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India
Class VIII

Simplify surd expressions involving powers or algebraic terms

Lesson

Ways to Manipulate Square Roots

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

 

Simplified Surds

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}$25 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.

 

Worked example

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}$52

 

Simplifying With Fractions

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

$\sqrt{\frac{a}{b}}$ab = $\frac{\sqrt{a}}{\sqrt{b}}$ab

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

 

Worked example

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}}$5213
  $=$= $\sqrt{\frac{52}{13}}$5213
  $=$= $\sqrt{4}$4
  $=$= $2$2

 

Squaring Surds

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!

 

Worked example

Simplify $\left(2\sqrt{6}\right)^2$(26)2.

Think: $\left(ab\right)^2$(ab)2 = $a^2b^2$a2b2

Do:

$\left(2\sqrt{6}\right)^2$(26)2 $=$= $2^2\times\left(\sqrt{6}\right)^2$22×(6)2
  $=$= $4\times6$4×6
  $=$= $24$24

 

Practice questions

Question 1

Simplify $\sqrt{150}$150.

Question 2

Simplify $5\sqrt{8}$58.

Question 3

Express $7\sqrt{2}$72 as an entire surd

Outcomes

8.NS.SSCC.2

Square roots using factor method and division method for numbers containing (a) no more than total 4 digits and (b) no more than 2 decimal places

8.NS.SSCC.4

Estimating square roots and cube roots. Learning the process of moving nearer to the required number.

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