In this post, we will learn to simplify the expression \(\sqrt{3 + 2\sqrt{2}}\, – \sqrt{3\, – 2\sqrt{2}}\). There are multiple approaches to simplify this expression. Here, I will give you one of those approaches.
Let \( x = \sqrt{3 + 2\sqrt{2}}\, – \sqrt{3\, – 2\sqrt{2}} \)
We square both sides to eliminate the square roots:
\( x^2 = \left( \sqrt{3 + 2\sqrt{2}}\, – \sqrt{3\, – 2\sqrt{2}} \right)^2 \)
Using the formula \((a – b)^2 = a^2 – 2ab + b^2\):
\( x^2 = \left( \sqrt{3 + 2\sqrt{2}} \right)^2 – 2 \cdot \sqrt{3 + 2\sqrt{2}} \cdot \sqrt{3\, – 2\sqrt{2}} + \left( \sqrt{3\, – 2\sqrt{2}} \right)^2 \\ \Rightarrow x^2 = (3 + 2\sqrt{2})\, – 2 \cdot \sqrt{(3 + 2\sqrt{2})(3\, – 2\sqrt{2})} + (3\, – 2\sqrt{2}) \dots (1) \\ \)Using the formula \((a + b)(a-b) = a^2 – b^2\):
\( (3 + 2\sqrt{2})(3\, – 2\sqrt{2}) \\ = 3^2 – (2\sqrt{2})^2 \\ = 9\, – 8 = 1 \\ \)Thus: \( \sqrt{(3 + 2\sqrt{2})(3\, – 2\sqrt{2})} = \sqrt{1} = 1 \)
Now, substitute back into the equation for \( x^2 \) in (1):
\( x^2 = (3 + 2\sqrt{2})\, – 2 \cdot 1 + (3\, – 2\sqrt{2}) \\ \Rightarrow x^2 = 6\, – 2 = 4 \\ \Rightarrow \quad x = \pm 2 \\ \)But the original expression is positive, so the result should be positive.
Therefore, \( \fbox{x = 2} \)
Please let me know in the comments if you find any error in this solution.