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Rationalisation of Surds

Rationalisation of surds involves the multiplication of a surd by its conjugate to get a rational number. This process requires us to not leave the denominator in the surd form, but as a rational number.

Let’s consider 3/sqrt{3} . As per the definition of rationalisation of surds, we should have a rational number in the denominator, and not have a surd there. So we’ll multiply both the numerator and denominator by sqrt{3} .

3/sqrt{3} = x sqrt{3}/sqrt{3}

= {3sqrt{3}}/3

= sqrt{3}

Examples of rationalisation of surds

Example 1: {2~+~sqrt{3}}/sqrt{5}

= {2~+~sqrt{3}}/sqrt{5} x sqrt{5}/sqrt{5}

= {2sqrt{5}~+~sqrt{15}}/5

Example 2: 2/{2~-~sqrt{3}}

= 2/{2~-~sqrt{3}} x {2~+~sqrt{3}}/{2~+~sqrt{3}}

= ${2(2~+~sqrt{3})}/{2^2~-~(sqrt{3})^2}$

= {2(2~+~sqrt{3})}/{4~-~3}

= 2(2~+~sqrt{3})

Example 3: {2~+~sqrt{3}}/{2~-~sqrt{3}}

= {2~+~sqrt{3}}/{2~-~sqrt{3}} x {2~+~sqrt{3}}/{2~+~sqrt{3}}

= {(2~+~sqrt{3})(2~+~sqrt{3})}/{4~-~3}

= ${(2~+~sqrt{3})^2}/1$

= $2^2~+~2*2*sqrt{3}~+~(sqrt{3})^2$

= 4 + 4sqrt{3} + 3

= 7 + 4sqrt{3}

Example 4: {2~+~sqrt{5}}/{-6~-~sqrt{5}}

= {2~+~sqrt{5}}/{-6~-~sqrt{5}} x {-6~+~sqrt{5}}/{-6~+~sqrt{5}}

= {(2~+~sqrt{5})(-6~-~sqrt{5})}/{(-6~+~sqrt{5})(-6~+~sqrt{5})}

= ${2*-6~+~2sqrt{5}~-~6sqrt{5}~+~(sqrt{5})^2}/{(-6)^2~-~(sqrt{5})^2}$

= {-12~+~2sqrt{5}~-~6sqrt{5}~+~5}/{36~-~5}

= {-7~-~4sqrt{5}}/31