Simplify $\sqrt{5-2\sqrt{6}}$

Question

So the given expression $\sqrt{5-2\sqrt{6}}$ has the same answer as $\sqrt{3}-\sqrt{2}$ Which is equal to $0.31783724...$

How do I simplify the 1st equation into the 2nd equation?

Answer 1

Here's another method that works to find the square roots of $\alpha \pm \beta \sqrt{\gamma}$.

Let

(i)$$\delta = \sqrt{\alpha^2 - \beta^2 \gamma}$$ (ii)$$\sigma = \sqrt{\frac{\alpha + \delta}{2}} \quad \text{and} \quad \tau = \sqrt{\frac{\alpha - \delta}{2}}$$

Then

$$\sqrt{\alpha \pm \beta \sqrt{\gamma}} = \sigma \pm \tau$$

For your example, to find the square root of $5 - 2\sqrt{6}$:

(i)$$\delta = \sqrt{5^2 - (-2)^2 (6)} = 1$$

(ii)$$\sigma = \sqrt{\frac{5 + 1}{2}} = \sqrt{3} \quad \text{and} \quad \tau = \sqrt{\frac{5 - 1}{2}} = \sqrt{2}$$

So

$$\sqrt{5 - 2\sqrt{6}} = \sqrt{3} - \sqrt{2}$$

Answer 2

The answer has been given in the comments, but I'll add an explanation just in case you didn't see where it came from.

When you see $\sqrt{5-2\sqrt{6}}$, your goal should be to find what $5-2\sqrt{6}$ is the square of, in which case you can get rid of the square root surrounding it.

We know that $(a-b)^2=a^2-2ab+b^2$, so we can capitalize on the $-2$ in front of the $\sqrt{6}$ and say that $ab=\sqrt{6}$. We can consequently deduce that $a^2+b^2=5$. From there, we can see that $a=\sqrt{3}$ and $b=\sqrt{2}$ will give us what we're looking for. This means that

$$5-2\sqrt{6}=(\sqrt{3}-\sqrt{2})^2$$

and, as a result:

$$\sqrt{5-2\sqrt{6}}=\sqrt{(\sqrt{3}-\sqrt{2})^2} =\sqrt{3}-\sqrt{2}$$