Surds and Indices Shortcuts, Tricks, PDF and Formulas

Important Formulas - Surds and Indices

  • An integer is a whole number (positive, negative or zero). A rational number is one that can be expressed as a fraction \frac{a}{b}, where a and b are integers. All integers, fractions and terminating or recurring decimals are rational.
  • An irrational number cannot be expressed in the form \frac{a}{b}, where a and b are integers. Examples of irrational numbers are \pi,\sqrt{2},\sqrt{3}\;\;and\:4\sqrt{5}\:(4\sqrt{5}\:means\:4\times \sqrt{5}).
  • Real numbers are numbers that can be represented by points on the number line. Real numbers include both rational and irrational number.
  • A surd is an irrational number involving a root. The numbers  \sqrt{3},4\sqrt{5}\:and\:\sqrt[3]{7} are examples of surds. Numbers such as  \sqrt{16}and \sqrt[3]{8} are not surds because they are equal to rational numbers.  \sqrt{16} =4 and root of ‘3’ or ‘root 3’. Note that we cannot take the square root of a negative number.
  • Like surds involve the square root of the same number. Only like surds can be added or subtracted. For example,  3\sqrt{2}+4\sqrt{2}=7\sqrt{2}\:but\: 3\sqrt{2}+4\sqrt{3}=3\sqrt{2}+4\sqrt{3}.
  • Surds of the form \sqrt{x} can be simplified if the number beneath the square root sign has a factor that is a perfect square. For example, \sqrt{8}=\sqrt{4\times 2}=\sqrt{4}\times \sqrt{2}=2\sqrt{2}.
  • The following rules can be used when multiplying or dividing surds.
    \sqrt{x}\times \sqrt{y}=\sqrt{xy}
  • Rationalising the denominator of a surd means changing the denominator so that is a rational number. To rationalize the denominator of a surd such as\frac{\sqrt{2}}{\sqrt{3}} we use the result that(\sqrt{x})^{2}=xso if we multiply the denominator by\sqrt{3}it will be rational. if we multiply the demoniator by\sqrt{3}we must also multiply the numerators by\sqrt{3}so that\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{\sqrt{3}}
  • Fractional indices may be used to express roots.x^{\frac{1}{n}}=\sqrt[n]{x}\:\:and\:\:x^{\frac{m}{n}}=(\sqrt[n]{x})^{m}
  • Make sure you can use your calculator to find powers and roots. Thex^{\frac{1}{y}}or \sqrt[x]{\:}button allows you to find roots.
    For example:  \sqrt[5]{7776}=(7776)^{\frac{1}{5}}


  • When the powers are fraction , then we can compare the indices in following manner.
    for example :find which is greather 2^{\frac{1}{4}} or 3^{\frac{1}{5}}
    step-1 : find the L.C.M of the denominator of the fraction i.e 4,5 = 20
    step-2 : find powers with the L.C.M 2^{\frac{1}{4}\times 20} OR 3^{\frac{1}{5}\times 20} = 2^{5} AND 3^{4}
    step-3 : compare the result obtained in setp-2 : i.e as 32 < 81 we can say 2^{\frac{1}{4}} < 3^{\frac{1}{5}}



  1. Any none - zero real number raised to the power 0 equals 1
  2. I raised to any power is always 1.
  3. a^{m}=a\times a\times a …….m times
  4. a^{-m}= \frac{1}{a\times a\times a......m times}
  5. a^{\frac{m}{n}}=n^{th} root of (a^{m})=n\sqrt{a^{m}}
  6. If a\neq 1; A\neq0; a\neq-1 and  a^{x}=a^{y}, then x = y
  7. if a\neq 0;b\neq 0 and a^{x}=b^{x}, then
    1. if x is odd, a=b
    2. if x is even, a=\pmb
  8. a^{m}\times a^{n}=a^{m+n}
  9. a^{m} \div a^{n}=a^{m-n}
  10. (a^{m})^{n}=a^{mn}
  11. (a^{m})^{n}=a^{mn}


Important Examples - Surds and Indices

Question 1.What is the value of (512^{\frac{1}{3}})+(512^{\frac{2}{9}})+(512^{\frac{1}{9}})

(512^{\frac{1}{9}})=2^{9}\times ^{\frac{1}{9}}=2
\therefore The answer is 8+4+2=14

Question 2.The product of 6^{2^{3}}\times (6^{2})^{^{3}}\times (6^{6}) is equal to

Solution:6^{8}\times 6^{6}\times 6^{6}=6^{20}

Question 3.The value of (5^{-1}+9^{-1})^{2}\div (5^{-1}-9^{-1})^{2} is equal to

Solution:\left ( \frac{1}{5} +\frac{1}{9}\right )^{2}\div \left ( \frac{1}{5}-\frac{1}{9} \right )^{2}
=\left ( \frac{14}{45} \right )^{2}\times \left ( \frac{4}{45} \right )^{-2}
=\left ( \frac{14}{45} \right )^{2}\times \left ( \frac{45}{4} \right )^{2}=\frac{14\times 14}{4\times 4}=\frac{49}{4}

Question 4.If x=-5 and y=-6, then what is the value of (x-y)^{y-x}\div (y-x)^{x-y} ?

Solution:(-5+6)^{-6+5}\div (-6+5)^{-5+6}=(1)^{-1}\times (-1)^{1}
1\times -1=-1

Question 5.In a cricket match, the number of runs scored by any team is equal to a power of the number of batsman playing in the team. Six batsman played in team A and eleven batsman played in team B. If team A won by 95 runs, then find the run scored by team A.

Solution: Let the power of the terms be x and y.
6^{x}-11^{y} =95
Put x =3, y= 2 (By trial and error)
6^{3}-11^{2} =95
Hence satisfied
Score of team A=6^{3} = 216

Note : These type of questions can be solved through answer choices.


  1. The conjugate surd of  \sqrt{a}+\sqrt{b} is \pm (\sqrt{a}-\sqrt{b})
  2. To rationalize  \frac{1}{\sqrt{a}+\sqrt{b}},multiply it by \frac{(\sqrt{a}-\sqrt{b})}{\sqrt{a}-\sqrt{b}}or\frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}-a}
  3. If  a+\sqrt{b}=c+\sqrt{d}, then a=c and b=d
  4. To find  \sqrt{(a+\sqrt{b})} Write it in the form m+n+ 2\sqrt{mn}, such that m+n=a and 4mn=b, then \sqrt{(a+\sqrt{b})}=\pm (\sqrt{m}+\sqrt{n})
  5. \sqrt{a.\sqrt{a.\sqrt{a.....\infty }}}=a
  6. \sqrt{a.\sqrt{a.\sqrt{a.\sqrt{a........ n times}}}}=a^{1-\frac{1}{2^{n}}}
  7. If \sqrt{a+\sqrt{a+\sqrt{a............\infty }}}=p, then p(p-1)=a
  8. The rationalizing factor of a^{\frac{1}{3}}+b^{\frac{1}{3}} is a^{\frac{2}{3}}-(ab)^{\frac{1}{3}}+b^{\frac{2}{3}}
  9. The rationalize factor of a^{\frac{1}{3}}-b^{\frac{1}{3}} is a^{\frac{2}{3}}+(ab)^{\frac{1}{3}}+b^{\frac{2}{3}}
  10. The rationalize factor of \sqrt{a}+\sqrt{b} is \sqrt{a}-\sqrt{b}
  11. The rationalize factor of \sqrt{a}-\sqrt{b} is \sqrt{a}+\sqrt{b}
  12. The rationalize factor of a+\sqrt{b} is a-\sqrt{b}


SOLVED EXAMPLES - Surds and Indices

Question 1.The value of  \sqrt{4^{4}\sqrt{4^{2}\sqrt{4\sqrt{4...........\infty }}}}.

Solution : The given expression is in the form of
 4^{2\times 4^{\frac{1}{2}}}\times 4^{\frac{1}{8}}\times 4^{\frac{1}{32}}.....\infty =4^{2}+^{\frac{1}{2}}+^{\frac{1}{8}}+^{\frac{1}{32}}+......\infty
The exponent i.e.,  2+\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+......\infty is in form of an infinite G.P
 S_{\infty }=\frac{a}{1-r}=\frac{2}{1-\frac{1}{4}}=\frac{8}{3}
The given expression=4^{\frac{8}{3}}

Question 2.Simplify \frac{\sqrt{17}-\sqrt{11}}{\sqrt{17}-\sqrt{11}}

Solution :By rationalizing \frac{(\sqrt{17}-\sqrt{11})(\sqrt{17}-\sqrt{11})}{(\sqrt{17}+\sqrt{11}){(\sqrt{17}-\sqrt{11}})}



Question 3.If \frac{6+\sqrt{6}}{6-2\sqrt{6}}=a+b\sqrt{6}, then find the value of (a+2b)

Solution : \frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{6}(\sqrt{6}-2)}




\Rightarrow a=4,b=1.5

\Rightarrow a+2b=4+2\times 1.5=7

Question 4.What is the positive square root of 75+4\sqrt{176}?

Solution : 75+2\sqrt{704}=64+11+2\sqrt{64}\times 11=(\sqrt{64}+\sqrt{11})^{2}

=(8+\sqrt{11})^{2}\therefore square root is 8+\sqrt{11}

Question 5.Which of the following is the great ?

Solution :By squaring the values



Since 25 is common to all the value, 2\sqrt{154} is the largest value.

Hence \sqrt{11}+\sqrt{14} has the largest value.


Surds and Indices Questions from Previous Year Exams

This test will cover Square, Cube, Indices and Surds syllabus of Bank Clerk Exam.


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Please comment on Surds and Indices Shortcuts, Tricks, PDF and Formulas


  1. Frank

    Dear all, I really do like your little quiz, but unfortunately the answer you are suggesting to question #10 ("Bank PO 1990") is not correct. Instead of 4 - as suggested - the correct answer is 2. (Maybe you forgot to calculate SQRT(4) during simplification 🙂 )

  2. Ratnareddy

    Superb Knowledge is available with you but there are some blunders.Hope you will correct,Thank you so much.

  3. Govind Singh

    nice pdf

  4. Leah Wambui

    help solve 5+2 root 6 all this into a square root =root2+root3


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