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Middle Years

2.04 Binomial expansions with surds

Lesson

We know how to multiply surds together using the rule $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$a×b=ab, and we have also looked at how to expand binomial products  using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD

Now, there is nothing special about $A$A, $B$B, $C$C or $D$D. They can be any terms, and needn't be a variable or an integer. This means we can combine the two concepts together allowing us to expand binomial expressions involving surds.

 

Worked examples

Example 1

Expand $\left(\sqrt{2}+4\right)\left(\sqrt{2}-8\right)$(2+4)(28)

Think: We can use the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.

 

Do:

$\left(\sqrt{2}+4\right)\left(\sqrt{2}-8\right)$(2+4)(28) $=$= $\sqrt{2}\times\sqrt{2}+\sqrt{2}\times\left(-8\right)+4\times\sqrt{2}+4\times\left(-8\right)$2×2+2×(8)+4×2+4×(8)

Expanding the brackets

  $=$= $\sqrt{2}^2-8\sqrt{2}+4\sqrt{2}-32$2282+4232

Simplifying the products

  $=$= $2-8\sqrt{2}+4\sqrt{2}-32$282+4232

Evaluating $\left(\sqrt{2}\right)^2$(2)2 using the rule $\left(\sqrt{a}\right)^2=a$(a)2=a

  $=$= $-30-4\sqrt{2}$3042

Collecting the like surds and adding the integer terms

 

Reflect: After using the rule we can then simplify the expression using any of the algebraic rules that we have learned.

 
Example 2

Expand $\left(\sqrt{5}+4\right)^2$(5+4)2

Think: Since $\left(\sqrt{5}+4\right)^2=\left(\sqrt{5}+4\right)\left(\sqrt{5}+4\right)$(5+4)2=(5+4)(5+4) we can still use the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.

 

Do:

$\left(\sqrt{5}+4\right)^2$(5+4)2 $=$= $\left(\sqrt{5}+4\right)\left(\sqrt{5}+4\right)$(5+4)(5+4)

Since $\left(\sqrt{5}+4\right)^2=\left(\sqrt{5}+4\right)\left(\sqrt{5}+4\right)$(5+4)2=(5+4)(5+4)

  $=$= $\sqrt{5}\times\sqrt{5}+\sqrt{5}\times4+4\times\sqrt{5}+4\times4$5×5+5×4+4×5+4×4

Expanding the brackets

  $=$= $\left(\sqrt{5}\right)^2+4\sqrt{5}+4\sqrt{5}+16$(5)2+45+45+16

Simplifying the products

  $=$= $5+4\sqrt{5}+4\sqrt{5}+16$5+45+45+16

Evaluating $\left(\sqrt{5}\right)^2$(5)2 using the rule $\left(\sqrt{a}\right)^2=a$(a)2=a

  $=$= $21+8\sqrt{5}$21+85

Collecting the like surds and adding the integer terms

 

Reflect: This method will work with any expression squared. But this example is a special case and we could have also solved it using the rule for perfect square expansions

$\left(A+B\right)^2=A^2+AB+BA+B^2=A^2+2AB+B^2$(A+B)2=A2+AB+BA+B2=A2+2AB+B2

 

Summary

We can expand the product of two binomial expressions using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.

There are two special cases of expanding binomials:

  • $\left(A+B\right)^2=A^2+2AB+B^2$(A+B)2=A2+2AB+B2 (called a perfect square)
  • $\left(A+B\right)\left(A-B\right)=A^2-B^2$(A+B)(AB)=A2B2 (called a difference of two squares)

Practice questions

Question 1

Expand and simplify: $\left(5-\sqrt{13}\right)\left(7-\sqrt{11}\right)$(513)(711)

Question 2

Expand and simplify: $\left(8\sqrt{5}-6\right)\left(8\sqrt{5}+6\right)$(856)(85+6)

Question 3

Expand and simplify: $\left(4\sqrt{2}-\sqrt{13}\right)^2$(4213)2

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