# How do you find a one-decimal place approximation for root3 99?

##### 1 Answer
Oct 21, 2015

$4.6$ and $4.7$.

#### Explanation:

If $\sqrt{99}$ is between $a$ and $b$, then the following must hold:

${a}^{3} < 99 < {b}^{3}$.

So, we must look for two numbers $a$ and $b$ with this property, and they must be less than $\frac{1}{10}$ apart.

First of all, let's focus on the nearest integer: we only need to make some calculation and go on with trials and errors: we'll begin listing the first cubes:

• ${1}^{3} = 1$;
• ${2}^{3} = 8$;
• ${3}^{3} = 27$;
• ${4}^{3} = 64$
• ${5}^{3} = 125$.

From this list, we deduce that the third root of $99$ is between $4$ and $5$.

Now we can simply list the numbers between $4$ and $5$ with one decimal digit, and compute their cubes (which may be boring, but we can easily do it without a calculator, so it's not "cheating"):

• $4 , {1}^{3} = 68.921$;
• $4 , {2}^{3} = 74.088$;
• $4 , {3}^{3} = 79.507$;
• $4 , {4}^{3} = 85.184$;
• ${4.5}^{3} = 91.125$;
• ${4.6}^{3} = 97.336$;
• ${4.7}^{3} = 103.823$.

There we go. The third root of $99$ is surely between $4.6$ and $4.7$.

Just to confirm our calculations, the calculator gave me back $4.626065009182741793092 \ldots$, so the answer is correct.