Surds and Indices (Simplification hack to score 10/10 )

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[3]{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.

Difference Between Rational and Irrational Numbers

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}

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[3]{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[3]{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.

Note: Click here for Quick Calculation Tips and Tricks.

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

(i) \sqrt[3]{\sqrt[4]{a^{3}}}

(ii) \sqrt[3]{\sqrt[5]{a^{2}}}

(iii) \sqrt{\sqrt[3]{a}}

(iv) \sqrt{\sqrt[5]{a^{3}}}

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

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

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

\sqrt{\sqrt[5]{a^{3}}}=(\sqrt[5]{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[16]{10^x}

Find the value of x?

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

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

The value of x=30

Frequently Asked Questions

Classification of Numbers: Click Here

Multiple Choice Questions (MCQ)

35 Questions with solutions, based on the concepts of Surds and Indices.

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