Home/quantitative-aptitude/quantitative-aptitude

Surds & Indices

Master powers and roots. Learn comparison techniques, infinite series shortcuts ($\sqrt{12+\sqrt{12...}}$) and rationalization without pen.

1. Laws of Indices

Basic rules governing powers.

$a^m \times a^n = a^{m+n}$
$a^m / a^n = a^{m-n}$
$(a^m)^n = a^{mn}$
$a^{-n} = 1/a^n$
$a^0 = 1$ (for $a \neq 0$ )

Example:

Q: Simplify

(2^3)^2 imes 2^{-4}

Solution:

2^{3 \times 2} \times 2^{-4} = 2^6 \times 2^{-4} = 2^{6-4} = 2^2 = 4

2. Infinite Series Shortcuts (The Ladder)

Solve infinite nested roots in 2 seconds.

Addition: $\sqrt{x + \sqrt{x + ...}} =$ Large Factor of x. (where diff of factors is 1).
Subtraction: $\sqrt{x - \sqrt{x - ...}} =$ Small Factor of x.
Multiplication: $\sqrt{x \sqrt{x ...}} = x$ .

Example:

Q: Value of

\sqrt{12 + \sqrt{12 + ...\infty}}

Solution: Factors of 12 are 4 and 3 (diff 1).
Sign is '+', so Answer is Larger Factor: 4.

3. Comparison of Surds

To compare $\sqrt[a]{x}$ and $\sqrt[b]{y}$ :
1. Take LCM of power indices ( $a, b$ ).
2. Raise numbers to the power of LCM.
3. Compare resulting integers.

Example:

Q: Which is larger:

\sqrt{2}

\sqrt[3]{3}

Solution: Indices: 2, 3. LCM = 6.

(\sqrt{2})^6 = 2^3 = 8

(\sqrt[3]{3})^6 = 3^2 = 9

9 > 8

, so

\sqrt[3]{3}

is larger.

4. Rationalization Strategy

Eliminate roots from denominator by multiplying with Conjugate.

Conjugate of $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$ .
Formula: $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = a-b$ .

Example:

Q: Simplify

1/(\sqrt{5}-\sqrt{3})

Solution: Multiply top/bottom by

\sqrt{5}+\sqrt{3}

.
Num:

\sqrt{5}+\sqrt{3}

. Denom:

5-3=2

.
Ans:

(\sqrt{5}+\sqrt{3})/2

5. Square Root of Surds

To find $\sqrt{A \pm 2\sqrt{B}}$ , find two numbers $x, y$ such that $x+y=A$ and $xy=B$ . Then answer is $\sqrt{x} \pm \sqrt{y}$ .

Golden Rule: Ensure coefficient of inner root is 2. If not, multiply/divide or adjust.

Example:

Q: Find

\sqrt{7 + 4\sqrt{3}}

Solution: 1. Convert to form

A + 2\sqrt{B}

7 + 2(2\sqrt{3}) = 7 + 2\sqrt{12}

.
2. Find factors of 12 sum to 7: 4 and 3.
3. Ans:

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