UK Secondary (7-11)
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Exact values or decimal approximations
Lesson

We've already had a look at how to approximate a surd, and we know that at the end of the day it is still an estimation, and that is why we usually leave surds in exact form (with the root) if we want to be accurate. However when dealing with exact forms we also want it in a form that is easier to understand, we that can be done through simplifying surds.

For example let's take a look at $\sqrt{24}$24. Just looking at this surd it is very hard to understand what values it might be around. Simplifying it gives us $2\sqrt{6}$26. $\sqrt{6}$6 is a smaller surd than $\sqrt{24}$24 so it is easier to approximate. Using techniques we already know, we know that $2\sqrt{6}$26 is a little more than $2\times2=4$2×2=4. Of course if we know our squares really well we can see that $24$24 is a little under $25$25 so $\sqrt{24}$24 must be a little under $\sqrt{25}=5$25=5. Finally using a calculator we see that $\sqrt{24}=4.9$24=4.9 which means all our evaluating makes sense.

Examples

Question 1

Which of the following can be written as an exact value without a root sign?

A) $\sqrt{2}+\sqrt{3}$2+3     B) $\sqrt{485}$485    C) $\sqrt{400}$400    D) $10\sqrt{83}$1083

Think: Is the number inside the root sign a perfect square?

Do 

A) can not be simplified any further as $2$2 and $3$3 are both not perfect squares

B) $485=5\times97$485=5×97 (two prime factors) so not a perfect square either

C) If we break $400$400 down into its prime factors we get $400=2\times2\times10\times10$400=2×2×10×10$=$=$20\times20$20×20 so it is a perfect square and $\sqrt{400}$400 can be rewritten as $20$20

D) $83$83 is a prime number so can not be a perfect square either

Therefore the correct answer is C).

 

Question 2

For the surd $\sqrt{360}$360:

a) Express in the simplest exact form

Think: Can $360$360 be broken down into any factors that are perfect squares?

Do:

$\sqrt{360}$360 $=$= $\sqrt{4\times90}$4×90
  $=$= $2\sqrt{90}$290
  $=$= $2\sqrt{9\times10}$29×10
  $=$= $2\times3\sqrt{10}$2×310
  $=$= $6\sqrt{10}$610

b) Without using a calculator, find the integer value of $\sqrt{360}$360

Think: Figure out if you can approximate $\sqrt{10}$10 and then use that to approximate $\sqrt{360}$360

Do:

$\sqrt{10}$10 is between $\sqrt{9}$9 and $\sqrt{16}$16, and is a bit closer to $\sqrt{9}=3$9=3.

Therefore the integer value of $\sqrt{360}=6\sqrt{10}$360=610 is $6\times3=18$6×3=18.

c) Evaluate $\sqrt{360}$360 to $2$2 decimal places

Think: The fastest way to get an exact value is by using the calculator

Do:

$\sqrt{360}$360$18.97$18.97

 

Question 3

For the value $10\sqrt{145}$10145:

  1. Write in simplest exact form.

  2. Evaluate to 2 decimal places.

 

Question 4

Between which two consecutive integers does $\sqrt{38}$38 lie?

  1. $\editable{}<\sqrt{38}<\editable{}$<38<

 

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