Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
3. Real Numbers and the Number Line
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Exercise 63 Page 21

The square root of 136 can be approximated by comparing it to two nearby perfect squares that are less than 136 and greater than 136.

≈ 12 feet

Practice makes perfect
We have a square patio with an area of 136ft^2, and are asked to approximate the side length to the nearest foot. First, we need to remember that the area of a square is found by squaring the side length s. A=s^2 By substituting the area into the formula we can find the side length. Notice that a length cannot be negative, so we only consider positive solutions.
A=s^2
136=s^2
s^2=136
s=± sqrt(136)

s > 0

s=sqrt(136)
Since 136 is not a perfect square, and we are not asked for an exact answer, we can find an approximation by comparing 136 to the nearest perfect squares. We will use perfect squares less than 136 and greater than 136. Let's try 121 and 144.
sqrt(121)
11
Now we know that the side length must be some number between 11 and 12. To decide which value is our answer, we need to think about the relationship between 121, 136, and 144. 121<13_(15)6<144_8 The difference between 121 and 136 is 15, while the difference between 136 and 144 is 8. Therefore, 136 is closer to 144, and consequently, sqrt(136) is closer to sqrt(144). We conclude that the side length of the patio is approximately 12 feet.