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Surds Questions with Detailed Solutions

This article is about surds questions and follow the link surds in math if you want to know about surds. Here in this article, you can find surds problem-solving questions.

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. $\sqrt {2^4} + \sqrt [3] {64} + \sqrt [4] {2^8}$
Answers

(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
Answers

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

$x = 3 + 2 \sqrt 2$
$x = 1 + (\sqrt 2)^2 + 2 \sqrt 2$
$x = (1+ \sqrt 2)^2$
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

$a = \frac {1}{7 + 4 \sqrt 3}$
$a= \frac { 7 – 4 \sqrt 3}{49 – 48}= 7 – 4 \sqrt 3$
$b= \frac {1}{7 – 4 \sqrt 3}$
$a= \frac { 7 + 4 \sqrt 3}{49 – 48}= 7 + 4 \sqrt 3$

$\frac {1}{a+1} + \frac {1}{b+1}$
$= \frac {1}{8 – 4 \sqrt 3} + \frac {1}{8 + 4 \sqrt 3}$
$ = \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

Further References