Simplifying the Radical Expression

The question asks us to find the value of the radical expression \(\sqrt{9-4\sqrt{5}}\).

To simplify this, we look for a way to express the term inside the square root, \(9-4\sqrt{5}\), as a perfect square. A perfect square often looks like \((a-b)^2 = a^2 + b^2 - 2ab\) or \((a+b)^2 = a^2 + b^2 + 2ab\).

Let's compare \(9-4\sqrt{5}\) with the form \(a^2 + b^2 - 2ab\).

• We need to match the term \(4\sqrt{5}\) with \(2ab\).
• We can rewrite \(4\sqrt{5}\) as \(2 \times 2 \times \sqrt{5}\).
• This suggests that maybe \(a=2\) and \(b=\sqrt{5}\) (or vice versa).
• Now, let's check if \(a^2 + b^2\) equals 9.
• If \(a=2\), then \(a^2 = 2^2 = 4\).
• If \(b=\sqrt{5}\), then \(b^2 = (\sqrt{5})^2 = 5\).
• Adding these squares: \(a^2 + b^2 = 4 + 5 = 9\).

This matches the constant term in our expression! So, we can rewrite \(9-4\sqrt{5}\) using \(a=2\) and \(b=\sqrt{5}\):

\(9 - 4\sqrt{5} = (2^2 + (\sqrt{5})^2) - (2 \times 2 \times \sqrt{5})\)

\(9 - 4\sqrt{5} = (\sqrt{5})^2 + 2^2 - 2(\sqrt{5})(2)\)

This is exactly the form \((a-b)^2\), where \(a=\sqrt{5}\) and \(b=2\).

So, \(9 - 4\sqrt{5} = (\sqrt{5} - 2)^2\).

Evaluating the Square Root

Now we can substitute this back into the original expression:

\(\sqrt{9-4\sqrt{5}} = \sqrt{(\sqrt{5} - 2)^2}\)

Remember that \(\sqrt{x^2} = |x|\) (the absolute value of x). Therefore:

\(\sqrt{(\sqrt{5} - 2)^2} = |\sqrt{5} - 2|\)

To find the absolute value, we need to know if \(\sqrt{5} - 2\) is positive or negative.

• We know that \((\sqrt{5})^2 = 5\) and \(2^2 = 4\).
• Since 5 is greater than 4, \(\sqrt{5}\) must be greater than \(\sqrt{4}\), which is 2.
• So, \(\sqrt{5} > 2\), which means \(\sqrt{5} - 2\) is a positive number.

Since \(\sqrt{5} - 2\) is positive, its absolute value is just the number itself:

\(|\sqrt{5} - 2| = \sqrt{5} - 2\)

Thus, the simplified value of the expression \(\sqrt{9-4\sqrt{5}}\) is \(\sqrt{5} - 2\).

This matches Option 2.