# How do you find the principal square root of 8?

Mar 26, 2015

For a non-negative real number, $n$ (like $8$) the principal square root is the non-negative solution to ${x}^{2} = n$

The symbol $\sqrt{n}$ is used for the principal square root of $n$.

Thw pricipal square root of $8$, denoted $\sqrt{8}$ is the number whose sguare is $8$. There is no easy method for finding this number. We use successive approximation (or other iterative techniques) to get increasingly accurate approximations.

We do, however write $\sqrt{8}$ in "simpler" form $2 \sqrt{2}$

$\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$. Perhaps this is what you meant by "find"?

Mar 26, 2015

If, by "find" you mean get a decimal approximation, you could use: start with a number you know is close ${3}^{2} = 9$ and ${2}^{2} = 4$, so we'll start with $3$

divide $8$ by your last estimate: $8 \div 3 = 2.6667$ (to 4 decimal places)
Average your previous estimate and the quotient:
$\frac{3 + 2.6667}{2} = 2.8334$ (rounding)

Repeat:
Divide: $8 \div 2.8334 = 2.8235$
Average: $\frac{2.8334 + 2.8235}{2} = 2.8284$

Repeat
Divide: $8 \div 2.8284 = 2.8284$
Average $= 2.8284$

$\sqrt{8} \approx 2.8284$ this estimate is accurate to 4 decimal places.
If you need more accuracy, start again and keep more places when rounding.