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C1 - Simplifying a fraction that has a root on the denominator

(7+√5)/(√5-1) 

In this example we are given (√5-1) as our denominator, and we want to simplify this complex fraction into the form a + b√5. 

In order to do this our first step is to elimate the denominator. To do this, we use the rule associated with the Difference of Two Squares. For example, if you were told to factorise and simplify: x 2 - 16, you would get (x-4)(x+4) since when you expand the '-4x' and '+4x' would cancel each other out to leave x2 - 16. 

To return to our question therefore, we need to multiply our fraction by something to eliminate the √5 on the denominator. This something is (√5 +1) as the opposite signs will mean the middle part of our expanded equation +√5-√5 will cancel each other out, and multiplying √5 by itself will eliminate the square root. But we must remember, that whatever we do to the bottom of the fraction we must also do to the top, therefore we must multiply both the numerator and the denominator by (√5 + 1)

So:  (7+√5)/(√5-1) x (√5+1)/(√5+1)

Multiplying the numerators (top of the fractions) together, we get: (7x√5) + (7x1) + (√5x√5) + (√5x1), which equals: 7√5 + 7 + 5 + √5    

By collecting like terms this simplifies to: 8√5 + 12

Multiplying the denominators (bottom of the fractions) together, we get: (√5x√5) + (√5x1) - (1x√5) - (1x1)     which equals: 5 + √5 - √5 - 1      

Again, by collecting like terms this simplifies to: 4

Therefore, we can also write (7+√5)/(√5-1) as:

(8√5 + 12) / 4

But this can be simplified further as both 8, and 12 are divisible by 4, and therefore this can finally be written as: 

2√5 + 3     or     3 + 2√5

SOLUTION:     a = 3      and      b = 2

Claudia M. GCSE Maths tutor, A Level Maths tutor, A Level Economics t...

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