Rationalising denominator

2-√3 In this case multiplying the bottom and top by √3wouldn't work as we would still get surd at the bottom To rationalise denominator we have to multiply the topand the bottom of the fraction by 2+√3.√3 Rationalise the denominator of Example 1:2-√3 √3 x(2+√3) x(2+√3) x√3 x√3 = = (2-√3)(2+√3) (2-√3) x √3 √3 x √3 √3(2+√3) = = 4+2√3-2√3-3 2√3-3 3 2√3+3 It's not good as we still have surd in denominator, so we have to multiply by different thing to get rid of the surd in denominator. 2-√3 √3 = 2√3+31+√2 1-√2 In the denominator we have 1+√2, so we have to multiply the top and the bottom of the fraction by 1-√2. We always use the two numbers from the bottom, butwe have to change the sign + to - or - to +. Example 2:Rationalise the denominator of x(1-√2) x(1-√2) = (1+√2)(1-√2) (1-√2)(1-√2) = 1-√2+√2-2 1-√2-√2+2 = 3-2√2 1+√2 1-√2 -1 = -3+2√2= √5+1 Rationalise the denominator of 5-√5+√5-1 3 √ x(√5-1) x(√5-1) = (√5+1) - 3( √ = ( √ - √ √5+1 - ) 3 - ) = |

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