# 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$(\sqrt{x})^{2}=x$so 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. The$x^{\frac{1}{y}}$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$\neq$0; a$\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=$\pm$b
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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