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.
Surd Easy Questions
Simplify the following
- √21×√7
- √180
- √14×√6
- 8√6×4√6
- 3√ab2c2×3√a2bc
- 3√32×3√250
- (√2+√3)(√2−√3)
- (\sqrt {10} – 2)(1 + \sqrt {10})
- \frac {2}{1 – \sqrt 3}
- \sqrt{252}-\sqrt {112}
- 6 \sqrt 3+3 \sqrt 3
- \frac {2\sqrt 7}{\sqrt 11}
- 6 \sqrt {27}-4 \sqrt 3
- ( 1 + \sqrt 2)^2
- \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 -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
- \sqrt {1} + \sqrt {2} = \sqrt {1+2}
- \sqrt {3} = \sqrt {-1 \times -3}= \sqrt {-1} . \sqrt {-3}
- \sqrt [3] {8} \div \sqrt {4} is a rational number
- (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 Surds Questions
Simplify the following
- \frac {1+ \sqrt 5}{1 – \sqrt 5} + \frac {1 – \sqrt 5}{1 + \sqrt 5}
- \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 }
- \sqrt {31 + 8 \sqrt {15}}
- (\sqrt {72} + \sqrt {108})(\sqrt {\frac {1}{3}} – \sqrt {\frac {2}{9}})
- \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
- 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}
- 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
- 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}
- 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}
- 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
First section – question 8: 2 x 2 = 4 not 2. So the correct answer is 6 not 8.
Thanks for the feedback. It is fixed
I want such difficult and tough numerical questions based on surds and radicals…..