APPROXIMATING RADICALS

Here, you will practice with radicals that don't come out ‘nicely’.

EXAMPLES:

Find the two closest integers between which the given radical lies.
Do not use the square root or cube root keys on a calculator.

Question: The number [beautiful math coming... please be patient] $\,\sqrt{7}\,$ lies between   ?   and   ?  
Solution: We need a nonnegative number which, when squared, gives $\,7\,$.
[beautiful math coming... please be patient] $2^2 = 4\,$   ($\,2\,$ is too small)
$3^2 = 9\,$   ($\,3\,$ is too big)
Thus, [beautiful math coming... please be patient] $\,\sqrt{7}\,$ lies between $\,2\,$ and $\,3\,$.
Question: The number [beautiful math coming... please be patient] $\,\root 3\of{29}\,$ lies between   ?   and   ?  
Solution: We need a number which, when cubed, gives [beautiful math coming... please be patient] $\,29\,$.
$3^3 = 27\,$   ($\,3\,$ is a bit too small)
$4^3 = 64\,$   ($\,4\,$ is too big)
Thus, [beautiful math coming... please be patient] $\,\root 3\of{29}\,$ lies between $\,3\,$ and $\,4\,$.
Question: The number [beautiful math coming... please be patient] $\,\root 3\of{-12}\,$ lies between   ?   and   ?  
Solution: We need a number which, when cubed, gives [beautiful math coming... please be patient] $\,-12\,$.
The answer will be negative.
Since it's easier to work with positive numbers, we first investigate [beautiful math coming... please be patient] $\,\root 3\of{12}\,$.
$2^3 = 8\,$   ($\,2\,$ is too small)
$3^3 = 27\,$   ($\,3\,$ is too big)
Thus, [beautiful math coming... please be patient] $\,\root 3\of{12}\,$ lies between $\,2\,$ and $\,3\,$,
and [beautiful math coming... please be patient] $\,\root 3\of{-12}\,$ lies between $\,-3\,$ and $\,-2\,$.
For this web exercise, you MUST must list the integers from least (farthest left) to greatest (farthest right).
EXAMPLE:
Question: Estimate [beautiful math coming... please be patient] $\,\sqrt{130}\,$ to the nearest tenth.
Use a non-calculator approach.
Solution:
To round to the tenths place, we must know if the digit in the hundredths place is $\,5\,$ or greater, or less than $\,5\,$.
As above, first determine that [beautiful math coming... please be patient] $\,\sqrt{130}\,$ is between $\,11\,$ and $\,12\,$.
Then, use long multiplication:
$\,11.5^2 = 132.25\,$, so $\,11.5\,$ is a bit too big
$\,11.4^2 = 129.96\,$, so $\,11.4\,$ is a bit too small
Thus, [beautiful math coming... please be patient] $\,\sqrt{130}\,$ lies between $\,11.4\,$ and $\,11.5\,$.
Again using long multiplication,
$\,11.45^2 = 131.1025\,$, so $\,11.45\,$ is a bit too big.
Thus, the digit in the hundredths place must be less than $\,5\,$, and so the square root is closer to $\,11.4\,$.
Thus, [beautiful math coming... please be patient] $\,\sqrt{130}\approx 11.4\,$.
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Renaming Square Roots

 
 
The number
lies between
and
    
(an even number, please)

 
 
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5
AVAILABLE MASTERED IN PROGRESS

(MAX is 5; there are 5 different problem types.)