To solve problems involving expressions like (\sqrt{a \pm \sqrt{b}}), the key is to rewrite the inner term as a square of a sum/difference of square roots. Here's the step-by-step approach:
General Method
We aim to express (\sqrt{a \pm \sqrt{b}}) as (\sqrt{m} \pm \sqrt{n}) (where (m > n > 0)).
Expanding ((\sqrt{m} \pm \sqrt{n})^2 = m + n \pm 2\sqrt{mn}), we equate to (a \pm \sqrt{b}):
- (m + n = a)
- (2\sqrt{mn} = \sqrt{b} \implies 4mn = b)
Solve for (m) and (n) using the quadratic equation (t^2 - a t + \frac{b}{4} = 0).
Example Problem
Suppose the problem is to compute (\sqrt{7 + 4\sqrt{3}} + \sqrt{7 - 4\sqrt{3}}):
-
For (\sqrt{7 + 4\sqrt{3}}):
(a = 7), (b = (4\sqrt{3})^2 = 48).
(mn = \frac{48}{4} = 12), (m + n =7).
The roots are (m=4), (n=3), so (\sqrt{7+4\sqrt{3}} = \sqrt{4} + \sqrt{3} = 2+\sqrt{3}). -
For (\sqrt{7 -4\sqrt{3}}):
(\sqrt{7-4\sqrt{3}} = 2 - \sqrt{3}).
Sum: ((2+\sqrt{3}) + (2-\sqrt{3}) = 4).
Answer: (\boxed{4}) (assuming this is the problem's context).
If the problem differs, apply the same method to simplify the expression!
(\boxed{4})
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