⇒7×(30−37)⇒30−3....(ii)\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} = \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} × \dfrac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} \\[1.5em] \Rightarrow\dfrac{(7\sqrt{3})(\sqrt{10} - \sqrt{3}) }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em]\Rightarrow\dfrac{7\sqrt{3} × \sqrt{10} - 7\sqrt{3} × \sqrt{3} }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow\dfrac{7\sqrt{3 × 10} - 7\sqrt{3 × 3} }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow\dfrac{7\sqrt{30}-(7×3)}{10-3} \\[1.5em] \Rightarrow\Big(\dfrac{7\sqrt{30}-(7×3)}{7}\Big) \\[1.5em] \Rightarrow 7 ×\Big(\dfrac{\sqrt{30} - 3}{7}\Big) \\[1.5em] \bold{\Rightarrow{\sqrt{30}-3}} \qquad \text{....(ii)} \\[1.5em]

256+5\dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}}

Let us rationalize its denominator,

256+5=256+5×6−56−5⇒(25)(6−5)(6)2−(5)2⇒25×6−25×5(6)2−(5)2⇒25×6−25×56−5⇒230−101⇒230−10....(iii)\dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} = \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} × \dfrac{\sqrt6 - \sqrt{5}}{\sqrt{6} - \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2\sqrt{5})(\sqrt{6} - \sqrt{5}) }{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{2\sqrt{5} × \sqrt{6} - 2\sqrt{5} ×\sqrt{5} }{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{2\sqrt{5 × 6} - 2\sqrt{5 × 5} }{6 - 5} \\[1.5em] \Rightarrow\dfrac{2\sqrt{30} - 10}{1} \\[1.5em] \bold{\Rightarrow{2\sqrt{30} - 10}} \qquad \text{....(iii)} \\[1.5em]

3215+32\dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}

Let us rationalize its denominator,

3215+32=3215+32×15−3215−32⇒(32)(15−32)(15)2−(32)2⇒32×15−32×32(15)2−(32)2⇒330−1815−18⇒330−18−3⇒[−3(−30+6)−3]⇒−(−30+61)⇒6−30....(iv)\dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} = \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} × \dfrac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(3\sqrt{2})(\sqrt{15} - 3\sqrt{2}) }{(\sqrt{15})^2 - (3\sqrt{2})^2} \\[1.5em] \Rightarrow\dfrac{3\sqrt{2} × \sqrt{15} - 3\sqrt{2} × 3\sqrt{2} }{(\sqrt{15})^2 - (3\sqrt{2})^2} \\[1.5em] \Rightarrow\dfrac{3\sqrt{30} - 18}{15 - 18} \\[1.5em] \Rightarrow\dfrac{3\sqrt{30} - 18}{-3} \\[1.5em] \Rightarrow\Big[\dfrac{-3(-\sqrt{30} + 6)}{-3}\Big] \\[1.5em] \Rightarrow-\Big(\dfrac{-\sqrt{30} + 6}{1}\Big) \\[1.5em] \bold{\Rightarrow{6 - \sqrt{30}}} \qquad \text{....(iv)} \\[1.5em]

Using (ii) , (iii) , (iv) in equation (i):

7310+3−256+5−3215+32=(30−3)−(230−10)−(6−30)\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} =(\sqrt{30} - 3)-(2\sqrt{30} - 10) - (6 -\sqrt{30}) \\[1.5em] ⇒30−3−230+10−6+30\Rightarrow \sqrt{30} - 3 - 2\sqrt{30} + 10 - 6 + \sqrt{30} \\[1.5em] ⇒230−230−3+10−6\Rightarrow 2\sqrt{30} - 2\sqrt{30} -3 + 10 - 6 \\[1.5em] ⇒1\bold{\Rightarrow 1}

Question 4

Simplify : 14+5+15+6+16+7+17+8+18+9\dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt{5} + \sqrt{6}} + \dfrac{1}{\sqrt{6} + \sqrt{7}} + \dfrac{1}{\sqrt{7} + \sqrt{8}} +\dfrac{1}{\sqrt{8} + \sqrt{9}}.

Answer

14+5+15+6+16+7+17+8+18+9....(i)\dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt{5} + \sqrt{6}} + \dfrac{1}{\sqrt{6} + \sqrt{7}} + \dfrac{1}{\sqrt{7} + \sqrt{8}} +\dfrac{1}{\sqrt{8} + \sqrt{9}} \qquad \text{....(i)}

Simplifying each term individually,

14+5\dfrac{1}{\sqrt{4} + \sqrt{5}}

Let us rationalise its denominator,

Then,

14+5=14+5×4−54−5⇒4−5(4)2−(5)2⇒4−54−5⇒−(4−5)⇒(5−4)....(ii)\dfrac{1}{\sqrt{4} + \sqrt{5}} = \dfrac{1}{\sqrt{4} + \sqrt{5}} × \dfrac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{4} - \sqrt{5}}}{(\sqrt{4})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{4} - \sqrt{5}}}{4 - 5} \\[1.5em] \Rightarrow{-(\sqrt{4} - \sqrt{5})} \\[1.5em] \Rightarrow{(\sqrt{5} - \sqrt{4})} \qquad \text{....(ii)} \\[1.5em]

15+6\dfrac{1}{\sqrt{5} + \sqrt{6}}

Let us rationalise its denominator,

Then,

15+6=15+6×5−65−6⇒5−6(5)2−(6)2⇒5−65−6⇒−(5−6)⇒(6−5)....(iii)\dfrac{1}{\sqrt{5} + \sqrt{6}} = \dfrac{1}{\sqrt{5}+\sqrt{6}}×\dfrac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{6}}}{(\sqrt{5})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{6}}}{5 - 6} \\[1.5em] \Rightarrow{-(\sqrt{5} - \sqrt{6})} \\[1.5em] \Rightarrow{(\sqrt{6} - \sqrt{5})} \qquad \text{....(iii)} \\[1.5em]

16+7\dfrac{1}{\sqrt{6} + \sqrt{7}}

Let us rationalise its denominator,

Then,

16+7=16+7×6−76−7⇒6−7(6)2−(7)2⇒6−76−7⇒−(6−7)⇒(7−6)....(iv)\dfrac{1}{\sqrt{6} + \sqrt{7}} = \dfrac{1}{\sqrt{6}+ \sqrt{7}} × \dfrac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{6} - \sqrt{7}}}{(\sqrt{6})^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{6} - \sqrt{7}}}{6-7} \\[1.5em] \Rightarrow{-(\sqrt{6} - \sqrt{7})} \\[1.5em] \Rightarrow{(\sqrt{7} - \sqrt{6})} \qquad \text{....(iv)} \\[1.5em]

17+8\dfrac{1}{\sqrt7+\sqrt8}

Let us rationalise its denominator,

Then,

17+8=17+8×7−87−8⇒7−8(7)2−(8)2⇒7−87−8⇒−(7−8)⇒(8−7)....(v)\dfrac{1}{\sqrt{7} + \sqrt{8}} = \dfrac{1}{\sqrt{7}+ \sqrt{8}} × \dfrac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{7} - \sqrt{8}}}{(\sqrt{7})^2 - (\sqrt{8})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{7} - \sqrt{8}}}{7-8} \\[1.5em] \Rightarrow{-(\sqrt{7} - \sqrt{8})} \\[1.5em] \Rightarrow{(\sqrt{8} - \sqrt{7})} \qquad \text{....(v)} \\[1.5em]

14+5\dfrac{1}{\sqrt{4} + \sqrt{5}}

Let us rationalise its denominator,

Then,

18+9=18+9×8−98−9⇒8−9(8)2−(9)2⇒8−98−9⇒−(8−9)⇒(9−8)....(vi)\dfrac{1}{\sqrt{8} + \sqrt{9}} = \dfrac{1}{\sqrt{8}+ \sqrt{9}} × \dfrac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}} \\[1.5em] \Rightarrow\dfrac{{\sqrt {8} - \sqrt{9}}}{(\sqrt{8})^2 - (\sqrt{9})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{8} - \sqrt{9}}}{8 - 9} \\[1.5em] \Rightarrow{-(\sqrt{8} - \sqrt{9})} \\[1.5em] \Rightarrow{(\sqrt{9}-\sqrt{8})} \qquad \text{....(vi)} \\[1.5em]

Using (ii) , (iii) , (iv) , (v) , (vi) in equation (i):

14+5+15+6+16+7+17+8+18+9=5−4+6−5+7−6+8−7+9−8=9−4=3−2=1\dfrac{1}{\sqrt4+\sqrt5} + \dfrac{1}{\sqrt5+\sqrt6} + \dfrac{1}{\sqrt6+\sqrt7} + \dfrac{1}{\sqrt7+\sqrt8} +\dfrac{1}{\sqrt8+\sqrt9} \\[1.5em] = \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8} \\[1.5em] = \sqrt{9} - \sqrt{4} \\[1.5em] = \bold{3 - 2 = 1 } \\[1.5em]

Question 5

Given a and b are rational numbers. Find a and b if :

(i)3−53+25=−1911+a5(ii)2+332−23=a−b6(iii)7+57−5−7−57+5=a+711b5(iv)22−322+23=a+b24\begin{matrix} \text{(i)} & \dfrac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = -\dfrac{19}{11} + a\sqrt{5} \\[1.5em] \text{(ii)} & \dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2}-2\sqrt{3}} = a - b\sqrt{6} \\[1.5em] \text{(iii)} & \dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} - \dfrac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \dfrac{7}{11}b\sqrt{5} \\[1.5em] \text{(iv)} & \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} = a + b\sqrt{24} \end{matrix}

Answer

(i) Since, it is given that

3−53+25\dfrac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = −1911+a5-\dfrac{19}{11} + a\sqrt{5}

On solving,

(3−5)(3+25)×(3−25)(3−25)⇒9−65−35+1032−(25)2⇒19−959−20⇒−95+19−11⇒−1911+9511∴−1911+9511=−1911+a5\dfrac{(3 - \sqrt{5})}{(3 + 2\sqrt{5})} × \dfrac{(3-2\sqrt{5})}{(3-2\sqrt{5})} \\[1.5em] \Rightarrow \dfrac{9 - 6\sqrt{5} - 3\sqrt{5} + 10 }{3^2 - (2\sqrt{5})^2} \\[1.5em] \Rightarrow \dfrac{19 - 9{\sqrt{5}}}{9 - 20} \\[1.5em] \Rightarrow \dfrac{-9{\sqrt{5}} + 19}{-11} \\[1.5em] \Rightarrow -\dfrac{19}{11} + \dfrac{9\sqrt{5}}{11} \\[1.5em] \therefore -\dfrac{19}{11} + \dfrac{9\sqrt5}{11} = -\dfrac{19}{11} + a\sqrt{5}

Hence, a = 911\dfrac{9}{11}.

(ii) Since, it is given that

2+332−23=a−b6\dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a - b\sqrt6

On solving,

2+332−23×(32+23)(32+23)⇒6+26+36+6(32)2−(23)2⇒12+5618−12⇒12+566⇒126+566⇒2+566∴2−(−56)6=a−b6\dfrac{\sqrt{2} + \sqrt3}{3\sqrt{2} - 2\sqrt{3}} × \dfrac{(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} + 2\sqrt{3})} \\[1.5em] \Rightarrow \dfrac{6 + 2\sqrt{6} + 3\sqrt6 + 6 }{(3\sqrt{2})^2 -(2\sqrt{3})^2} \\[1.5em] \Rightarrow \dfrac{12 + 5{\sqrt6}}{18-12} \\[1.5em] \Rightarrow \dfrac{12 + 5{\sqrt{6}}}{6} \\[1.5em] \Rightarrow \dfrac{12}{6} + \dfrac{5\sqrt{6}}{6} \\[1.5em] \Rightarrow 2 + \dfrac{5\sqrt{6}}{6} \\[1.5em] \therefore 2 - \dfrac{(-5\sqrt{6})}{6} = a - b\sqrt{6}

Hence, a = 2 and b = −56\dfrac{-5}{6}.

(iii) Since, it is given that

(7+5)(7−5)−(7−5)(7+5)=a+711b5\dfrac{(7 + \sqrt5)}{(7 - \sqrt{5})} - \dfrac{(7 - \sqrt5)}{(7 + \sqrt{5})} = a+\dfrac{7}{11}b\sqrt{5}

On solving,

(7+5)(7+5)−(7−5)(7−5)(7−5)(7+5)(7+5)2−(7−5)2(7−5)(7+5)⇒(72+2×7×5+(5)2)−(72−2×7×5+(5)2)72−(5)2⇒(49+2×75+5)−(49−2×75+5)72−(5)2⇒49+145+5−49+145−572−(5)2⇒28549−5⇒28544⇒7511∴a+711b5=7511⇒a+711b5=0+711×1×5\dfrac{(7 + \sqrt{5})(7 + \sqrt{5}) - (7-\sqrt{5})(7 - \sqrt{5})}{(7 - \sqrt{5})(7 + \sqrt{5})} \\[1.5em] \dfrac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7+\sqrt{5})} \\[1.5em] \Rightarrow \dfrac{(7^2 + 2 × 7 × \sqrt5 + (\sqrt{5})^2)-(7^2 - 2 × 7 × \sqrt5 + (\sqrt{5})^2)}{7^2-(\sqrt5)^2} \\[1.5em] \Rightarrow \dfrac{(49 + 2 × 7\sqrt{5} + 5)-(49 - 2 × 7\sqrt{5}+ 5)}{7^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow \dfrac{49+14\sqrt{5} + 5 - 49 + 14\sqrt{5} - 5}{7^2 - (\sqrt5)^2} \\[1.5em] \Rightarrow \dfrac{28{\sqrt{5}}}{49 - 5} \\[1.5em] \Rightarrow \dfrac{28{\sqrt{5}}}{44} \\[1.5em] \Rightarrow \dfrac{7{\sqrt{5}}}{11} \\[1.5em] \therefore a+\dfrac{7}{11}b\sqrt{5} = \dfrac{7{\sqrt5}}{11} \\[1.5em] \Rightarrow a + \dfrac{7}{11}b\sqrt{5} = 0 + \dfrac{7}{11} × 1 × \sqrt{5}

Hence, value of a = 0 and b = 1.

(iv) Since, it is given that

22−322+23=a+b24\dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} = a + b\sqrt{24}

Rationalizing, 22−322+23\dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}, we get :

⇒22−322+23×22−2322−23⇒(22−3)(22−23)(22)2−(23)2⇒8−46−26+68−12⇒14−66−4⇒−144+664⇒−72+326⇒−72+362×22⇒−72+36×224⇒−72+3244.\Rightarrow \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} \times \dfrac{2\sqrt{2} - 2\sqrt{3}}{2\sqrt{2} - 2\sqrt{3}} \\[1em] \Rightarrow \dfrac{(2\sqrt{2} - \sqrt{3})(2\sqrt{2} - 2\sqrt{3})}{(2\sqrt{2})^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{8 - 4\sqrt{6} - 2\sqrt{6} + 6}{8 - 12} \\[1em] \Rightarrow \dfrac{14 - 6\sqrt{6}}{-4} \\[1em] \Rightarrow -\dfrac{14}{4} + \dfrac{6\sqrt{6}}{4} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3}{2}\sqrt{6} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3\sqrt{6}}{2} \times \dfrac{2}{2} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3\sqrt{6 \times 2^2}}{4} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3\sqrt{24}}{4}.

Comparing −72+3244-\dfrac{7}{2} + \dfrac{3\sqrt{24}}{4} with a + b24b\sqrt{24}, we get :

a = −72,b=34-\dfrac{7}{2}, b = \dfrac{3}{4}.

Hence, a = −72,b=34-\dfrac{7}{2}, b = \dfrac{3}{4}.

Question 6

If 7+353+5−7−353−5=p+q5\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = p + q\sqrt{5} , find the values of p and q where p and q are rational numbers.

Answer

Since, it is given that

(7+35)3+5\dfrac{(7 + 3\sqrt{5})}{3 + \sqrt{5}} - (7−35)(3−5)\dfrac{(7 - 3\sqrt{5})} {(3 - \sqrt{5})} = p + q5\sqrt{5}

On solving,

(7+35)(3−5)−(7−35)(3+5)(3+5)(3−5)⇒(21−75+95−15)−(21+75−95−15)32−(5)2⇒(21−75+95−15)−(21+75−95−15)9−5⇒21−75+95−15−21−75+95+159−5⇒185−1454⇒4×54=5∴5=p+q5⇒0+1×5=p+q5\dfrac{(7 + 3\sqrt{5})(3 - \sqrt{5}) - (7 - 3\sqrt{5})(3 + \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} \\[1.5em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - {(\sqrt{5})}^2} \\[1.5em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{9-5} \\[1.5em] \Rightarrow \dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9-5} \\[1.5em] \Rightarrow \dfrac{18{\sqrt{5}} - 14{\sqrt{5}}}{4} \\[1.5em] \Rightarrow \dfrac{4 × {\sqrt{5}}}{4} = \sqrt{5} \\[1.5em] \therefore\sqrt{5} = p + q{\sqrt{5}} \\[1.5em] \Rightarrow 0 + 1 × \sqrt{5} = p + q{\sqrt{5}}

Hence, value of p = 0 and q = 1.

Question 7

Rationalise the denominator of the following and hence evaluate by taking 2\sqrt{2} = 1.414 and 3\sqrt{3} = 1.732 , upto three places of decimal:

(i)22+2(ii)13+2\begin{matrix} \text{(i)} & \dfrac{\sqrt2}{2+\sqrt2} \\[1.5em] \text{(ii)} & \dfrac{1}{\sqrt3+\sqrt2} \\[1.5em] \end{matrix}

Answer

(i) Rationalise the denominator ,

22+2=22+2×2−22−2⇒2×(2−2)22−(2)2⇒22−24−2⇒(2−1)\dfrac{\sqrt{2}}{2 + \sqrt{2}} = \dfrac{\sqrt{2}}{2+ \sqrt{2}} × \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}} \\[1.5em] \Rightarrow \sqrt{2} × \dfrac{(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{2{\sqrt{2}} - 2}{4 - 2} \\[1.5em] \Rightarrow (\sqrt{2} - 1) \\[1.5em]

Since, 2\sqrt{2} = 1.414

⇒1.414−1⇒0.414\Rightarrow {1.414 - 1} \\[1.5em] \Rightarrow\bold{ 0.414} \\[1.5em]

(ii)(\text{ii}) Rationalise the denominator ,

13+2=13+2×3−23−2⇒(3−2)(3)2−(2)2⇒3−23−2⇒(3−2)\dfrac{1}{\sqrt{3} + \sqrt{2}} = \dfrac{1}{\sqrt{3} + \sqrt{2}} × \dfrac{{\sqrt{3} - \sqrt{2}}}{{\sqrt{3} - \sqrt{2}}} \\[1.5em] \Rightarrow \dfrac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 -(\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2}}{3-2} \\[1.5em] \Rightarrow (\sqrt{3} - \sqrt{2}) \\[1.5em]

Since, 3\sqrt{3} = 1.732

⇒1.732−1.414⇒0.318\Rightarrow {1.732 - 1.414} \\[1.5em] \Rightarrow\bold{0.318} \\[1.5em]

Question 8

If a = 2 + 3\sqrt{3} , then find the value a−1aa - \dfrac{1}{a}.

Answer

Given,

a=2+3∴1a=12+3×2−32−3⇒2−322−(3)2⇒2−34−3⇒(2−3)∴a−1a=2+3−2+3⇒a−1a=23a = 2 + \sqrt{3} \\[1.5em] \therefore \dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3}} ×\dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \\[1.5em] \Rightarrow \dfrac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow \dfrac{2 - \sqrt{3}}{4 - 3} \\[1.5em] \Rightarrow (2 - \sqrt{3}) \\[1.5em] \therefore a - \dfrac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3} \\[1.5em] \Rightarrow\bold{ a - \dfrac{1}{a} = 2\sqrt{3}} \\[1.5em]

Question 9

If x=1−2x = 1-\sqrt{2}, find the value of (x−1x)4{\Big(x-\dfrac{1}{x}\Big)}^4.

Answer

Given,

x=1−2∴1x=11−2×1+21+2⇒1+21−(2)2⇒1+21−2⇒−(1+2)∴x−1x=1−2+1+2⇒x−1x=2∴(x−1x)4=24=16x = 1 - \sqrt{2} \\[1.5em] \therefore \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}} × \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1.5em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - 2} \\[1.5em] \Rightarrow - (1 +\sqrt{2}) \\[1.5em] \therefore x - \dfrac{1}{x} = 1 - \sqrt{2} + 1 + \sqrt{2} \\[1.5em] \Rightarrow x - \dfrac{1}{x} = 2 \\[1.5em] \therefore \bold{{\Big(x-\dfrac{1}{x}\Big)}^4 = 2 ^4 = 16}

Question 10

If x = 5−265 - 2\sqrt{6}, find the value of x2+1x2x^2 + \dfrac{1}{x^2}.

Answer

Given x = 5−265 - 2\sqrt{6}

∴1x=15−26=15−26×5+265+26⇒5+26(5)2−(26)2⇒5+2625−24=5+261⇒1x=5+26∴(x+1x)=(5−26)+(5+26)=10....(i)\therefore \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} = \dfrac{1}{5 - 2\sqrt{6}} × \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{25 - 24} = \dfrac{5 + 2\sqrt{6}}{1} \\[1.5em] \Rightarrow\dfrac{1}{x} = 5+2\sqrt6 \\[1.5em] \therefore (x + \dfrac{1}{x}) = (5 - 2\sqrt6) + (5 + 2\sqrt6) = 10 \qquad \text{....(i)}

We know that (x+1x)2=x2+1x2+2{\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2

⇒x2+1x2=(x+1x)2−2⇒x2+1x2=102−2.....using(i)⇒x2+1x2=100−2x2+1x2=98\Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 10^2 - 2 ..... \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 100 - 2 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 98}

Question 11

If p = 2−52+5\dfrac{2-\sqrt{5}}{2+\sqrt{5}} and q = 2+52−5\dfrac{2+\sqrt{5}}{2-\sqrt{5}} find the values of :

(i) p + q

(ii) p - q

(iii) p2 + q2

(iv) p2 - q2

Answer

(i)

p+q=2−52+5+2+52−5⇒(2−5)2+(2+5)2(2−5)(2+5)⇒4+5−45+4+5+45(2)2−(5)2p+q=−18....(i)p+q = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} +\dfrac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2-\sqrt{5})^2 + (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\dfrac{4 + 5 - 4\sqrt{5} + 4 + 5 + 4\sqrt{5}}{(2)^2 -(\sqrt{5})^2} \\[1.5em] \bold{p+q = -18} \qquad \text{....(i)} \\[1.5em]

(ii)(\text{ii})

p−q=2−52+5−2+52−5⇒(2−5)2−(2+5)2(2−5)(2+5)⇒4+5−45−4−5−45(2)2−(5)2⇒p−q=85....(ii)p-q = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} -\dfrac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2-\sqrt{5})^2 - (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\dfrac{4 + 5 - 4\sqrt{5} - 4 - 5 - 4\sqrt{5}}{(2)^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\bold {p - q = 8\sqrt{5}} \qquad \text{....(ii)}

(iii)(\text{iii})

(p+q)2=(p)2+(q)2+2pq⇒(p)2+(q)2=(p+q)2−2pq....(iii)⇒pq=2−52+5×2+52−5=1....(iv)(p+q)^2 = (p)^2 + (q)^2 +2pq \\[1.5em] \Rightarrow(p)^2+(q)^2 =(p+q)^2 -2pq \qquad \text{....(iii)} \\[1.5em] \Rightarrow pq = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} ×\dfrac{2+\sqrt{5}} {2-\sqrt{5}} = 1 \qquad \text{....(iv)} \\[1.5em]

substituting value of (i) and (iv) in (iii) :

p2+q2=(−18)2−2=324−2=322\bold{p^2+q^2 = (-18)^2 - 2 = 324-2 = 322}

(iv)(\text{iv})

(p)2−(q)2=(p+q)(p−q)....(v)(p)^2 - (q)^2 = (p+q)(p-q) \qquad \text{....(v)}

Using (i) and (ii) in (v) we get,

p2−q2=−18×85=−1445\bold{p^2 - q^2 = -18 × 8\sqrt{5} = -144\sqrt{5}}

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