# Approximating Radicals

Need some basic understanding of radicals first? Practice with 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.

Solution: We need a nonnegative number which, when squared, gives $\,7\,.$

$2^2 = 4\,$ ($\,2\,$ is too small)

$3^2 = 9\,$ ($\,3\,$ is too big)

Thus,
$\,\sqrt{7}\,$ lies between $\,2\,$ and $\,3\,.$

Solution: We need a number which, when cubed, gives $\,29\,.$

$3^3 = 27\,$ ($\,3\,$ is a bit too small)

$4^3 = 64\,$ ($\,4\,$ is too big)

Thus,
$\,\root 3\of{29}\,$ lies between $\,3\,$ and $\,4\,.$

Solution: We need a number which, when cubed, gives $\,-12\,.$ The answer will be negative. Since it's easier to work with positive numbers, we first investigate $\,\root 3\of{12}\,.$

$2^3 = 8\,$ ($\,2\,$ is too small)

$3^3 = 27\,$ ($\,3\,$ is too big)

Thus,
$\,\root 3\of{12}\,$ lies between $\,2\,$ and $\,3\,,$
and
$\,\root 3\of{-12}\,$ lies between $\,-3\,$ and $\,-2\,.$

For this web exercise, you *must* list the integers from least (farthest left)
to greatest (farthest right).

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 $\,\sqrt{130}\,$ 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,
$\,\sqrt{130}\,$ 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, $\,\sqrt{130}\approx 11.4\,.$