Surds

UK Secondary (7-11)

Simplify surd expressions involving powers or algebraic terms

Lesson

A surd is easily transformed due to some interesting properties of roots. Let's take a look at square roots for example.

We know that $\sqrt{36}$√36 can be computed to be $6$6, but we can also reach this solution another way. $\sqrt{36}$√36 can be written as $\sqrt{4\times9}$√4×9, and as we'll learn soon this can also be written as $\sqrt{4}\times\sqrt{9}$√4×√9. We know that $\sqrt{4}$√4 = $2$2 and $\sqrt{9}$√9 = $3$3 so the answer to $\sqrt{36}$√36 can be rewritten as $2\times3$2×3 = $6$6. We can do this because of a special property:

$\sqrt{a\times b}$√`a`×`b` = $\sqrt{a}\times\sqrt{b}$√`a`×√`b`

Sometimes $\sqrt{a}$√`a` or $\sqrt{b}$√`b` are surds and can not be expressed as a rational number, and we can use this to help transform larger surds into smaller ones. For example, $\sqrt{20}$√20 can be rewritten as $\sqrt{4\times5}$√4×5 = $\sqrt{4}\times\sqrt{5}$√4×√5 = $2\times\sqrt{5}$2×√5. The last part can be rewritten as $2\sqrt{5}$2√5 and this is called a **simplified surd**, as we can not break it down any more into even smaller surds.

The key to simplifying surds is to find factors of $a$`a` or $b$`b` that are square numbers. So in the example above, I found that the square number $4$4 goes into $20$20. However, if I chose other factors to split $20$20 into, it would not work. So I can write $\sqrt{20}$√20 as $\sqrt{2\times10}$√2×10 = $\sqrt{2}\times\sqrt{10}$√2×√10 which does not simplify down to anything.

Simplify** **$\sqrt{50}$√50.

**Think: **Find a square number that is also a factor of $50$50

**Do: **$\sqrt{50}$√50 = $\sqrt{25\times2}$√25×2, $25$25 is the square factor here

$\sqrt{50}$√50 | $=$= | $\sqrt{25}\times\sqrt{2}$√25×√2 |

$=$= | $5\times\sqrt{2}$5×√2 | |

$=$= | $5\sqrt{2}$5√2 |

Just as there is a square root property involving multiplication, there is also a similar property involving division:

$\sqrt{\frac{a}{b}}$√`a``b` = $\frac{\sqrt{a}}{\sqrt{b}}$√`a`√`b`

Again we can use this to help us transform and simplify surds. For example $\frac{\sqrt{8}}{\sqrt{2}}$√8√2 can be rewritten as $\sqrt{\frac{8}{2}}$√82 = $\sqrt{4}$√4 which has an answer of $2$2.

Simplify** **$\sqrt{52}\div\sqrt{13}$√52÷√13.

**Think: **remember that a division operation can be rewritten as a fraction

**Do: **

$\sqrt{52}\div\sqrt{13}$√52÷√13 | $=$= | $\frac{\sqrt{52}}{\sqrt{13}}$√52√13 |

$=$= | $\sqrt{\frac{52}{13}}$√5213 | |

$=$= | $\sqrt{4}$√4 | |

$=$= | $2$2 |

If we were to look at $\sqrt{36}$√36 again, and we decided to square it, what answer would we get? $\left(\sqrt{36}\right)^2$(√36)2 = $6^2$62 = $36$36. We've come full circle and back to $36$36! This brings us to the third property of square roots:

$\left(\sqrt{a}\right)^2$(√`a`)2 = $a$`a`

This make sense as the square root and square operations are 'opposite' to each other like addition and subtraction or multiplication and division, so they cancel each other out and we're just left with $a$`a`!

Simplify** **$\left(2\sqrt{6}\right)^2$(2√6)2.

**Think: **$\left(ab\right)^2$(`a``b`)2 = $a^2b^2$`a`2`b`2

**Do:**

$\left(2\sqrt{6}\right)^2$(2√6)2 | $=$= | $2^2\times\left(\sqrt{6}\right)^2$22×(√6)2 |

$=$= | $4\times6$4×6 | |

$=$= | $24$24 |

Simplify $\sqrt{150}$√150.

Simplify $5\sqrt{8}$5√8.

Express $7\sqrt{2}$7√2 as an entire surd