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KC Sinha Mathematics Solution Class 9 Chapter 2 Operations on Real Numbers and Laws of Exponents for Real Numbers exercise 2.1


Page 2.20

Exercise 2.1

Type 1

Question 1

Classif‌y the following numbers as rational or irrational,

(i) 3+√5

Sol :Irrational

 

(ii) 7√5

Sol :Irrational

 

(iii) \dfrac{7}{\sqrt{5}}

Sol :Irrational

 

(iv) 5+√5-√5

Sol : Rational

 

(v) √2+21

Sol : Irrational

 

(vi) π-2

Sol :Irrational

 

(vii) √2 + √3

Sol : Irrational

 


Question 2

Solution of which of the following represent rational numbers ?

(i) x2=5

Sol :

x=±√5

Which is not a rational number


(ii) x^2=\dfrac{16}{9}

Sol :

x=\sqrt{\dfrac{16}{9}}

x=\pm \dfrac{4}{3}

Which is a Rational number


(iii) (x+√2)(x-√3)=0

Sol :

x=-√2 , x=-√3

Which is not a rational number


(iv) 3x2=9

Sol :

x^2=\dfrac{9}{3}

x2=3

x=±√3

Which is not a rational number


(v) \sqrt{3}x=\dfrac{3}{4}

Sol :

x=\dfrac{3}{4\sqrt{3}}

x=\dfrac{\sqrt{3}}{4}

Which is not a rational number


(vi) x^2=\dfrac{25}{49}

Sol :

x=\pm \sqrt{\dfrac{25}{49}}

\pm \dfrac{5}{7}

Which is a rational number


(vii) (x+1)2=6

Sol :

⇒x+1=±√6

⇒x=±(√6)-1

Which is a rational number


(viii) (x+√5)(x-√3)=0

Sol :

x=-√5 , +√5

Which is a rational number


Page 2.21

Question 3

Solution of which of the following represent irrational numbers :

(i) x2=5

Sol :

⇒x=±√5

Irrational number


(ii) x^2=\dfrac{16}{9}

Sol :

x=\sqrt{\dfrac{16}{9}}

x=\pm \dfrac{4}{3}

Rational number


(iii) (x-1)^2=\dfrac{49}{16}

Sol :

x-1=\sqrt{\dfrac{49}{16}}

x-1=\pm \dfrac{7}{4}

x-1= +\dfrac{7}{4} and x-1=-\dfrac{7}{4}

x=\dfrac{7}{4}+1 and x=-\dfrac{7}{4}+1

x=\dfrac{7+4}{4} and x=\dfrac{-7+4}{4}

x=\dfrac{11}{4} and x=\dfrac{-3}{4}

Rational numbers


(iv) (x+1)(x-1)=0

Sol :

⇒x2-1=0

⇒x=±√1

Rational numbers


(v) x^2=\dfrac{19}{29}

Sol :

x=\sqrt{\dfrac{19}{29}}

Irrational number


(vi) (x-1)=5

Sol :

⇒x=5+1

⇒x=6


Question 4

For each of the following give example of two irrational numbers such that their :

(i) sum is a rational number

Sol :

1+√2 , 1-√2


(ii) sum is an irrational number

Sol :

\sqrt{2}+1,\sqrt{2}-1


(iii) difference is a rational number

Sol :

\sqrt{2}+1,\sqrt{2}-1


(iv) difference is an irrational number

Sol :

1+\sqrt{2}+1,1-\sqrt{2}


(v) product is a rational number

Sol :

3+\sqrt{2},3-\sqrt{2}


(vi) product is an irrational number

Sol :

2\sqrt{2},2\sqrt{3}


(vii) quotient is a rational number

Sol :

4\sqrt{2},\sqrt{2}


(viii) quotient is an irrational number

Sol :

4\sqrt{2},\sqrt{3}


Question 5

Give example of a rational number and an irrational number such that their product is a rational number.

Sol :

Rational number=0 ,

Irrational number=√2

0×√2=0 (rational number)

If you multiply any irrational number by the rational number zero, the result will be zero, which is rational


Type 2

Question 6

Simplify each of the following :

(i) (5+√5)(5-√5)

Sol :

Using identity:

(a+b)(a-b)=a2-b2

⇒(5)2-(√5)2

⇒25-5

⇒20


(ii) (5+√7)(2+√5)

Sol :

⇒5(2+√5)+√7(2+√5)

⇒10+5√5+2√7+√35


(iii) (√11-√7)(√11+√7)

Sol :

Using identity:

(a+b)(a-b)=a2-b2

⇒(√11)2-(√7)2

⇒11-7

⇒4


(iv) (11+√11)(11-√11)

Sol :

⇒11(11-√11)+√11(11-√11)

⇒121-11√11+11√11-11

⇒110


(v) (3+√2)(3-√2)

Sol :

⇒3(3-√2)+√2(3-√2)

⇒9-3√2+3√2-2

⇒7


(vi) (√3+√7)2

Sol :

Using identity:

(a+b)2=a2+b2+2ab

⇒(√3)2+(√7)2+2(√3)(√7)

⇒3+7+2√21

⇒10+2√21


Question 7

Simplify each of the following :

(i) 5√2+4√2

Sol :

⇒√2(5+4)

⇒9√2


(ii) 3√7+2√7

Sol :

⇒√7(3+2)

⇒5√7


(iii) 8√3-5√3

Sol :

⇒√3(8-5)

⇒3√3


(iv) 4√7+5√7-3√7

Sol :

⇒√7(4+5-3)

⇒6√7


(v) 8\sqrt[3]{5}+7\sqrt[3]{5}-13\sqrt[3]{5}

Sol :

\sqrt[3]{5}(8+7-13)

2\sqrt[3]{5}


(vi) 5√3+2√27

Sol :

⇒5√3+2√3×3×3

⇒ 5√3+2×3√3

⇒ 5√3+6√3

⇒√3(6+5)

⇒11√3


Question 8

Simplify each of the following :

(i) 4√3-3√2+2√75

Sol :

⇒4√3-3√2+2√5×5×3

⇒4√3+10√3-3√2

⇒√3(4+10)-3√2

14\sqrt{13}-3\sqrt{2}


(ii) √8+√32-√2

Sol :

⇒ √2×2×2+√2×2×2×2×2-√2

⇒2√2+4√2-√2

⇒2√2+3√2

⇒5√2


(iii) \sqrt{192}-\dfrac{1}{2}\sqrt{48}-\sqrt{75}

Sol :

\sqrt{8\times 8\times 3}-\dfrac{1}{2}\sqrt{4\times 4\times 3}-\sqrt{5\times 5\times 3}

8\sqrt{3}-\dfrac{1}{2}\times 4\sqrt{3}-5\sqrt{3}

\sqrt{3} \left(8-\dfrac{4}{2}-5\right)

⇒√3(8-2-5)

⇒√3


Type 3

Problems based on rationalization of the denominator.

WORKING RULE:

1. First of all f‌ind the rationalizing factor (R.F) of the denominator :

(i) R.F of the monomial surd a√b is √b

(ii) R.F of the binomial surd a+√b is a-√b

(iii) R.F of the binomial surd a-√b is a+√b

(iv) R.F of the binomial surd a±c√b is a∓c√b

(v) R.F of the binomial surd √a±√b is √a∓√b

2. Multiply numerator and denominator of the given surd by R.F of denominator and simplify.

3. If denominator of the quadratic polynomial is trinomial surd then taking two terms together and use working rules 1 and 2

4.Use the following algebraic formulae whichever is required:

(a+b)2=a2+b2+2ab

(a-b)2=a2+b2-2ab

a2-b2=(a+b)(a-b)


Question 9

Write the simplest rationalizing factor (R.F.) for each of the following :

(i) 5√2

Sol :

⇒5√2 × √2

⇒10 [rational number]

Rationalizing factor is √2


(ii) 2√2

Sol :

⇒2√2 × √2

⇒4

Rationalizing factor is √2


(iii) √7

Sol :

⇒√7 × √7

⇒7

Rationalizing factor is √7


(iv) √15

Sol :

⇒√15 × √15

⇒15

Rationalizing factor is √15

 


Question 10

If a,b,c are rational numbers , then write the R.F of 

(i) \sqrt[5]{a^2b^3c^4}

Sol :

Note: product of two rational numbers are rational number

Let x be another rational number and on multiplying with rational number gives rational number abc

\left(\sqrt[5]{a^2b^3c^4}\right)\times \left(x\right)=abc

x=\dfrac{abc}{a^{\frac{2}{5}} b^{\frac{3}{5}} c^{\frac{4}{5}}}

x=a^{1-\frac{2}{5}} b^{1-\frac{3}{5}} c^{1-\frac{4}{5}}

x=a^{\frac{5-2}{5}} b^{\frac{5-3}{5}} c^{\frac{5-4}{5}}

x=a^{\frac{3}{5}} b^{\frac{2}{5}} c^{\frac{1}{5}}

x=\sqrt[5]{a^3 b^2 c^1}

 


(ii) \sqrt[9]{a^2b^4c^8}

Sol :

Note: product of two rational numbers are rational number

Let x be another rational number and on multiplying with rational number gives rational number abc

\left(\sqrt[9]{a^2b^4c^8}\right)\times \left(x\right)=abc

x=\dfrac{abc}{a^{\frac{2}{9}} b^{\frac{4}{9}} c^{\frac{8}{9}}}

x=a^{1-\frac{2}{9}} b^{1-\frac{4}{9}} c^{1-\frac{8}{9}}

x=a^{\frac{9-2}{9}} b^{\frac{9-4}{9}} c^{\frac{9-8}{9}}

x=a^{\frac{7}{9}} b^{\frac{5}{9}} c^{\frac{1}{9}}

x=\sqrt[9]{a^{7} b^{5} c^{1}}

 


Question 11

Rationalize the denominator of the following :

(i) \dfrac{1}{\sqrt{2}}

Sol :

\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}

\dfrac{\sqrt{2}}{2}

 


(ii) \dfrac{1}{\sqrt{12}}

Sol :

\dfrac{1}{\sqrt{12}}\times \dfrac{\sqrt{12}}{\sqrt{12}}

\dfrac{\sqrt{12}}{12}

\dfrac{\sqrt{2×2×3}}{12}

\dfrac{2\sqrt{3}}{12}

\dfrac{\sqrt{3}}{6}

 


(iii) \dfrac{2\sqrt{7}}{\sqrt{11}}

Sol :

\dfrac{2\sqrt{7}}{\sqrt{11}}\times \dfrac{\sqrt{11}}{\sqrt{11}}

\dfrac{2\sqrt{77}}{11}

 


(iv) \dfrac{2}{\sqrt{17}}

Sol :

\dfrac{2}{\sqrt{17}}\times \dfrac{\sqrt{17}}{\sqrt{17}}

\dfrac{2\sqrt{17}}{17}

 


Question 12

Fill up the blanks after rationalizing the denominator :

(i) \dfrac{1}{\sqrt{2}+1}=\dots

Sol :

\dfrac{1}{\sqrt{2}+1}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}

\dfrac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{\sqrt{2}-1}{(\sqrt{2})^2-1^2)}

\dfrac{\sqrt{2}-1}{2-1)}

⇒(√2)-1

 


(ii) \dfrac{1}{2-\sqrt{3}}=\dots

Sol :

\dfrac{1}{2-\sqrt{3}}\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2}

\dfrac{2+\sqrt{3}}{4-3}

⇒2+(√3)

 


(iii) \dfrac{3}{\sqrt{5}+\sqrt{3}}=\dots

Sol :

\dfrac{3}{\sqrt{5}+\sqrt{3}}\times \dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}

\dfrac{3(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{3(\sqrt{5}-\sqrt{3})}{5-3}

\dfrac{3}{2}(\sqrt{5}-\sqrt{3})

 


(iv) \dfrac{7}{\sqrt{5}-\sqrt{3}}=\dots

Sol :

\dfrac{7}{\sqrt{5}-\sqrt{3}}\times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}

\dfrac{7(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}

\dfrac{7(\sqrt{5}+\sqrt{3})}{5-3}

\dfrac{7}{2}(\sqrt{5}+\sqrt{3})

 


Question 13

Write the following surds in the simplest form :

(i) √48

Sol :

⇒√2×2×2×2×3

⇒4√3

 


(ii) √175

Sol :

⇒√5×5×7

⇒5√7

 


(iii) \sqrt[3]{72}

Sol :

\sqrt[3]{2×2×2×3×3}

2\sqrt[3]{9}

 


(iv) √125

Sol :

⇒√5×5×5

⇒5√5

 


(v) \sqrt[3]{54}

Sol :

\sqrt[3]{2\times 3 \times 3 \times 3 }

3\sqrt[3]{2}

 


(vi) \sqrt[3]{144}

Sol :

\sqrt[3]{\underline{2\times 2\times 2} \times 2 \times 3\times 3}

2\sqrt[3]{18}

 


(vii) \sqrt[5]{320}

Sol :

\sqrt[5]{\underline{2\times 2\times 2\times 2 \times 2} \times 2 \times 5}

2\sqrt[5]{2\times 5}

2\sqrt[5]{10}

 


(viii) \sqrt{\dfrac{125}{63}}

Sol :

\sqrt{\dfrac{\underline{5\times 5}\times 5}{\underline{3\times 3}\times 7}}

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

On rationalizing

5\sqrt{5}\times \dfrac{1}{3\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}

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

\dfrac{5}{21}\sqrt{35}

 


Question 14

Rationalize the denominator in each of the following :

(i) \dfrac{1}{2+\sqrt{3}}

Sol :

\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{2+\sqrt{3}}{2^2-{\sqrt{3}}^2}

\dfrac{2+\sqrt{3}}{4-3}

⇒2-√3

 


(ii) \dfrac{1}{7+3\sqrt{2}}

Sol :

\dfrac{1}{7+3\sqrt{2}}\times \dfrac{7-3\sqrt{2}}{7-3\sqrt{2}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{7-3\sqrt{2}}{(7)^2-(3\sqrt{2})^2}

\dfrac{7-3\sqrt{2}}{49-(9\times 2)}

\dfrac{7-3\sqrt{2}}{49-18}

\dfrac{7-3\sqrt{2}}{31}

 


(iii) \dfrac{5}{\sqrt{3}-\sqrt{5}}

Sol :

\dfrac{5}{\sqrt{3}-\sqrt{5}}\times\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{5(\sqrt{3}+\sqrt{5})}{(\sqrt{3})^2-(\sqrt{5})^2}

\dfrac{5(\sqrt{3}+\sqrt{5})}{3-5}

\dfrac{5(\sqrt{3}+\sqrt{5})}{-2}

[but denominator can’t be negative]

-\dfrac{5}{2}(\sqrt{3}+\sqrt{5})

 


(iv) \dfrac{6}{3\sqrt{2}-2\sqrt{3}}

Sol :

\dfrac{6}{3\sqrt{2}-2\sqrt{3}}\times \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}

\dfrac{6(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^2-(2\sqrt{3})^2}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{6(3\sqrt{2}+2\sqrt{3})}{(9\times 2)-(4\times 3}

\dfrac{6(3\sqrt{2}+2\sqrt{3})}{18-12}

\dfrac{6(3\sqrt{2}+2\sqrt{3})}{6}

⇒3√2+2√3

 


(v) \dfrac{4}{\sqrt{5}+\sqrt{3}}

Sol :

\dfrac{4}{\sqrt{5}+\sqrt{3}}\times \dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{4(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}

\dfrac{4(\sqrt{5}-\sqrt{3})}{5-3}

\dfrac{4(\sqrt{5}-\sqrt{3})}{2}

⇒2(√5-√3)

 


(vi) \dfrac{5+\sqrt{6}}{5-\sqrt{6}}

Sol :

\dfrac{5+\sqrt{6}}{5-\sqrt{6}}\times \dfrac{5+\sqrt{6}}{5+\sqrt{6}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{5+\sqrt{6}\times (5+\sqrt{6})}{(5)^2-(\sqrt{6})^2}

\dfrac{5(5+\sqrt{6})+\sqrt{6}(5+\sqrt{6})}{25-6}

\dfrac{25+5\sqrt{6}+5\sqrt{6}+6}{19}

\dfrac{31+10\sqrt{6}}{19}

 


Question 15

Simplify the following:

(i) \dfrac{3}{5-\sqrt{3}}+\dfrac{2}{5+\sqrt{3}}

Sol :

\dfrac{3(5+\sqrt{3})+2(5-\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{15+3\sqrt{3}+10-2\sqrt{3}}{(5)^2-(\sqrt{3})^2}

\dfrac{25+\sqrt{3}}{25-3}

\dfrac{25+\sqrt{3}}{22}

 


(ii) \dfrac{\sqrt{5}-2}{\sqrt{5}+2}-\dfrac{\sqrt{5}+2}{\sqrt{5}-2}

Sol :

\dfrac{(\sqrt{5}-2)(\sqrt{5}-2)-(\sqrt{5}+2)(\sqrt{5}+2)}{(\sqrt{5}+2)(\sqrt{5}-2)}

Using identity:

(a+b)(a-b)=a2-b2

and

(a-b)(a-b)=(a-b)2=a2+b2-2ab

and

(a+b)(a+b)=(a+b)2=a2+b2+2ab

\dfrac{[(\sqrt{5})^2+(2)^2-2(\sqrt{5})(2)]-[(\sqrt{5})^2+(2)^2+2(\sqrt{5})(2))]}{(\sqrt{5})^2-(2)^2}

\dfrac{5+4-4\sqrt{5}-[5+4+4\sqrt{5}}{5-4}

⇒9-4√5-9-4√5

⇒-8√5

 


(iii) \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}

Sol :

\dfrac{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})+(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}

Using identity:

(a+b)(a-b)=a2-b2

and

(a-b)(a-b)=(a-b)2=a2+b2-2ab

and

(a+b)(a+b)=(a+b)2=a2+b2+2ab

\dfrac{[(\sqrt{5})^2+(\sqrt{3})^2+2(\sqrt{5})(\sqrt{3})]+[(\sqrt{5})^2+(\sqrt{3})^2-2(\sqrt{5})(\sqrt{3})]}{(\sqrt{5})^2-(\sqrt{3})^2}

\dfrac{[5+3+2(\sqrt{5})(\sqrt{3})]+[5+3-2(\sqrt{5})(\sqrt{3})]}{5-3}

\dfrac{[8+2\sqrt{15}+8-2\sqrt{15}}{2}

\dfrac{16}{2}

⇒8

 


Question 16

Simplify the following :

(i) \dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}+\sqrt{35}

Sol :

\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\times \dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\sqrt{35}

Using identity:

(a+b)(a-b)=a2-b2

and

(a-b)(a-b)=(a-b)2=a2+b2-2ab

\dfrac{(\sqrt{7})^2+(\sqrt{5})^2-2(\sqrt{7})(\sqrt{5})}{(\sqrt{7})^2-(\sqrt{5})^2}+\sqrt{35}

\dfrac{7+5-2\sqrt{35}}{7-5}+\sqrt{35}

\dfrac{12-2\sqrt{35}}{2}+\sqrt{35}

\dfrac{12-2\sqrt{35}+2(\sqrt{35})}{2}

\dfrac{12-2\sqrt{35}+2\sqrt{35}}{2}

\dfrac{12}{2}

⇒6

 


(ii) \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+2\sqrt{6}

Sol :

\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+2\sqrt{6}

Using identity:

(a+b)(a-b)=a2-b2

and

(a-b)(a-b)=(a-b)2=a2+b2-2ab

\dfrac{(\sqrt{3})^2+(\sqrt{2})^2-2(\sqrt{3})(\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}+2\sqrt{6}

\dfrac{3+2-2\sqrt{6}}{3-2}+2\sqrt{6}

⇒5-2√6+2√6

⇒5

 


Question 17

(i) If a=3+√8 , find the value of a^2+\dfrac{1}{a^2}

Sol :

⇒a=3+√8 and \dfrac{1}{a}=\dfrac{1}{3+\sqrt{8}}

\dfrac{1}{a}=\dfrac{1}{3+\sqrt{8}}\times \dfrac{3-\sqrt{8}}{3-\sqrt{8}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{1}{a}=\dfrac{3-\sqrt{8}}{(3)^2-(\sqrt{8})^2}

\dfrac{1}{a}=\dfrac{3-\sqrt{8}}{9-8}

\dfrac{1}{a}=3-\sqrt{8}

\left(a+\dfrac{1}{a}\right)=3+\sqrt{8}+3-\sqrt{8}

\left(a+\dfrac{1}{a}\right)=6

[Squaring both sides]

\left(a+\dfrac{1}{a}\right)^2=6^2

a^2+\left(\dfrac{1}{a}\right)^2+2(a)\left(\dfrac{1}{a}\right)=36

a^2+\left(\dfrac{1}{a}\right)^2+2=36

a^2+\left(\dfrac{1}{a}\right)^2=36-2

⇒34


(ii) If a=\dfrac{\sqrt{2}+1}{\sqrt{2}-1} , b=\dfrac{\sqrt{2}-1}{\sqrt{2}+1} , prove that a2+ab+b2=35

Sol :

a^2=\left(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2

Using identity:

(a+b)2=a2+b2+2ab

(a-b)2=a2+b2-2ab

a^2=\left(\dfrac{(\sqrt{2})^2+(1)^2+2(\sqrt{2})(1)}{(\sqrt{2})^2-(1)^2-2(\sqrt{2})(1)}\right)

a^2=\left(\dfrac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}\right)

a^2=\left(\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}\right)..(i)

 

b^2=\left(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\right)^2

Using identity:

(a+b)2=a2+b2+2ab

(a-b)2=a2+b2-2ab

b^2=\left(\dfrac{(\sqrt{2})^2+(1)^2-2(\sqrt{2})(1)}{(\sqrt{2})^2+(1)^2+2(\sqrt{2})(1)}\right)

b^2=\left(\dfrac{2+1-2\sqrt{2}}{2+1+2\sqrt{2}}\right)

b^2=\left(\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}\right)..(ii)

 

putting (i) and (ii) in a2+b2+ab

\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}+\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}+\left(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\right)\left(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\right)

\dfrac{(3+2\sqrt{2})(3+2\sqrt{2})+(3-2\sqrt{2})(3-2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}+1

Using identity:

(a+b)2=a2+b2+2ab

(a-b)2=a2+b2-2ab

\dfrac{[(3)^2+(2\sqrt{2})^2+2(3)(2\sqrt{2})]+[(3)^2+(2\sqrt{2})^2-2(3)(2\sqrt{2})}{(3)^2-(2\sqrt{2})^2}+1

\dfrac{9+8+12\sqrt{2}+9+8-12\sqrt{2}}{9-8}+1

\dfrac{17+17}{1}+1

⇒34+1

⇒35

Hence proved

 


(iii) If a=2+√3 , find the value of 3^3+\dfrac{1}{a^3}

Sol :

⇒a=2+√3 and \dfrac{1}{a}=\dfrac{1}{2+\sqrt{3}}

\dfrac{1}{a}=\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}

\dfrac{1}{a}=\dfrac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}

\dfrac{1}{a}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}

Now ,

\left(a+\dfrac{1}{a}\right)=2+\sqrt{3}+2-\sqrt{3}

\left(a+\dfrac{1}{a}\right)^3=(4)^3 [taking cube both sides]

a^3+\dfrac{1}{a^3}+3(a)\times \dfrac{1}{a}\left(a+\dfrac{1}{a}\right)=64 [squaring both sides]

\left(a^3+\dfrac{1}{a^3}\right)+3\left(\dfrac{a^2+1}{a}\right)=64

\left(a^3+\dfrac{1}{a^3}\right)=64-3\left(\dfrac{(2+\sqrt{3})^2+1}{2+\sqrt{3}}\right)

64-3\left(\dfrac{(2)^2+(\sqrt{3})^2+2(2)(\sqrt{3})+1}{2+\sqrt{3}}\right)

64-3\left(\dfrac{4+3+4\sqrt{3}+1}{2+\sqrt{3}}\right)

64-3\left(\dfrac{8+4\sqrt{3}}{2+\sqrt{3}}\right)

64-\left(\dfrac{24+12\sqrt{3}}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}\right)

64-\left(\dfrac{(24+12\sqrt{3})(2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}\right)

64-\left(\dfrac{48+24\sqrt{3}-24\sqrt{3}-36}{4-3}\right)

⇒64-(12)

⇒52


(iv) If a=\dfrac{\sqrt{3}+1}{\sqrt{3}-1} and b=\dfrac{\sqrt{3}-1}{\sqrt{3}-1} , find the value of a2+ab-b2

Sol :

a=\dfrac{\sqrt{3}+1}{\sqrt{3}-1} and b=\dfrac{1}{a}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}

⇒To find a2-b2+ab

\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2-\left(\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right)^2+\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\right)\left(\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right)

Using identity:

(a+b)2=a2+b2+2ab

(a-b)2=a2+b2-2ab

\left(\dfrac{(\sqrt{3})^2+(1)^2+2(\sqrt{3})(1)}{(\sqrt{3})^2+(1)^2-2(\sqrt{3})(1)}\right)-\left(\dfrac{(\sqrt{3})^2+(1)^2-2(\sqrt{3})(1)}{(\sqrt{3})^2+(1)^2+2(\sqrt{3})(1)}\right)+1

\left(\dfrac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}\right)-\left(\dfrac{3+1-2\sqrt{3}}{3+1+2\sqrt{3}}\right)+1

\left(\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}\right)-\left(\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}\right)+1

\left(\dfrac{(4+2\sqrt{3})(4+2\sqrt{3})-(4-2\sqrt{3})(4-2\sqrt{3})}{(4-2\sqrt{3}})(4+2\sqrt{3})\right)+1

\dfrac{12+12+16\sqrt{3}-(16+12-16\sqrt{3})}{16+8\sqrt{3}-8\sqrt{3}-12}+1

\dfrac{32\sqrt{3}}{4}+1

⇒1+8√3


Type 4

Problems based on properties of binomial surds

WORKING RULE:

1. Equate the rational and irrational parts of both sides of the given equality

2. If a+√b=c+√d , then a=c and √b=√d

If a-√b=c-√d , then a=c and √b=√d

where a,b,c,d are rational and √b and √c are irrational numbers

 


Question 18

If a and b be two rational numbers , find the value of a and b in the following equalities:

(i) \dfrac{3+\sqrt{7}}{3-\sqrt{7}}=a+b\sqrt{7}

Sol :

\dfrac{3+\sqrt{7}}{3-\sqrt{7}}\times \dfrac{3+\sqrt{7}}{3+\sqrt{7}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{(3+\sqrt{7})(3+\sqrt{7})}{(3)^2-(\sqrt{7})^2}

Using identity:

(a+b)(a+b)=(a+b)2=(a2+b2+2ab)

\dfrac{(3)^2+(\sqrt{7})^2+2(3)(\sqrt{7})}{(3)^2-(\sqrt{7})^2}

\dfrac{9+7+6\sqrt{7}}{9-7}

\dfrac{16+6\sqrt{7}}{2}

8+3\sqrt{7}=a+b\sqrt{7}

On comparing we get

⇒a=8 , b=3

 


(ii) \dfrac{4+2\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}

Sol :

\dfrac{4+2\sqrt{5}}{4-3\sqrt{5}}\times \dfrac{4+3\sqrt{5}}{4+3\sqrt{5}}

Using identity:

(a+b)(a-b)=a2-b2

\dfrac{(4+2\sqrt{5})(4+3\sqrt{5})}{(4)^2-(3\sqrt{5})^2}

\dfrac{4(4+3\sqrt{5})+2\sqrt{5}(4+3\sqrt{5})}{16-45}

\dfrac{16+12\sqrt{5}+8\sqrt{5}+30}{-29}

\dfrac{46+20\sqrt{5}}{-29}

\dfrac{-46}{29}+\dfrac{-20\sqrt{5}}{29}=a+b\sqrt{5}

On comparing , we get

a=\dfrac{-46}{29},b=\dfrac{-20}{29}


Question 19

Find the value of a and b in the following equalities:

(i) \dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{\sqrt{5}+1}{\sqrt{5}-1}=a+b\sqrt{5}

Sol :

\dfrac{(\sqrt{5}-1)(\sqrt{5}-1)+(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}+1)(\sqrt{5}-1)}

Using identity:

(a+b)(a-b)=a2-b2 and

(a+b)(a+b)=(a+b)2=(a2+b2+2ab) and

(a-b)(a-b)=(a-b)2=(a2+b2-2ab)

\dfrac{[(\sqrt{5})^2+(1)^2+2(\sqrt{5})(1)]+[(\sqrt{5})^2+(1)^2+2(\sqrt{5})(1)}{(\sqrt{5})^2-(1)^2}

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

\dfrac{12+4\sqrt{5}}{4}

⇒3+√5=a+b\sqrt{5}

On comparing , we get

⇒a=3 , b=0


(ii) \dfrac{7+\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}=a+7\sqrt{5}b

Sol :

\dfrac{(7+\sqrt{5})(7+\sqrt{5})-(7-\sqrt{5})(7-\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}

Using identity:

(a+b)(a-b)=a2-b2 and

(a+b)(a+b)=(a+b)2=(a2+b2+2ab) and

(a-b)(a-b)=(a-b)2=(a2+b2-2ab)

\dfrac{(7+\sqrt{5})(7+\sqrt{5})-(7-\sqrt{5})(7-\sqrt{5})}{(7)^2-(\sqrt{5})^2}

\dfrac{[(7)^2+(\sqrt{5})^2+2(7)(\sqrt{5})]-[(7)^2+(\sqrt{5})^2+2(7)(\sqrt{5})}{49-5}

\dfrac{[49+5+14\sqrt{5}-[49+5-14\sqrt{5}}{44}

\dfrac{54+14\sqrt{5}-54+14\sqrt{5}}{44}

\dfrac{28\sqrt{5}}{44}

\dfrac{7\sqrt{5}}{11}=a+7\sqrt{5}b

On comparing , we get

a=0,b=\dfrac{1}{11}


Type 5

Problems based on simplifying a number of the form of \sqrt[n]{a} i.e. (a)1/n

WORKING RULE:

1. Find that factor of a which is nth power of a positive rational number.

2. Write a in the form bnc, where c is not the nth power of a rational number.

3. \sqrt[n]{a}=(b^nc)^{1/n}=bc^{1/n}=b\sqrt[n]{c}

4. Use of the following laws of exponents whichever is required :

(i) a1/m . a1/n = a(1/m+1/n)

(ii) \dfrac{a^{1/m}}{a^{1/n}}=a(1/m-1/n)

(iii) a1/n . b1/n = (ab)1/n

(iv) \dfrac{a^{1/n}}{b^{1/n}}=\left(\dfrac{a}{b}\right)^{1/n}

(v) a^{-1/n}=\dfrac{1}{a^{1/n}}

(vi) (am)n=amn

 


Question 20

Find the value of the following as the power of a positive integer :

(i) 73 . 93

Sol :

Using property:

an . bn = (ab)n

⇒(7×9)3

⇒(63)3


(ii) 7-3 . (9)-3

Sol :

Using identity:

a^{-1/n}=\dfrac{1}{a^{1/n}}

\dfrac{1}{7}^3\times \dfrac{1}{9}^3

Using identity :

an.bn=(ab)mn

\left(\dfrac{1}{7}\times \dfrac{1}{9}\right)^3

\left(\dfrac{1}{63}\right)^3

⇒(63)-3


(iii) 172 . 175

Sol :

Using property:

am . an = (a)m+n

⇒(17)2+5

⇒(17)7


(iv) 172 . 17-5

Sol :

Using identity :

am.an=(a)m+n

⇒(17)2+(-5)

⇒(17)-3


(v) (52)7

Sol :

Using identity:

⇒(am)n=am×n

⇒(5)2×7

⇒(5)14


(vi) (52)-7

Sol :

Using identity:

⇒(am)n=am×n

⇒(5)2×-7

⇒(5)-14


(vii) \dfrac{23^{10}}{23^7}

Sol :

Using identity:

\dfrac{a^m}{a^n}=a^{m-n}

⇒2310-7

⇒(23)3


(viii) \dfrac{(23)^{-10}}{(23)^7}

Sol :

Using identity:

\dfrac{a^m}{a^n}=a^{m-n}

⇒(23)-10-7

⇒(23)-17


Question 21

Simplify :

(i) 22/3 . 21/3

Sol :

Using identity :

am. an=(a)m+n

(2)^{\frac{2}{3}+\frac{1}{3}}

(2)^{\frac{2+1}{3}}

(2)^{\frac{3}{3}}

⇒2

 


(ii) (31/5)4

Sol :

Using identity :

(am)n=am×n

(3)^{\frac{1}{5}\times 4}

⇒34/5

 


(iii) 131/5 . 171/5

Sol :

Using identity:

am.bm=(ab)m

⇒(13×17)1/5

⇒(221)1/5

 


(iv) \dfrac{7^{1/5}}{7^{1/3}}

Sol :

Using identity:

\dfrac{a^m}{a^n}=a^{m-n}

\dfrac{7^{1/5}}{7^{1/3}}=7^{\frac{1}{5}-\frac{1}{3}}

7^{\frac{3-5}{15}}

7^{\frac{-2}{15}} or

\dfrac{1}{7^{2/15}}

 


Question 22

Simplify the following :

(i) √15×√7

Sol :

⇒√3×5×7

⇒√105

 


(ii) \sqrt[3]{18}\times \sqrt[3]{15}

Sol :

\sqrt[3]{18\times 15}

\sqrt[3]{2\times \underline{3\times 3\times 3}\times 5}

3\sqrt[3]{10}

 


(iii) \sqrt[4]{5}\times \sqrt[4]{8}

Sol :

\sqrt[4]{5\times 8}

\sqrt[4]{40}

 


(iv) \sqrt[7]{9}\times \sqrt[7]{5}\times \sqrt[7]{2}

Sol :

\sqrt[7]{9\times {5}\times {2}}

\sqrt[7]{90}

 


(v) \sqrt[8]{12}\div \sqrt[8]{3}

Sol :

\sqrt[4]{\dfrac{12}{3}}

\sqrt[8]{4}

 


(vi) \sqrt[5]{24}\div \sqrt[5]{6}

Sol :

\sqrt[5]{\dfrac{24}{6}}

\sqrt[5]{4}

 


Question 23

Simplify the following :

(i) \sqrt[3]{2}\times \sqrt{5}

Sol :

Rewritten as

⇒21/3×51/2

L.C.M of 2 and 3 is 6

2^{\frac{1}{3}\times \frac{2}{2}}\times 5^{\frac{1}{2}\times \frac{3}{3}}

⇒22/6×53/6

\sqrt[6]{2^2}\times \sqrt[6]{5^3}

\sqrt[6]{4}\times \sqrt[6]{125}

\sqrt[6]{4\times 125}

\sqrt[6]{500}

 


(ii) \sqrt[3]{7}\times \sqrt{2}

Sol :

Rewritten as

⇒71/3×21/2

L.C.M of 3 and 2 is 6

7^{\frac{1}{3}\times \frac{2}{2}}\times 2^{\frac{1}{2}\times \frac{3}{3}}

⇒72/6×23/6

\sqrt[6]{7^2}\times \sqrt[6]{2^3}

\sqrt[6]{49}\times \sqrt[6]{8}

\sqrt[6]{49\times 8}

\sqrt[6]{392}

 


(iii) \sqrt[3]{5}\times \sqrt{3}

Sol :

Rewritten as

⇒51/3×31/2

L.C.M of 3 and 2 is 6

5^{\frac{1}{3}\times \frac{2}{2}}\times 3^{\frac{1}{2}\times \frac{3}{3}}

⇒52/6×33/6

\sqrt[6]{5^2}\times \sqrt[6]{3^3}

\sqrt[6]{25}\times \sqrt[6]{27}

\sqrt[6]{27\times 25}

\sqrt[6]{675}

 


(iv) \sqrt[3]{7}\times \sqrt[4]{3}

Sol :

Rewritten as

⇒71/3×31/4

L.C.M of 3 and 4 is 12

7^{\frac{1}{3}\times \frac{4}{4}}\times 3^{\frac{1}{4}\times \frac{3}{3}}

⇒74/12×33/12

\sqrt[6]{7^4}\times \sqrt[6]{3^3}

\sqrt[6]{2401}\times \sqrt[6]{27}

\sqrt[6]{2401\times 27}

\sqrt[12]{64827}

 


(v) \sqrt{2}.\sqrt[3]{3}.\sqrt[4]{4}

Sol :

Rewritten  as

⇒21/2×31/3×41/4

L.C.M of 2,3 and 4 is 12

2^{\frac{1}{2}\times \frac{6}{6}} \times 3^{\frac{1}{3}\times \frac{4}{4}}\times 4^{\frac{1}{4}\times \frac{3}{3}}

⇒26/12×34/12×43/12

\sqrt[12]{2^6 \times 3^4 \times 4^3}

\sqrt[12]{2^6 \times 3^4 \times (2^2)^3}

\sqrt[12]{2^6 \times 3^4 \times 2^6}

\sqrt[12]{2^12 \times 3^4 }

2\times\sqrt[12]{3^4 }

2\times 3^{\frac{4}{12}}

⇒2×31/3

2\sqrt[3]{3}

 


(vi) \sqrt{3}.\sqrt[3]{4}.\sqrt[4]{5}

Sol :

Rewritten as

⇒31/2×41/3×51/4

L.C.M of 2,3,4 is 12

3^{\frac{1}{2}\times \frac{6}{6}} \times 4^{\frac{1}{3}\times \frac{4}{4}}\times 5^{\frac{1}{4}\times \frac{3}{3}}

⇒36/12×44/12×53/12

\sqrt[12]{3^6 \times 4^4 \times 5^3}

\sqrt[12]{729 \times 256 \times 125}

\sqrt[12]{23328000}

 


(vii) 24\div \sqrt[3]{200}

Sol :

\dfrac{24}{\sqrt[3]{2\times 2\times 2\times 5\times 5}}

\dfrac{2\times 2\times 2\times 3}{2\sqrt[3]{25}}

\dfrac{12}{\sqrt[3]{25}}

 


(viii) \sqrt[4]{36}\div\sqrt[3]{6}

Sol :

⇒(36)1/4 × (6)-1/3

LCM of 3,4 is 12

(36)^{\frac{1}{4}\times \frac{3}{3}} \times (6)^{-\frac{1}{3}\times \frac{4}{4}}

(36)^{\frac{3}{12}} \times (6)^{-\frac{4}{12}}

\sqrt[12]{36^3\times 6^{-4}}

\sqrt[12]{\dfrac{36^3}{6^{4}}}

\sqrt[12]{\dfrac{36\times 36\times 36}{6\times 6\times 6\times 6}}

\sqrt[12]{6\times 6}

\sqrt[12]{2^2\times 3^2}

\sqrt[12]{6^2}

⇒62/12

⇒61/6

\sqrt[6]{6}

 


Question 24

If √2=1.414 , √3=1.732 , √5=2.236 and √10=3.162 , find the value of the following:

(i) \dfrac{1}{\sqrt{2}}

Sol :

\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}

\dfrac{\sqrt{2}}{2}

\dfrac{1.414}{2}

⇒0.707

 


(ii) \dfrac{3}{\sqrt{10}}

Sol :

\dfrac{3}{\sqrt{10}}\times \dfrac{\sqrt{10}}{\sqrt{10}}

\dfrac{3\sqrt{10}}{10}

\dfrac{3\times 3.162}{10}

⇒3×0.316

⇒0.948

 


(iii) \dfrac{2+\sqrt{3}}{3}

Sol :

\dfrac{2+1.732}{3}

⇒1.244

 


(iv) \dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}

Sol :

\dfrac{3.162+\sqrt{3\times 5}}{\sqrt{2}}

\dfrac{3.162+ 1.732 \times 2.236}{\sqrt{2}}

\dfrac{3.162+ 3.872}{1.414}

\dfrac{7.034}{1.414}

⇒4.975

 


Question 25

If m and n be the two rational numbers such that mn=25 , then the value of nm is :

(i) 4

(ii) 10

(iii) 32

(iv) 16

Sol :

⇒mn=25

⇒mn=(5)2

⇒n=2 and m=5

⇒nm=(2)5

⇒nm=32

 


Question 26

\sqrt{10}\times \sqrt{15} is equal to :

(i) 5√6

(ii) 6√5

(iii) √30

(iv) √25

 

Sol :

\sqrt{10\times \sqrt{15}}

\sqrt{10\times \sqrt{15}}

⇒√150

⇒√2×3×5×5

⇒5√2×3

⇒5√6


Question 27

\sqrt[5]{6}\times \sqrt[5]{6^0} is equal to

(i) \sqrt[5]{36}

(ii) \sqrt[5]{6\times 0}

(iii) \sqrt[5]{6}

(iv) \sqrt[5]{12}

 

Sol :

\sqrt[5]{6} \times \sqrt[5]{6^0}

\sqrt[5]{6\times 6^0}

[60=1]

\sqrt[5]{6\times 1}

\sqrt[5]{6}

 


Question 28

\sqrt[3]{8^2} is equal to:

(i) 82/3

(ii) 83/2

(iii) 4×42/3

Sol :

\sqrt[3]{8^2}

Can be rewritten as

(8^2)^{\frac{1}{3}}

⇒82/3

 


Type 6

Question 29

Prove that √2 + √3 is an irrational number

Sol :

Let,√2 + √3 is rational number

We can find co-primes a and b (b≠0) such that

⇒√2 + √3=a/b

Squaring on both sides

⇒(√2 + √3)2=(a/b)2

⇒ 2+3+2(√2)(√3)= a2/b2

[∵ (a+b)2=a2+b2+2ab]

⇒ 6+2√6=a2/b2

⇒ 2√6=a2/b2-6

Since a and b are integers, a2/b2-6 is rational and 2√6 is irrational

But this contradicts our assumption. Hence, √2+√3 is irrational number

 


Question 30

Prove that √2 + √5 is an irrational number

Sol :

Let,√2 + √5 is rational number

We can find co-primes a and b (b≠0) such that

⇒√2 + √5=a/b

Squaring on both sides

⇒(√2 + √5)2=(a/b)2

⇒ 2+5+2(√2)(√5)= a2/b2

[∵ (a+b)2=a2+b2+2ab]

⇒ 7+2√10=a2/b2

⇒ 2√10=a2/b2-7

Since a and b are integers, a2/b2-7 is rational and 2√10 is irrational

But this contradicts our assumption. Hence, √2+√3 is irrational number


 

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