Simplifying Nested Radicals: Square Root of $(8-4\sqrt{3})$

This problem asks us to find the square root of the expression $(8-4\sqrt{3})$. This type of expression, involving a square root inside another square root, is called a nested radical. We can simplify it by rewriting the expression inside the square root in the form of a perfect square, specifically $(a-b)^2 = a^2 - 2ab + b^2$.

Step 1: Rewrite the Expression

Our expression is $\sqrt{8-4\sqrt{3}}$. We want to express the term $4\sqrt{3}$ in the form $2ab$. Let's rewrite $4\sqrt{3}$ as $2 \times (2\sqrt{3})$. So, the expression becomes $\sqrt{8 - 2 \times 2\sqrt{3}}$. Now, we want this to match the form $\sqrt{a^2 + b^2 - 2ab}$. We already have $2ab = 2 \times 2\sqrt{3}$. This suggests that $a$ and $b$ might be related to $2$ and $\sqrt{3}$.

Step 2: Identify Components for $(a-b)^2$

We are looking for two numbers, let's call them $x$ and $y$, such that the expression inside the square root, $8 - 4\sqrt{3}$, can be written as $(x - y)^2$. We know that $(x-y)^2 = x^2 + y^2 - 2xy$. Comparing $8 - 4\sqrt{3}$ with $x^2 + y^2 - 2xy$, we need:

• $x^2 + y^2 = 8$
• $2xy = 4\sqrt{3}$

From $2xy = 4\sqrt{3}$, we get $xy = 2\sqrt{3}$. Now we need to find two numbers whose sum of squares is 8 and whose product is $2\sqrt{3}$. Let's consider the components of $2\sqrt{3}$. We can think of this product as coming from $x = \sqrt{6}$ and $y = \sqrt{2}$, or $x = 2$ and $y = \sqrt{3}$. Let's test these pairs.

Step 3: Test Potential Values for $x$ and $y$

Test 1: $x=2$ and $y=\sqrt{3}$

• Check product: $xy = 2 \times \sqrt{3} = 2\sqrt{3}$. This matches.
• Check sum of squares: $x^2 + y^2 = 2^2 + (\sqrt{3})^2 = 4 + 3 = 7$. This does NOT match 8.

Test 2: $x=\sqrt{6}$ and $y=\sqrt{2}$

• Check product: $xy = \sqrt{6} \times \sqrt{2} = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$. This matches.
• Check sum of squares: $x^2 + y^2 = (\sqrt{6})^2 + (\sqrt{2})^2 = 6 + 2 = 8$. This matches!

So, we have found our $x$ and $y$: $x = \sqrt{6}$ and $y = \sqrt{2}$.

Step 4: Rewrite the Expression as a Perfect Square

Since $x = \sqrt{6}$ and $y = \sqrt{2}$, we can rewrite $8 - 4\sqrt{3}$ as $x^2 + y^2 - 2xy$, which is $(x-y)^2$. Therefore, $8 - 4\sqrt{3} = (\sqrt{6} - \sqrt{2})^2$.

Step 5: Calculate the Square Root

Now we can find the square root:

$\sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2}$

The square root of a squared term is the absolute value of the term: $\sqrt{A^2} = |A|$.

So, $\sqrt{(\sqrt{6} - \sqrt{2})^2} = |\sqrt{6} - \sqrt{2}|$.

Since $\sqrt{6}$ is approximately $2.45$ and $\sqrt{2}$ is approximately $1.41$, $\sqrt{6}$ is greater than $\sqrt{2}$. Therefore, $\sqrt{6} - \sqrt{2}$ is positive.

Thus, $|\sqrt{6} - \sqrt{2}| = \sqrt{6} - \sqrt{2}$.

Conclusion

The square root of $(8-4\sqrt{3})$ is $(\sqrt{6}-\sqrt{2})$.

Verification (Optional)

Let's square the result $(\sqrt{6}-\sqrt{2})$ to confirm:

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

$= 6 - 2\sqrt{12} + 2$

$= 8 - 2\sqrt{4 \times 3}$

$= 8 - 2(2\sqrt{3})$

$= 8 - 4\sqrt{3}$

This matches the original expression, confirming our answer.