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})^{2}=\sqrt{x^{2}}=x
    \sqrt{x}\times \sqrt{y}=\sqrt{xy}
    \frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}
  • 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}}

INDICES

  • 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}}

 

POINT TO REMEMBER

  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}})

Solution:(512^{\frac{1}{3}})=(2^{9})^{\frac{1}{3}}=2^{3}=8
(512^{\frac{2}{9}})=(2^{9})^{\frac{3}{9}}=2^{2}=4
(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.

POINTS TO REMEMBER

  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}})}

=\frac{(\sqrt{17}-\sqrt{11})^{2}}{(\sqrt{17})^{2}-(11)^{2}}=\frac{17+11+-2\sqrt{187}}{17-11}

=\frac{28-2\sqrt{187}}{6}=14-\sqrt{\frac{187}{3}}

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)}

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

=\frac{6+3\sqrt{6}+2}{6-4}

=4+\frac{3}{2}\sqrt{6}

\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 ?
\sqrt{15}+\sqrt{10})^{2},\sqrt{11}+\sqrt{14},\sqrt{20}+\sqrt{15},\sqrt{8}+\sqrt{17}

Solution :By squaring the values

=(\sqrt{15}+\sqrt{10}=25+2\sqrt{150};(\sqrt{11}+\sqrt{14})^{2}=25+2\sqrt{154}

=(\sqrt{20}+\sqrt{5})=25+2\sqrt{100};(\sqrt{8}+\sqrt{17})^{2}=25+2\sqrt{130}

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

3 Comments

  1. Ratnareddy

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

    Reply
  2. Govind Singh

    nice pdf

    Reply
  3. Leah Wambui

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

    Reply

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