How do you simplify sqrt(68)?

May 23, 2016

$\sqrt{68} = 2 \sqrt{17}$

Explanation:

If $a , b \ge 0$ then:

$\sqrt{a b} = \sqrt{a} \sqrt{b}$

The prime factorisation of $68$ is:

$68 = 2 \times 2 \times 17$

So we have:

$\sqrt{68} = \sqrt{{2}^{2} \cdot 17} = \sqrt{{2}^{2}} \cdot \sqrt{17} = 2 \sqrt{17}$

$\sqrt{17}$ cannot be simplified any further, but since $17 = {4}^{2} + 1$ is of the form ${n}^{2} + 1$ it has a very simple continued fraction expansion:

sqrt(17) = [4;bar(8)] = 4+1/(8+1/(8+1/(8+1/(8+1/(8+1/(8+...))))))

You can terminate this expansion at any point to give a rational approximation. For example:

sqrt(17) ~~ [4;8,8] = 4+1/(8+1/8) = 268/65 ~~ 4.1231