Page 2.20
Exercise 2.1
Type 1Question 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 :
$x=\dfrac{3}{4\sqrt{3}}$
$x=\dfrac{3}{4\sqrt{3}}\times \dfrac{\sqrt3}{\sqrt3}$
$x=\dfrac{\sqrt{3}}{4}$
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
⇒x=+√6-1 , -√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√3-3√2+10√3
⇒4√3+10√3-3√2
⇒√3(4+10)-3√2
⇒$14\sqrt{3}-3\sqrt{2}$
(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 numberSol :
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 numberSol :
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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