1.09 Binomial expansions with surds
We know how to multiply surds together using the rule $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$√a×√b=√ab, so we can now combine this with the rule on how to expand binomial products: $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(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)(√2−8)
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)(√2−8) | $=$= | $\sqrt{2}\times\sqrt{2}+\sqrt{2}\times\left(-8\right)+4\times\sqrt{2}+4\times\left(-8\right)$√2×√2+√2×(−8)+4×√2+4×(−8) |
Expanding the brackets |
| $=$= | $\sqrt{2}^2-8\sqrt{2}+4\sqrt{2}-32$√22−8√2+4√2−32 |
Simplifying the products |
|
| $=$= | $2-8\sqrt{2}+4\sqrt{2}-32$2−8√2+4√2−32 |
Evaluating $\left(\sqrt{2}\right)^2$(√2)2 using the rule $\left(\sqrt{a}\right)^2=a$(√a)2=a |
|
| $=$= | $-30-4\sqrt{2}$−30−4√2 |
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+4√5+4√5+16 |
Simplifying the products |
|
| $=$= | $5+4\sqrt{5}+4\sqrt{5}+16$5+4√5+4√5+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+8√5 |
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
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)(A−B)=A2−B2 (called a difference of two squares)
Practice questions
Question 1
Expand and simplify: $\left(5-\sqrt{13}\right)\left(7-\sqrt{11}\right)$(5−√13)(7−√11)
Question 2
Expand and simplify: $\left(8\sqrt{5}-6\right)\left(8\sqrt{5}+6\right)$(8√5−6)(8√5+6)
Question 3
Expand and simplify: $\left(4\sqrt{2}-\sqrt{13}\right)^2$(4√2−√13)2