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Surds & Indices

Surds are less common in MBA entrance tests, including CAT. However, the concept of surds is quite simple and could be applied in other calculations. Note that we may not have direct questions, but these concepts might have to be applied while solving other algebraic questions.

Questions on Indices are quite common in MBA entrance tests. The basics of indices is already covered in the lesson on Number Theory.

1. Surds

A surd is an irrational number which includes the root of an integer. Surds can also be expressed as the sum of a rational number and an irrational number. The following are examples of surds:
$\sqrt{5}, 2 + \sqrt{5}, 5^{\frac{1}{3}} + 6^{\frac{2}{3}}, \sqrt[7]{56} + \sqrt[9]{67}$

Surds where the highest power is $\dfrac{1}{2}$ are called quadratic surds and where the highest power is $\dfrac{1}{3}$ are called cubic surds.

From the perspective of entrance tests, we will be primarily tested on quadratic surds, and not cubic or other lower power surds.

Where $a, b, c$ and $d$ are integers and $b$ and $d$ are not perfect squares, if $a + \sqrt{b} = c + \sqrt{d}$, then $a = c$ and $b = d$.

For instance, if $x + \sqrt{y} = 4 + 2 \sqrt{5}$

$\implies x + \sqrt{y} = 4 + \sqrt{2^{2} \times 5}$
$\implies x + \sqrt{y} = 4 + \sqrt{20}$

∴ $x = 4, y = 20$

To summarise, if two surds are equal, then their rational parts are equal and their irrational parts are equal.

1.1 Conjugate of surds

Quadratic surds can be eliminated if each of its terms are squared.

As $(a + b)(a - b) = a^{2} - b^{2}$, for the term $\bold{(a + b)}$, the conjugate is $\bold{(a - b)}$ and vice versa.

If a surd in the denominator has to be removed, then we multiply and divide by the conjugate of the denominator.

∴ $\dfrac{4}{\sqrt{5} + \sqrt{3}} = \dfrac{4}{\sqrt{5} + \sqrt{3}} \times \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \dfrac{4 \times (\sqrt{5} - \sqrt{3})}{(\sqrt{5})^{2} - (\sqrt{3})^{2}}$

$= \dfrac{4 \times (\sqrt{5} - \sqrt{3})}{5 - 3} = 2 \times (\sqrt{5} - \sqrt{3})$

Likewise, $\dfrac{2 + \sqrt{3}}{5 - 2 \sqrt{5}} = \dfrac{2 + \sqrt{3}}{5 - 2 \sqrt{5}} \times \dfrac{5 + 2 \sqrt{5}}{5 + 2 \sqrt{5}} = \dfrac{10 + 5 \sqrt{3} + 4 \sqrt{5} + 2 \sqrt{15}}{5}$

Example 1

Where $a$ and $b$ are rational numbers, if $\dfrac{3 + \sqrt{5}}{3 - \sqrt{5}} = a + \sqrt{b}$, then $a + 2b =$

Solution

$a + \sqrt{b} = \dfrac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3 + \sqrt{5}}{3 + \sqrt{5}} = \dfrac{9 + 5 + 6 \sqrt{5}}{9 - 5}$

$\implies a + \sqrt{b} = \dfrac{7 + 3 \sqrt{5}}{2} = \dfrac{7}{2} + \sqrt{\dfrac{3^{2} \times 5}{2^{2}}} = \dfrac{7}{2} + \sqrt{\dfrac{45}{4}}$

∴ $a = \dfrac{7}{2}$ and $b = \dfrac{45}{4}$

$a + 2b = \dfrac{7}{2} + \dfrac{45}{2} = 26$

Answer: 26