Quantitative Aptitude - SPLessons

Surds – Indices

Chapter 10

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Surds – Indices

shape Introduction

Surds – Indices deals with problems on surds and indices involved with laws of surds and laws of indices respectively.


shape Methods

Surds:

Surds are the  irrational numbers that contain the radical sign \(\sqrt{}\). The set of irrational numbers contain numbers such as \(\sqrt{2}\), \(\sqrt[3]{2}\), \(\pi\), etc.


  • Surds can be added or subtracted only if they are like surds.
    Example: \(\sqrt{2}\) + 5\(\sqrt{2}\) = 6\(\sqrt{2}\)

  • Surds can be multiplied using \(\sqrt{x}\) x \(\sqrt{x}\)= \(x\) and \(\sqrt{x}\) x \(\sqrt{y}\) = \(\sqrt{xy}\)

  • To rationalize the denominator of a fraction, we need to convert it into equivalent fraction.
    Example: \(\frac{1}{\sqrt{7}}\) = \(\frac{1}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{\sqrt{7}}{7}\)

  • A surd can be expressed in index form as a fractional index.
    Example: \(\sqrt[n]{a}\) = \(a^{(\frac{1}{n})}\)
    here \(\sqrt[n]{a}\) is in surd form and \(a^{(\frac{1}{n})}\) is in index form.


Indices

Indices are a useful way expressing large numbers with the help of powers or also known as indices. Exponent is another commonly used term for a power. In the below example the power/exponent of 4 is 6.

Example: 4096 is obtained by multiplying 4 by itself for 6 times and so 4096 can be expresses as:\(4^6\)  where 4 x 4 x 4 x 4 x 4 x 4 = 4096


  • Laws of indices can be applied only to the expressions having equal bases.
    Example: \(2^6\), \(2^4\), \(2^8, etc\)


Example 1:
Show that for any positive real number p, the expression \(a^{-p}\) is equivalent to \(\frac{1}{a^{p}}\).

Solution:

    We proceed with the following manipulation – \(a^{-p}\) = \(a^{0 – p}\)

    Using Law 2 i.e. \(\frac{a^{m}}{a^{n}}\) = \(a^{(m – n)}\), we can rewrite the above expression as –

    \(\frac{a^{0}}{a^{p}}\)

    = \(\frac{1}{a^{p}}\), which is the required result.


Example 2:
Simplify and evaluate \((\frac{16}{81})^{-(\frac{3}{4})}\)

Solution:

    Using the laws of indices and some manipulation –

    \((\frac{16}{81})^{-(\frac{3}{4})}\) = \(\frac{1}{(\frac{16}{81})^{\frac{3}{4}}}\)

    = \((\frac{81}{16})^{\frac{3}{4}}\)

    = \(((\frac{81}{16})^{\frac{1}{4}})^{3}\)

    = \((\frac{81^{\frac{1}{4}}}{16^{\frac{1}{4}}})^{3}\)

    = \((\frac{3}{2})^{3}\)

    = \(\frac{3^{3}}{2^{3}}\)

    = \(\frac{27}{8}\)


Example 3:
Simplify the expression \(y = x^{a – b} \times x^{b – c} \times x^{c – a} \times x^{-a – b}\)

Solution:

    Using the laws of indices:

    \(y = x^{a – b} \times x^{b – c} \times x^{c – a} \times x^{-a – b}\)

    \(y = x^{(a – b) + (b – c) + (c – a) + (-a – b)}\)

    \(y = x^{-a – b}\)

    \(y = \frac{1}{x^{a + b}}\)


Laws of Surds/ Rules of Surds:


Example:
Simplify \(\sqrt{18}\)

Solution:

    Since 18 = 9 x 2 = \(3^{2} \times 2\), as 9 is the largest perfect square factor of 18.

    \(\sqrt{18}\) = \(\sqrt{3^{2} \times 2}\)

    = \(\sqrt{3^{2}}\) x \(\sqrt{2}\) (Using the rule \(\sqrt{(a \times b)}\) = \(\sqrt{a}\) x \(\sqrt{b}\))

    = 3\(\sqrt{2}\)


Example:
Simplify \(\sqrt{\frac{12}{121}}\)

Solution:

    \(\sqrt{\frac{12}{121}}\) = \(\frac{\sqrt{12}}{\sqrt{121}}\) (Using the rule \(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\))

    = \(\frac{\sqrt{2^{2} \times 3}}{11}\) (Since 4 is the largest perfect square factor of 12)

    = \(\frac{\sqrt{2^{2} \times \sqrt{3}}}{11}\) (Using the rule \(\sqrt{(a \times b)}\) = \(\sqrt{a}\) x \(\sqrt{b}\))

    = \(\frac{2\sqrt{3}}{11}\)


Example:
Rationalise \(\frac{5}{\sqrt{7}}\)

Solution:

    \(\frac{5}{\sqrt{7}}\) = \(\frac{5}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) (Multiply both numerator and denominator by \(\sqrt{7}\))

    = \(5\frac{\sqrt{7}}{7}\)


Example:
Simplify \(5\sqrt{6} + 4\sqrt{6}\)

Solution:

    \(5\sqrt{6} + 4\sqrt{6}\) = (5 + 4)\(\sqrt{6}\) (Using the rule \(a\sqrt{c} \pm b\sqrt{c} = (a \pm b)\sqrt{c}\))

    = \(9\sqrt{6}\)


Following this rule enables to rationalise the denominator.

Example:
Rationalise \(\frac{3}{2 + \sqrt{2}}\)

Solution:

    \(\frac{3}{2 + \sqrt{2}}\) = \(\frac{3}{2 + \sqrt{2}}\) x \(\frac{2 – \sqrt{2}}{2 – \sqrt{2}}\) (Multiply the numerator and denominator by \(2 – \sqrt{2}\))

    = \(\frac{6 – 3\sqrt{2}}{4 – 2}\)

    = \(\frac{6 – 3\sqrt{2}}{2}\)


Following this rule enables to rationalise the denominator.

Example:
Rationalise \(\frac{3}{2 – \sqrt{2}}\)

Solution:

    \(\frac{3}{2 – \sqrt{2}}\) = \(\frac{3}{2 – \sqrt{2}}\) x \(\frac{2 + \sqrt{2}}{2 + \sqrt{2}}\) (Multiply the numerator and denominator by \(2 + \sqrt{2}\))

    = \(\frac{6 + 3\sqrt{2}}{4 – 2}\)

    = \(\frac{6 + 3\sqrt{2}}{2}\)

shape Formulae

1. Laws of Indices/ Laws of Exponents:


  • \(a^m\) x \(a^n\) = \(a^{m + n}\)

  • \(\frac{a^m}{a^n}\) = \(a^{m – n}\)

  • \(({a^m})^{n}\) = \(a^{mn}\)

  • \((ab)^m\) = \(a^m\) x \(b^m\)

  • \((\frac{a}{b})^n\) = \(\frac{a^n}{b^n}\)

  • \(a^0\) = 1


2. Laws of Surds:


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

  • \(\sqrt[n]{ab}\) = \(\sqrt[n]{a}\) x \(\sqrt[n]{b}\)

  • \(\sqrt[n]{\frac{a}{b}}\) = \(\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\)

  • \((\sqrt[n]{a})^n\) = a

  • \(\sqrt[m]{\sqrt[n]{a}}\) = \(\sqrt[mn]{a}\)

  • \((\sqrt[n]{a})^m\) = \(\sqrt[n]{a}^m\)

shape Samples

1 Find \(\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8…..∝}}}}\) = ?

Solution:

    Assume \(\sqrt{{8}\sqrt{{8}{\sqrt{{8…..}}}}}\) = \( x \)

    Now, consider \(\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8…..∝}}}}\)

    ⇒\(\sqrt{8 x}\) = \( x \)

    ⇒\( x^2 \) = 8\( x \)

    ⇒\( x \) = 8

    Therefore, the value of \(\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8…..∝}}}}\) = 8


2. Simplify the value of \((125)^{(\frac{2}{3})}\)

Solution:

    Consider\((125)^{(\frac{2}{3})}\)

    =\((5^3)^{(\frac{2}{3})}\)

    =\(5^2\)

    =25

    Therefore, the value of \((125)^{(\frac{2}{3})}\) = 25


3. Find the value of [\((8)^{\frac{5}{3}}\) + \((8)^{\frac{-5}{3}}\)]?

Solution:

    Consider [\((8)^{\frac{5}{3}}\) + \((8)^{\frac{-5}{3}}\)]

    = [\((2^3)^{\frac{5}{3}}\) + \((2^3)^{\frac{-5}{3}}\)]

    = [\((2^5)\) + \((2^{-5})\)]

    = \((2^5)\) + \(\frac{1}{(2^5)}\)

    = \(\frac{2^5 + 1}{2^5}\)

    = \(\frac{1024 + 1}{32}\)

    = \(\frac{1025}{32}\)

    Therefore, [\((8)^{\frac{5}{3}}\) + \((8)^{\frac{-5}{3}}\)] = \(\frac{1025}{32}\)


4. What will be the quotient if \((x^{-1}-1)\) is divided by \( (x – 1)\)?

Solution:

    Given that

    Dividend = \((x^{-1}-1)\)

    Divisor = \( (x – 1)\)

    Now, Consider \(\frac{(x^{-1}-1)}{(x – 1)}\)

    = \(\frac{\frac{1}{x} -1}{(x – 1)}\)

    = \(\frac{\frac{(1-x)}{x}}{(x – 1)}\)

    = \(\frac{(1 – x)}{x}\) x \(\frac{1}{(x – 1)}\)

    = \(\frac{(1-x)}{(-x)(1-x)}\)

    = –\(\frac{1}{x}\)

    Therefore, the required quotient = –\(\frac{1}{x}\)


5. Which is larger among \(\sqrt[4]{6}\), \(\sqrt[3]{4}\), \(\sqrt[2]{5}\) ?

Solution:

    Given surds are \(\sqrt[4]{6}\), \(\sqrt[3]{4}\), \(\sqrt[2]{5}\)

    Surds are in the order of 4, 3, 2 respectively.

    L.C.M. of 4, 3, 2 is 12

    Now, change the each given surd in the order of 12

    \(\sqrt[4]{6}\) = \({6}^{(\frac{1}{4})}\) = \({6}^{(\frac{1}{4} * \frac{3}{3})}\) = \({6}^{(\frac{3}{12})}\) = \({(6^3)}^{(\frac{1}{12})}\) = \({(216)}^{(\frac{1}{12})}\)

    \(\sqrt[3]{4}\) =\({4}^{(\frac{1}{3})}\) = \({4}^{(\frac{1}{3} * \frac{4}{4})}\) = \({4}^{(\frac{4}{12})}\) = \({(4^4)}^{(\frac{1}{12})}\) = \({(256)}^{(\frac{1}{12})}\)

    \(\sqrt[2]{5}\) =\({5}^{(\frac{1}{2})}\) = \({5}^{(\frac{1}{2} * \frac{6}{6})}\) = \({5}^{(\frac{6}{12})}\) = \({(5^6)}^{(\frac{1}{12})}\) = \({(3125)}^{(\frac{1}{12})}\)

    Hence, \({(3125)}^{(\frac{1}{12})}\) > \({(256)}^{(\frac{1}{12})}\) > \({(216)}^{(\frac{1}{12})}\)

    Therefore, \(\sqrt[2]{5}\) > \(\sqrt[3]{4}\) > \(\sqrt[4]{6}\)