Approximating Radicals

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