The estimates of the <em>square roots </em>and <em>cube roots using the attached process</em> are;
<u><em>1) √15 = 3.81</em></u>
<u><em>2) √44 = 6.6.</em></u>
<u><em>3) √81 = 9</em></u>
<u><em>4) ∛12 = 2.88.</em></u>
<u><em>5) ∛125 = 5</em></u>
<u><em>6) ∛320 = 6..8</em></u>
- The process we are to follow is shown in the attachment.
- The first thing we are to do is to list all the perfect squares from 1 to 100 and then do the same for all perfect cubes from 1 to 1000.
Perfect squares from 1 to 100; 4, 9, 16, 25, 36, 49, 64, 81, 100
Perfect cubes from 1 to 1000; 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000.
1) √15;
The nearest perfect squares to it are; 9 and 16
Thus; √9 < √15 < √16
⇒ 3 < √15 < 4
This means that √15 is between 3 and 4.
Since 15 is closer to 16 than 9, it means √15 is closer to 4 and thus, we can estimate √15 as 3.81.
2) √44;
The nearest perfect squares to it are; 36 and 49
Thus; √36 < √44 < √49
⇒ 6 < √44 < 7
This means that √44 is between 6 and 7.
Since 44 is closer to 16 than 9, it means √44 is closer to 6 and thus, we can estimate √44 as 6.6.
3) √81
This is a perfect square and the method used to estimate other non-perfect square doesn't apply here. Thus; √81 = 9
4) ∛12
The nearest perfect cubes to it are; 8 and 27
Thus; ∛8 < ∛12 < ∛27
⇒ 2 < ∛12 < 3
This means that √12 is between 2 and 3.
Since 12 is closer to 8 than 27, it means ∛12 is closer to 2 than 3 and thus, we can estimate ∛12 as 2.88.
5) ∛125
This is a perfect cube and the method used to estimate other non-perfect cubes doesn't apply here. Thus; ∛125 = 5
6) ∛320
The nearest perfect cubes to it are; 216 and 343
Thus; ∛216 < ∛320 < ∛343
⇒ 6 < ∛320 < 7
This means that ∛320 is between 6 and 7.
Since 320 is closer to 343 than 216, it means ∛320 is closer to 7 than 6 and thus, we can estimate ∛320 as 6..8
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