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Question
Rationalise the denominator of `1/[ √3 - √2 + 1]`
Solution
`1/[ √3 - √2 + 1]`
= `1/[(√3 - √2) + 1] xx [(√3 - √2) - 1]/[(√3 - √2) - 1]`
= `(√3 - √2 - 1)/[(√3 - √2)^2 - (1)^2]`
= `(√3 - √2 - 1)/[(√3)^2 - 2√6 + (√2)^2 - 1 ]`
= `(√3 - √2 - 1)/(3 - 2√6 + 2 - 1)`
= `(√3 - √2 - 1)/( 4 - 2√6 )`
= `[(√3 - √2) - 1]/[2( 2 - √6 )]`
= `[ √3 - √2 - 1 ]/[ 2( 2 - √6 ) ] xx [ 2 + √6 ]/[ 2 + √6 ]`
= `[ 2√3 - 2√2 - 2 + √18 - √12 - √6 ]/[ 2[ (2)^2 - ( √6)^2 ] ]`
= `[ 2√3 - 2√2 - 2 + 3√2 - 2√3 - √6 ]/[ 2[ 4 - 6] ]`
= `[ √2 - 2 - √6 ]/[ 2(-2) ]`
= `[ √2 - 2 - √6 ]/[ -4 ]`
= `1/4(2 + √6 - √2)`
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