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# Surds Questions with Detailed Solutions

## Easy Questions

Simplify the following

1. $\sqrt{21} \times \sqrt{7}$
2. $\sqrt{180}$
3. $\sqrt{14} \times \sqrt 6$
4. $8 \sqrt 6 \times 4 \sqrt 6$
5. $\sqrt [3]{ab^2c^2} \times \sqrt [3]{a^2bc}$
6. $\sqrt [3] {32} \times \sqrt [3] {250}$
7. $(\sqrt{2}+\sqrt 3)(\sqrt{2}-\sqrt 3)$
8. $(\sqrt {10} – 2)(1 + \sqrt {10})$
9. $\frac {2}{1 – \sqrt 3}$
10. $\sqrt{252}-\sqrt {112}$
11. $6 \sqrt 3+3 \sqrt 3$
12. $\frac {2\sqrt 7}{\sqrt 11}$
13. $6 \sqrt {27}-4 \sqrt 3$
14. $( 1 + \sqrt 2)^2$
15. $\sqrt {2^4} + \sqrt [3] {64} + \sqrt [4] {2^8}$

(1) $\sqrt{21} \times \sqrt{7}= \sqrt { 3 \times 7 \times 7} = 7 \sqrt 3$
(2) $\sqrt{180} = \sqrt {2^2 \times 3^2 \times 5} = 6 \sqrt 5$
(3) $\sqrt{14} \times \sqrt 6 = \sqrt {2 \times 7} \times \sqrt {2 \times 3} = 2 \sqrt {21}$
(4) $8 \sqrt 6 \times 4 \sqrt 6 = 32 \times 6 = 192$
(5) $\sqrt [3]{ab^2c^2} \times \sqrt [3]{a^2bc} = \sqrt [3] {a^3b^3c^3} = abc$
(6) $\sqrt [3] {32} \times \sqrt [3] {250}= \sqrt [3] { 32 \times 250} = 20$
(7) $(\sqrt{2}+\sqrt 3)(\sqrt{2}-\sqrt 3) = 2 -3 =-1$
(8) $(\sqrt {10} – 2)(2 + \sqrt {10})= 10 -2 = 8$
(9) $\frac {2}{1 – \sqrt 3} = \frac {2}{1 – \sqrt 3} \times \frac {1 + \sqrt 3}{1 + \sqrt 3} = -1 – \sqrt 3$
(10) $\sqrt{252}-\sqrt {112} = 6 \sqrt 7 – 4\sqrt 7 = 2 \sqrt 7$
(11) $6 \sqrt 3+3 \sqrt 3= 9 \sqrt 3$
(12) $\frac {2\sqrt 7}{\sqrt 11}= \frac {2 \sqrt {77}}{11}$
(13) $6 \sqrt {27}-4 \sqrt 3= 18 \sqrt 3 – 4 \sqrt 3 = 14 \sqrt 3$
(14) $( 1 + \sqrt 2)^2= 1 +2 + 2 \sqrt 2 = 3 + 2 \sqrt 2$
(15) $\sqrt {2^4} + \sqrt [3] {64} + \sqrt [4] {2^8} =2^2 + 4 + 2^2 = 12$

True and False statement

1. $\sqrt {1} + \sqrt {2} = \sqrt {1+2}$
2. $\sqrt {3} = \sqrt {-1 \times -3}= \sqrt {-1} . \sqrt {-3}$
3. $\sqrt [3] {8} \div \sqrt {4}$ is a rational number
4. $(1 + \sqrt 5) ( 1- \sqrt 5)$ is a irrational number

True and False statement
(1) $\sqrt {1} + \sqrt {2} = \sqrt {1+2}$ : False
(2) $\sqrt {3} = \sqrt {-1 \times -3}= \sqrt {-1} . \sqrt {-3}$ :False , we cannot have negative number inside the square root
(3) $\sqrt [3] {8} \div \sqrt {4}$ is a rational number : $2 \div 2 =1$, Hence Rational number. So True
(4) $(1 + \sqrt 5) ( 1- \sqrt 5)$ is a irrational number : $1 – 5 = -4, Hence Rational number, So False ## Moderate Questions Simplify the following 1.$\frac {1+ \sqrt 5}{1 – \sqrt 5} + \frac {1 – \sqrt 5}{1 + \sqrt 5}$2.$\frac {1}{1 + \sqrt 2} + \frac {1}{\sqrt 2+ \sqrt 3 } + \frac {1}{\sqrt 3+ \sqrt 4 } + \frac {1}{\sqrt 4+ \sqrt 5 } +….+ \frac {1}{\sqrt 8+ \sqrt 9 }$3.$\sqrt {31 + 8 \sqrt {15}}$4.$(\sqrt {72} + \sqrt {108})(\sqrt {\frac {1}{3}} – \sqrt {\frac {2}{9}})$5.$ \frac {7 \sqrt 3}{\sqrt {10} + \sqrt 3} – \frac {2 \sqrt 5}{\sqrt 6 + \sqrt 5} – \frac {3 \sqrt 2}{\sqrt {15} + 3 \sqrt 2}$Answers (1)$\frac {1+ \sqrt 5}{1 – \sqrt 5} + \frac {1 – \sqrt 5}{1 + \sqrt 5}= \frac { (1+ \sqrt 5)^ + (1- \sqrt 5)^2 }{1 -5} =-3$(2)$\frac {1}{1 + \sqrt 2} + \frac {1}{\sqrt 2+ \sqrt 3 } + \frac {1}{\sqrt 3+ \sqrt 4 } + \frac {1}{\sqrt 4+ \sqrt 5 } +….+ \frac {1}{\sqrt 8+ \sqrt 9 }=\frac {1}{1 + \sqrt 2} \frac {1 – \sqrt 2}{1 – \sqrt 2} + \frac {1}{\sqrt 2+ \sqrt 3 } \frac {\sqrt 2 – \sqrt 3}{\sqrt 2 – \sqrt 3} + \frac {1}{\sqrt 3+ \sqrt 4 } \frac {\sqrt 3 – \sqrt 4}{\sqrt 3 – \sqrt 4}+….+ \frac {1}{\sqrt 8+ \sqrt 9 }\frac {\sqrt 8 – \sqrt 9}{\sqrt 8 – \sqrt 9}= -1(1- \sqrt 2) -1(\sqrt 2 – \sqrt 3) – 1( \sqrt 3 – \sqrt 4) + ….-1(\sqrt 8 – \sqrt 9)= \sqrt 9 -1 = 2$(3)$\sqrt {31 + 8 \sqrt {15}}=\sqrt { 4^2 + 15 + 2 \times 4 \sqrt {15}} = \sqrt {(4+ \sqrt {15})^2} = 4 + \sqrt {15}$(4)$(\sqrt {72} + \sqrt {108})(\sqrt {\frac {1}{3}} – \sqrt {\frac {2}{9}})=\sqrt {24} -\sqrt {16} + \sqrt {36} -\sqrt {24}=6 -4 = 2$(5)$\frac {7 \sqrt 3}{\sqrt {10} + \sqrt 3} – \frac {2 \sqrt 5}{\sqrt 6 + \sqrt 5} – \frac {3 \sqrt 2}{\sqrt {15} + 3 \sqrt 2}=\frac {7 \sqrt 3( \sqrt {10} – \sqrt 3)}{7} – \frac { 2 \sqrt 5(\sqrt 6 – \sqrt 5)}{1} – \frac {3 \sqrt 2(\sqrt {15} – 3\sqrt 2}{-3}=\frac { 7 \sqrt {30} -21}{7} – \frac {2 \sqrt {30} -10}{1} – \frac {3 \sqrt {15} -18}{-3}=1$1. Find the value of a and b :$\frac {3 + \sqrt 7}{ 3 – 4 \sqrt 7} = a + b \sqrt 7$Answers LHS$\frac {3 + \sqrt 7}{ 3 – 4 \sqrt 7}= \frac {3 + \sqrt 7}{ 3 – 4 \sqrt 7} \frac {3 + 4 \sqrt 7}{ 3 + 4 \sqrt 7}= \frac { 9 + 12 \sqrt 7 + 3 \sqrt 7 + 28}{9 – 112}=\frac {37 + 15 \sqrt 7}{-103}= -\frac {37}{103} – \frac {15}{103} \sqrt 7$Comparing$a= -\frac {37}{103}$and$b=- \frac {15}{103}$1. If$x = 2 + \sqrt 3$Find the value (a)$x + \frac {1}{x}$(b)$x^2 + \frac {1}{x^2}$Answer (a)$ 2 + \sqrt 3 + \frac {1}{2 + \sqrt 3}=2 + \sqrt 3 -(2- \sqrt 3)=2 \sqrt 3 $(b)$x^2 + \frac {1}{x^2}=(x + \frac {1}{x})^2 – 2=(2 \sqrt 3)^2 -2=12 -2=10$1. Rationalize the Denominator of$\frac {1}{\sqrt 3 – \sqrt 2 -1 }$Answer$=\frac {1}{\sqrt 3 – 1 – \sqrt 2 }=\frac {1}{\sqrt 3 – 1 – \sqrt 2 } \frac {\sqrt 3 – 1 + \sqrt 2}{\sqrt 3 – 1 + \sqrt 2 }= \frac {\sqrt 3 – 1 + \sqrt 2}{ (\sqrt 3 -1)^2 – 2}=\frac {\sqrt 3 – 1 + \sqrt 2}{4 -2 \sqrt 3 -2}=\frac {\sqrt 3 – 1 + \sqrt 2}{2 -2 \sqrt 3 }=\frac {\sqrt 3 – 1 + \sqrt 2}{2 -2 \sqrt 3 } \frac {2 +2 \sqrt 3 }{2 +2 \sqrt 3 } =\frac {(\sqrt 3 – 1 + \sqrt 2)(2 +2 \sqrt 3)}{4 – 12}=\frac{ 2 \sqrt 3 + 6 -2- 2\sqrt 3 + 2\sqrt 2 + 2 \sqrt 6}{-8}=-\frac {4 + 2\sqrt 2 + 2 \sqrt 6}{8}= – \frac { 2 + \sqrt 2 + \sqrt 6}{4}$1. Convert into pure surds$2 \sqrt {5 \sqrt 7}$Answer$2 \sqrt {5 \sqrt 7} = \sqrt {2^2 \times 5 \times \sqrt 7} = \sqrt {\sqrt {20^2 \times 7}}= \sqrt {\sqrt {2800}}= \sqrt [4] {2800}$1. Which of the following has greatest value$\sqrt [3] {7}$,$\sqrt [6] {15}$,$\sqrt [4] {10}$Answer We need to convert into equal surd power to compare, LCM 3,6 and 4 is 12$\sqrt [3] {7} = \sqrt [12] {7^4} = \sqrt [12] { 2401}\sqrt [6] {15}=\sqrt [12] {15^2} = \sqrt [12] { 225}\sqrt [4] {10}=\sqrt [12] {10^3} = \sqrt [12] { 1000}$So greatest is$\sqrt [3] {7}$## Tough Surd Questions (1) Simplify the expression$\frac {\sqrt 3}{ \sqrt {19 + 8 \sqrt 3} – \sqrt {19 – 8 \sqrt 3}}$Answer$\frac {\sqrt 3}{ \sqrt {19 + 8 \sqrt 3} – \sqrt {19 – 8 \sqrt 3}}=\frac {\sqrt 3}{ \sqrt {19 + 8 \sqrt 3} – \sqrt {19 – 8 \sqrt 3}} \frac {\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}{\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}= \frac {\sqrt 3 (\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3})}{19 + 8 \sqrt 3 – 19 + 8 \sqrt 3}=\frac {\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}{16}$Now$19 + 8 \sqrt 3= 4^2 + (\sqrt 3)^2 + 8 \sqrt 3= (4 + \sqrt 3)^2$Therefore$=\frac {\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}{16}= \frac {\sqrt {(4 + \sqrt 3)^2} + \sqrt {(4 – \sqrt 3)^2}}{16}=\frac {4+ \sqrt 3 + 4 – \sqrt 3}{16} = \frac {1}{2}$(2) if$y= \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} – \sqrt {a-b}}$Find the value of$by^2 – 2ay + b$Answer$y= \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} – \sqrt {a-b}}y= \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} – \sqrt {a-b}} \times \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} + \sqrt {a-b}}y= \frac { a+b +a-b + 2 \sqrt {a^2 -b^2}}{a+b -a +b}y=\frac { 2a + 2 \sqrt {a^2 -b^2}}{2b}y=\frac {a + \sqrt {a^2 -b^2}}{b}by^2 – 2ay + b= b [\frac {a + \sqrt {a^2 -b^2}}{b}]^2 – 2a [\frac {a + \sqrt {a^2 -b^2}}{b}] + b=\frac {a^2 + a^2 -b^2 + 2a \sqrt {a^2 -b^2}}{b} – 2a [\frac {a + \sqrt {a^2 -b^2}}{b}] + b=\frac {a^2 + a^2 -b^2 + 2a \sqrt {a^2 -b^2} -2 a^2 -2a \sqrt {a^2 -b^2} + b^2}{b}=0$(3) if$x = 3 + 2 \sqrt 2$Then find the value of$( \sqrt x – \frac {1}{\sqrt x})$Answer$x = 3 + 2 \sqrt 2x = 1 + (\sqrt 2)^2 + 2 \sqrt 2x = (1+ \sqrt 2)^2$Now$( \sqrt x – \frac {1}{\sqrt x})= ( (1+ \sqrt 2) – \frac {1}{1 + \sqrt 2})= (1 + \sqrt 2 +(1- \sqrt 2))= 2$(4) if$a = \frac {1}{7 + 4 \sqrt 3}$and$b= \frac {1}{7 – 4 \sqrt 3}$Then find the value of$\frac {1}{a+1} + \frac {1}{b+1}$Answer$a = \frac {1}{7 + 4 \sqrt 3}a= \frac { 7 – 4 \sqrt 3}{49 – 48}= 7 – 4 \sqrt 3b= \frac {1}{7 – 4 \sqrt 3}a= \frac { 7 + 4 \sqrt 3}{49 – 48}= 7 + 4 \sqrt 3\frac {1}{a+1} + \frac {1}{b+1}= \frac {1}{8 – 4 \sqrt 3} + \frac {1}{8 + 4 \sqrt 3} = \frac { 16 }{64 – 48}=1$(5) if$y =\frac {\sqrt {3}}{2}$,then find the value of$\frac {\sqrt {1+y}}{1+ \sqrt {1+y}} + \frac {\sqrt {1-y}}{1+ \sqrt {1-y}}$Answer$y =\frac {\sqrt {3}}{2}1 +y = \frac { 2 + \sqrt 3}{2} = \frac { 4 + 2\sqrt 3}{4} = \frac {(\sqrt 3 + 1)^2}{4}$Therefore$\sqrt {1 + y} = \frac {\sqrt 3 + 1}{2}$Similarly$1 -y = \frac { 2 – \sqrt 3}{2} = \frac { 4 – 2\sqrt 3}{4} = \frac {(\sqrt 3 – 1)^2}{4}$Therefore$\sqrt {1 + y} = \frac {\sqrt 3 + 1}{2}\frac {\sqrt {1+y}}{1+ \sqrt {1+y}} + \frac {\sqrt {1-y}}{1+ \sqrt {1-y}}=\frac { (\sqrt 3 + 1)/2}{1 + (\sqrt 3 + 1)/2} + \frac { (\sqrt 3 – 1)/2}{1 – (\sqrt 3 -1 )/2}=\frac {\sqrt 3 + 1}{3 + \sqrt 3} + \frac {\sqrt 3 – 1}{3 – \sqrt 3} \frac {1 + \sqrt 3}{ \sqrt 3(1 + \sqrt 3)} + \frac { \sqrt 3 – 1}{\sqrt 3 ( \sqrt 3 – 1)} = \frac {2}{\sqrt 3}\$

I hope you like these surd questions with detailed solution and it helps in your preparation for the examination

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