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Binomial Products with Surds

Like the distributive law of numbers, binomial products with surds is given as:

(a + b)(c + d)  =  ac + ad + bc + bd

 

We use this procedure to look at a few examples of products of surds

Example 1: (sqrt{2}~+~5)(sqrt{2}~-~5)

=  sqrt{2}*sqrt{2}~~-~~5*sqrt{2}~~+~~5*sqrt{2}~~-~~5*5

=  2 – 25

=  -23

Example 2(3sqrt{2}~-~sqrt{5})(2sqrt{5}~+~3sqrt{2})

=  3*2*sqrt{2}*sqrt{5}~~+~~3*3*sqrt{2}*sqrt{2}~~-~~sqrt{5}*2*sqrt{5}~~-~~sqrt{5}*3*sqrt{2}

=  6sqrt{10}~~+~~18~~-~~2*5~~-~~3sqrt{10}

=  6sqrt{10}~~+~~18~~-~~10~~-~~3sqrt{10}

=  3sqrt{10}~~+~~8

 

Remember these rules/formulae from Algebra – they will be useful in expanding and simplifying surds

(a + b)2  =  a2 + 2ab + b2

(a – b)2  =  a2 – 2ab + b2

a2 – b2  =  (a – b)(a + b)

 

Example 3: (sqrt{2}~+~sqrt{3})^2

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

We can use the distributive law to solve the above equation, but instead we will use the formula (a + b)2  =  a2 + 2ab + b2 to solve it. So

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

=  2 + 2*sqrt{6} + 3

=  5 + 2sqrt{6}

Example 4(3sqrt{5}~-~3sqrt{3})^2

=  (3sqrt{5})^2  –  (2*3*3*sqrt{5}*sqrt{3})  +  (3sqrt{3})^2

Notice we have used the formula (a – b)2  =  a2 – 2ab + b2 above

=  9×5  –  18sqrt{15}  +  9×3

=  45 + 18 – 18sqrt{15}

=  63 – 18sqrt{15}

Example 5(sqrt{5}~-~sqrt{3})(sqrt{5}~+~sqrt{3})

=  (sqrt{5})^2 – (sqrt{3})^2  (using the formula a2 – b2  =  (a – b)(a + b))

=  5 – 3

=  2

sqrt{5}~-~sqrt{3} and sqrt{5}~+~sqrt{3} are called conjugate surds. Two surds are called conjugate surds when they multiply to give a rational number. This process of getting a rational number is called rationalisation of surds, which we will see in the next section.