 Surds and indices chapter plays a crucial role in simplification problems of arithmetic. Here we would learn about its basic concepts and tricks to handle these problems efficiently in competitive examinations.

Surds: A surd is a root of a rational number with an irrational value. In other words, when it is a root and irrational, it is surd. Surds can not be further simplified into whole numbers or integers.

Example: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{\frac{1}{7}}$, etc.

Note: An irrational number is a real number that can not be written as a simple fraction.

Example: $\pi, \sqrt{2},\sqrt{3}$, etc.

If N is an irrational number then $N\neq\frac{p}{q}$, where p and q are integers and $q\neq0$.

## Types of Surds

• Pure Surds: When surds have only one irrational number.

Example: $\sqrt{2}, \sqrt{11}, \sqrt{17}$, etc.

• Mixed Surds: When surds have both rational and irrational numbers.

Example: $2\sqrt{5}, 4\sqrt{7}, 8\sqrt{3}$, etc.

• Compound Surds: When there are two or more surds in one mathematical expression.

Example: $4+\sqrt{5}, \sqrt{5}+\sqrt{6}, 3+2\sqrt{6}$, etc.

Indices: The index (Plural: Indices) is the power (exponent) of a number.

Example: $2^{5}$ has an index of 5 and 2 is its base. Here exponent 5 defines, how many times 2 has been multiplied by itself.

## Rules of Surds and Indices

If a and b are any two real numbers, m and n are integers then

(i) $a^{m}\times a^{n}=a^{m+n}$

Example: $2^{3}\times 2^{5}=2^{8}=256$

(ii) $\frac{a^{m}}{a^{n}}=a^{m-n}$

Example: $\frac{5^{3}}{5^{4}}=5^{-1}=\frac{1}{5}$

(iii) $(ab)^{m}=a^{m}\times b^{m}$

(iv) $x^{m^{n}}=x^{(m^{n})}$

(v) $(a^{m})^{n}=a^{m\times n}$

(vi) $a^{-m}=\frac{1}{a^{m}}$

(vii) $a^{0}=1$

(viii) $\sqrt[n]{a}=a^{\frac{1}{n}}$

(ix) $\sqrt[n]{x^m}=x^{\frac{m}{n}}$

(x) If $x^{\frac{1}{n}}=a$ then $x=a^{n}$

(xi) $\sqrt{a}\displaystyle\pm \sqrt{b} \neq \sqrt{a\displaystyle\pm b}$

(xii) $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$

(xiii) $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$

## Rationalizing Surds

Rationalization is the process of changing the denominator of a fraction to a rational number.

Que: Rationalize $\frac{1}{\sqrt{5}+\sqrt{3}}$.

Solution: Multiply the conjugate of the denominator by both, the numerator and the denominator.

=$\frac{1\times (\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})\times (\sqrt{5}-\sqrt{3})}$

=$\frac{\sqrt{5}-\sqrt{3}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$

=$\frac{1}{2}\times (\sqrt{5}-\sqrt{3})$

## Surds and Indices problems

Que 1: Find the value of $\frac{(243)^\frac{n}{5}\times 3^{2n+1}}{9^{n}\times 3^{n-1}}$.

Solution: Numerator =$(3^{5})^{\frac{n}{5}}\times 3^{2n+1}=3^{n}\times 3^{2n+1}=3^{3n+1}$

Denominator=$3^{2n}\times 3^{n-1}=3^{3n-1}$

$\frac{3^{3n+1}}{3^{3n-1}}$$=3^{(3n+1)-(3n-1)}=3^{2}=9$

Que 2: If $x^{x\sqrt{x}}=(x\sqrt{x})^{x}$ then find the value of x?

Solution: $x^{x^{1}\times x^{\frac{1}{2}}}=(x^{1}\times x^{\frac{1}{2}})^{x}$.

$x^{x^{\frac{3}{2}}}=(x^{\frac{3}{2}})^{x}$.

So, $x^{\frac{3}{2}}=\frac{3x}{2}$.

After squaring both sides of the equation,

$x^{3}=\frac{9x^{2}}{4}$.

$x^{2}(x-\frac{9}{4})=0$.

Now, $x=\frac{9}{4}$ is the correct answer (x=0 would give an indeterminate value)

Que 3:  Find the value of $\frac{5^{n+3}-6\times 5^{n+1}}{9\times 5^{n}-5^{n}\times 2^{2}}$?

Solution: $\frac{5^{n}(5^{3}-6\times 5)}{5^{n}(9-4)}$.

=$\frac{125-30}{5}=19$

Que 4: Find the value of $[(\sqrt{256^{2}})^{\frac{3}{2}}]^{\frac{1}{4}}$?

Solution: $256=2^{8}$

=$[(2^{\frac{8\times 2}{3}})^{\frac{3}{2}}]^{\frac{1}{4}}$

=$(2^{8})^{\frac{1}{4}}=2^{2}=4$

Que 5:  Find the value of $\sqrt{4\frac{12}{125}}$?

Solution: $4\frac{12}{125}=\frac{512}{125}=(\frac{8}{5})^{3}$

So, The correct answer=$\frac{8}{5}=1\frac{3}{5}$

Que 6: If $\frac{\sqrt{4356\times \sqrt{x}}}{\sqrt{6084}}=11$, then find the value of x ?

Solution: After squaring both sides of the equation,

$\frac{4356\times\sqrt{x}}{6084}=121$.

$\sqrt{x}=169$.

$x=169^{2}=28561$.

Que 7: Arrange the following in ascending order if a>1

(i) $\sqrt{\sqrt{a^{3}}}$

(ii) $\sqrt{\sqrt{a^{2}}}$

(iii) $\sqrt{\sqrt{a}}$

(iv) $\sqrt{\sqrt{a^{3}}}$

Solution: $\sqrt{\sqrt{a^{3}}}=(\sqrt{a^{3}})^{\frac{1}{3}}=(a^{\frac{3}{4}})^{\frac{1}{3}}=a^{\frac{1}{4}}$.

$\sqrt{\sqrt{a^{4}}}=(\sqrt{a^{4}})^{\frac{1}{3}}=(a^{\frac{4}{5}})^{\frac{1}{3}}=a^{\frac{4}{15}}$.

$\sqrt{\sqrt{a}}=(a^{\frac{1}{3}})^{\frac{1}{2}}=a^{\frac{1}{6}}$.

$\sqrt{\sqrt{a^{3}}}=(\sqrt{a^{3}})^{\frac{1}{2}}=(a^{\frac{3}{5}})^{\frac{1}{2}}=a^{\frac{3}{10}}$.

Now compare exponents,

$(\frac{1}{4}, \frac{4}{15}, \frac{1}{6}, \frac{3}{10})\times 60=(15, 16, 10, 18)$.

Here, LCM(4,15,6,10)=60

Now, 10<15<16<18

So, $\frac{1}{6}< \frac{1}{4}<\frac{4}{15}<\frac{3}{10}$

Correct ascending order= (iii)<(i)<(ii)<(iv)

Que 8: Find the square root of $\sqrt{5+2\sqrt{6}}$ ?

Solution: $\sqrt{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2\times\sqrt{3}\times\sqrt{2}}$.

=$\sqrt{(\sqrt{3}+\sqrt{2})^{2}}$

=$\sqrt{3}+\sqrt{2}$

## Important results of Surds and Indices

1. If $N=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+......}}}}$

Then $N=\frac{1+\sqrt{1+4x}}{2}$

Proof: $N=\sqrt{x+N}$.

$N^{2}=x+N$.

$N^{2}-N-x=0$.

$N=\frac{1\displaystyle\pm\sqrt{1+4x}}{2}$.   (Sridharacharya Formula)

Note: N has a positive value so take the positive sign here.

2. If $N=\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-......}}}}$

Then $N=\frac{-1+\sqrt{1+4x}}{2}$

3. If $N=\sqrt{x\displaystyle\pm\sqrt{x\displaystyle\pm\sqrt{x\displaystyle\pm\sqrt{x\displaystyle\pm......}}}}$ then find the value of N?

Here x=k×(k+1), k is a Natural Number (Positive Integer)

Case 1: (+) sign

N=k+1

Case 2: (-) sign

N=k

Example: $N=\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+......}}}}=4$

$N=\sqrt{12-\sqrt{12-\sqrt{12-\sqrt{12-......}}}}=3$

Here, 12=3×4

4. If $N=\sqrt{x\sqrt{x\sqrt{x\sqrt{x......}}}}$ then N=x

Proof: $N=\sqrt{x.N}$.

$N^{2}=x.N$.

$N(N-x)=0$.

So, N=x    (For x>0, N can’t be 0)

5. If $K=\sqrt{x\sqrt{x\sqrt{x\sqrt{x.....n_{th}term}}}}$ then find the value of K ?

Number of terms= Number of square roots=n

$K=(x)^{\frac{2^{n}-1}{2^{n}}}$.

Example: $K=\sqrt{5\sqrt{5\sqrt{5\sqrt{5}}}}=(5)^{\frac{2^{4}-1}{2^{4}}}=(5)^{\frac{15}{16}}$.

Question: $\frac{\sqrt{100!\sqrt{100!\sqrt{100!\sqrt{100!}}}}}{\sqrt{99!\sqrt{99!\sqrt{99!\sqrt{99!}}}}}=\sqrt{10^x}$

Find the value of x?

Solution: Numerator=$(100!)^\frac{15}{16}$.

Denominator=$(99!)^\frac{15}{16}$.

$\sqrt{10^x}=(100!/99!)^{\frac{15}{16}}=(100)^{\frac{15}{16}}=\sqrt{10^{30}}$

The value of x=30

1. ### What is the difference between Surds and Indices?

Surds can be expressed as the nth root of a rational number while Indices refer to the power to which a number is raised.

2. ### Is surd an irrational number?

Yes, Surd is irrational because, in decimal form, it goes on without repetition.

3. ### How can we identify Surds?

A rational number that contains a radical sign ($\sqrt[n]{}$) and its solution is an irrational value, called Surd.

4. ### Can a Surd be expressed in Index form?

Yes, A surd can be expressed in the form of a fractional index.
Surd Form: $\sqrt[n]{a}$
Index Form: $a^{\frac{1}{n}}$

5. ### The square root of ‘pi’ is a Surd or not?

No, $\pi$ is an irrational number but a Surd is the nth root of a rational number.