Surds Questions with Detailed Solutions

(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) $(\sqrt {10} – 2)(2 + \sqrt {10})= 10 -4 = 6$
(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) $\sqrt {1} + \sqrt {2} = \sqrt {1+2}$ : 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

(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) $\frac {7 \sqrt 3}{\sqrt {10} + \sqrt 3} – \frac {2 \sqrt 5}{\sqrt 6 + \sqrt 5} – \frac {3 \sqrt 2}{\sqrt {15} + 3 \sqrt 2}$
$=\frac {7 \sqrt 3( \sqrt {10} – \sqrt 3)}{7} – \frac { 2 \sqrt 5(\sqrt 6 – \sqrt 5)}{1} – \frac {3 \sqrt 2(\sqrt {15} – 3\sqrt 2}{-3}$
$=\frac { 7 \sqrt {30} -21}{7} – \frac {2 \sqrt {30} -10}{1} – \frac {3 \sqrt {15} -18}{-3}$
$=1$

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 $b=- \frac {15}{103}$

(a) $ 2 + \sqrt 3 + \frac {1}{2 + \sqrt 3}$
$=2 + \sqrt 3 -(2- \sqrt 3)$
$=2 \sqrt 3 $
(b)
$x^2 + \frac {1}{x^2}$
$=(x + \frac {1}{x})^2 – 2$
$=(2 \sqrt 3)^2 -2$
$=12 -2=10$

$=\frac {1}{\sqrt 3 – 1 – \sqrt 2 }$
$=\frac {1}{\sqrt 3 – 1 – \sqrt 2 } \frac {\sqrt 3 – 1 + \sqrt 2}{\sqrt 3 – 1 + \sqrt 2 }$
$= \frac {\sqrt 3 – 1 + \sqrt 2}{ (\sqrt 3 -1)^2 – 2}$
$=\frac {\sqrt 3 – 1 + \sqrt 2}{4 -2 \sqrt 3 -2}$
$=\frac {\sqrt 3 – 1 + \sqrt 2}{2 -2 \sqrt 3 }$
$=\frac {\sqrt 3 – 1 + \sqrt 2}{2 -2 \sqrt 3 } \frac {2 +2 \sqrt 3 }{2 +2 \sqrt 3 } $
$=\frac {(\sqrt 3 – 1 + \sqrt 2)(2 +2 \sqrt 3)}{4 – 12}$
$=\frac{ 2 \sqrt 3 + 6 -2- 2\sqrt 3 + 2\sqrt 2 + 2 \sqrt 6}{-8}$
$=-\frac {4 + 2\sqrt 2 + 2 \sqrt 6}{8}$
$= – \frac { 2 + \sqrt 2 + \sqrt 6}{4}$

$2 \sqrt {5 \sqrt 7} = \sqrt {2^2 \times 5 \times \sqrt 7} = \sqrt {\sqrt {20^2 \times 7}}= \sqrt {\sqrt {2800}}= \sqrt [4] {2800}$

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 $\sqrt [3] {7}$

$\frac {\sqrt 3}{ \sqrt {19 + 8 \sqrt 3} – \sqrt {19 – 8 \sqrt 3}}$
$=\frac {\sqrt 3}{ \sqrt {19 + 8 \sqrt 3} – \sqrt {19 – 8 \sqrt 3}} \frac {\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}{\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}$
$= \frac {\sqrt 3 (\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3})}{19 + 8 \sqrt 3 – 19 + 8 \sqrt 3}$
$=\frac {\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}{16}$
Now $19 + 8 \sqrt 3= 4^2 + (\sqrt 3)^2 + 8 \sqrt 3= (4 + \sqrt 3)^2$
Therefore
$=\frac {\sqrt {19 + 8 \sqrt 3} + \sqrt {19 – 8 \sqrt 3}}{16}$
$= \frac {\sqrt {(4 + \sqrt 3)^2} + \sqrt {(4 – \sqrt 3)^2}}{16}$
$=\frac {4+ \sqrt 3 + 4 – \sqrt 3}{16} = \frac {1}{2}$

$y= \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} – \sqrt {a-b}}$
$y= \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} – \sqrt {a-b}} \times \frac { \sqrt {a+b} + \sqrt {a-b}}{ \sqrt {a+b} + \sqrt {a-b}}$
$y= \frac { a+b +a-b + 2 \sqrt {a^2 -b^2}}{a+b -a +b}$
$y=\frac { 2a + 2 \sqrt {a^2 -b^2}}{2b}$
$y=\frac {a + \sqrt {a^2 -b^2}}{b}$

$by^2 – 2ay + b$
$= b [\frac {a + \sqrt {a^2 -b^2}}{b}]^2 – 2a [\frac {a + \sqrt {a^2 -b^2}}{b}] + b$
$=\frac {a^2 + a^2 -b^2 + 2a \sqrt {a^2 -b^2}}{b} – 2a [\frac {a + \sqrt {a^2 -b^2}}{b}] + b$
$=\frac {a^2 + a^2 -b^2 + 2a \sqrt {a^2 -b^2} -2 a^2 -2a \sqrt {a^2 -b^2} + b^2}{b}$
$=0$

$x = 3 + 2 \sqrt 2$
$x = 1 + (\sqrt 2)^2 + 2 \sqrt 2$
$x = (1+ \sqrt 2)^2$
Now
$( \sqrt x – \frac {1}{\sqrt x})$
$= ( (1+ \sqrt 2) – \frac {1}{1 + \sqrt 2})$
$= (1 + \sqrt 2 +(1- \sqrt 2))= 2$

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

$\frac {1}{a+1} + \frac {1}{b+1}$
$= \frac {1}{8 – 4 \sqrt 3} + \frac {1}{8 + 4 \sqrt 3}$
$ = \frac { 16 }{64 – 48}=1$

$y =\frac {\sqrt {3}}{2}$
$1 +y = \frac { 2 + \sqrt 3}{2} = \frac { 4 + 2\sqrt 3}{4} = \frac {(\sqrt 3 + 1)^2}{4}$
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)} $
$= \frac {2}{\sqrt 3}$