• Revision:Indices and surds

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Indices are used to describe the general term for ^2 in say  x^2 . There are a few laws to know when manipulating expressions involving indices.

Contents

Laws of Indices

 a^m \times a^n = a^{m+n}

 a^m \div a^n = a^{m-n}

 (a^m)^n = a^{mn}

 a^{\frac{1}{m}} = \sqrt[m]{a}

 a^{-m} = \frac{1}{a^m}

 a^{\frac{m}{n}} = \sqrt[n]a^m

 a^0 = 1

Examples

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

 5^3 \times 5^4 = 5^7

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

Surds

Surds are basically an expression involving a root, squared or cubed etc...

There are some basic rules when dealing with surds

 \sqrt{a} + \sqrt{a} = 2\sqrt{a}

 6\sqrt{a} - 2\sqrt{a} = 4\sqrt{a}

 \sqrt{a} \times \sqrt{b} = \sqrt{ab}

 \sqrt{\frac{a}{b}} = \frac{\sqrt a}{\sqrt b}

Also notice the special case

 \sqrt{a} \times \sqrt{a} = a^{\frac{1}{2}} \times a^{\frac{1}{2}} = a

Difference of Two Squares

 x^2 - y^2 = (x + y)(x - y) This is called the difference of two squares

Rationalising Surds

When you have a fraction where both the nominator and denominator are surds, rationalising the surd is the process of getting rid of the surd on the denominator.

To rationalise a surd you multiply top and bottom by fraction that equals one. Take the example shown below

 \frac{1}{\sqrt{2}}

To rationalise this multiply by effectively 1

 \frac{1}{\sqrt2} \times \frac{\sqrt 2}{\sqrt{2}}

Can you see why  \frac{\sqrt2}{\sqrt{2}} was chosen? This is because  \sqrt{2} \times \sqrt{2} = 2 so the denominator becomes surd free.

For a more complex term

 \frac{1+\sqrt{2}}{1 - \sqrt{5}}

First of all, we need to get rid of the surd expression on the bottom, you should remember the difference of two squares formula.

 a^2 - b^2 = (a + b)(a - b)

suppose a = 1 and b =  \sqrt{5}

 1^2 - 5 = (1+\sqrt{5})(1 - \sqrt{5}) = -4

So to get rid of the denominator surd we multiply  \frac{1+\sqrt{2}}{1 - \sqrt{5}} by  \frac{{1+\sqrt{5}}}{1+\sqrt{5}} like so.

 \frac{1+\sqrt{2}}{1 - \sqrt{5}} \times \frac{{1+\sqrt{5}}}{1+\sqrt{5}}

 = \frac{(1+\sqrt{2})(1+\sqrt{5})}{(1-\sqrt{5})(1+\sqrt{5})}

 = \frac{1 + \sqrt{5} + \sqrt{2} + \sqrt{10}}{1^2 - 5}

 = -\frac{1}{4} - \frac{\sqrt5}{4} - \frac{\sqrt2}{4} - \frac{\sqrt10}{4}

In general

  • Fractions in the form  \sqrt{\frac{1}{a}} multiply top and bottom by  \sqrt{a}
  • Fractions in the form  \frac{1}{a + \sqrt{b}} multiply the top and bottom by  a - \sqrt{b}
  • Fractions in the form  \frac{1}{a - \sqrt{b}} multiply the top and bottom by  a + \sqrt{b}
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