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Posts Tagged ‘roots’

## How to use Surds – Add, Multiply and Rationalise

Surds are roots to some power (usually square roots) that are left in their “root” form as opposed to being calculated as a decimal. In a way they are the power equivalent of fractions and are used when decimal value of a root isn’t needed, such as in the middle of a calculation, so it is better to leave the result more precisely as a surd.

Surds usually take the form a+b√c where b and c are non-zero

eg) √2
4+2√3

Simplifying Surds

It is easier to deal with surds if you can get the value in the square root to its lowest possible value by taking any square factors outside the root.

For example √50 = √(25×2) = 5√2

To do this I looked for any factors of 50 which are square, eg 25=5×5, can took these outside the square root – when taking a number outside the square root you obviously have to take the square root of the number so when you take out 25 you get a 5 left outside.

So in general

√(a2b) =a√b

This has the benefit of when multiple surds are in a calculation al the surds will be in their lowest terms so they will be easier to group

Surds can easily be added just by considering then as another variable, so just add up all the coefficients.

eg) 2√3 +6√3 – 3√3 = 5√3

Note– only roots of the same number can be added and subracted in this way, roots of different numbers must be left as they are. This makes it important to simplify all surds as it means you will be able to spot all the surds that can be added or subracted.

eg) 2√3 + 4√5 cannot by simplified

but

3√2 +2√50 = 3√2 +2×5√2=13√2

Multiplying and Dividing Surds

To multiply or divide surds just multiply or divide the values within the roots with each other and the coefficents of the roots with each other. When surds contain roots and non root parts multiply as if they were brackets

eg) 3√2 x 4√5 = 12√10

3√2 ÷ 4√5 = ¾ √(2/5)

(2+4√3) x (3+2√5) = 6 +4√5+12√3+8√15

Note – you may be able to further simplify the surd after you multiply

eg) 2√6 x √3 = 2√18=2×3√2 = 6√2

In general

(a+b√c) x (d + e√f) = ad + ae√f + db√c + be√fc

Rationalising and Conjugates of Surds

When simplifying fractions all surds are usually removed from the bottom. In order to do this both the top and the bottom of the fraction are multiplied by the conjugate of the botom. If there is only a root on the bottom you can simply multiply the top and bottom by the root on the bottom.

The conjugate of a surd is simply the surd with the sign of the coefficient of the root reversed so

a+b√c becomes a-b√c

So to rationalise the fraction

(2+√3)/(3+√4)

we multiply the top and bottom by 3-√4 to get

(3-√4)(2+√3)/(3+√4)(3-√4)

which once the brackets are expanded becomes

(6+3√3-2√4-√12)/5

as the √4 x √4 becomes √16 = 4 and the bottom is the difference of two squares.

By David Woodford

Categories: maths, surds

## Complex/Imaginary numbers

Most of the time in maths you are working with “real” numbers ie) all the rational numbers so we have 1,3,3.5567,4/7,root 2 etc, however sometimes we will need to extend our field of numbers to imaginary numbers. This is when we want to find the root of a negative number eg root -9, as (-3)2 and 32 both equal 9 we are unable to find the square root of -9. In order to solve this number we can use the “imaginary” number i such that
i = √-1

We then say that √-9 = 3i since √ab = √a √b and therefore √-9=√9 √-1 = 3i . This allows us to solve all equations with negative roots. It is also useful to know what in gives as shown below

i2 = -1
i3=-i
i4=1
i5=i

Complex numbers are made up of both real and imaginary parts,
eg z =a+ib
is a complex number. When comparing complex numbers we need only compare the real and imaginary parts, by equation them to each other,

ie) if the complex number z=a+ib
and 2z = 6+10i

a=3 and b=5

To solve complex equations with fractions we often have to rationalise the fractions, like we would with surd’s, so that there are no imaginary parts on the bottom and we can compare the coefficient’s, this is done by multiplying by the conjugate.

note// the conjugate of z=a+ib is z’ = a-ib

eg) (2+3i)/(4+5i) = (2+3i)(4-5i)/(4+5i)(4-5i) = (23+2i)/41

Complex numbers can be represented on an argand diagram when the y axis is the imaginary part and the x axis is the real part, so that the number a+ib is plotted at (a,b), and a line is draw between this and the origin to represent the number. The length of the line is called the moduls and the angle between the real axis and the line is the argument.

Categories: maths Tags: , , , , , ,