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If x=sqrt(1+sqrt(3)/(2))-sqrt(1-sqrt(3)/...

If `x=sqrt(1+sqrt(3)/(2))-sqrt(1-sqrt(3)/(2))`, then the value of `(sqrt(2)-x)/(sqrt(2)+x)` will be closet to :

A

1.4

B

1.2

C

0.17

D

0.12

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we start with the given expression for \( x \): \[ x = \sqrt{1 + \frac{\sqrt{3}}{2}} - \sqrt{1 - \frac{\sqrt{3}}{2}} \] ### Step 1: Simplify \( x \) First, we simplify the terms inside the square roots. 1. Calculate \( 1 + \frac{\sqrt{3}}{2} \): \[ 1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \] 2. Calculate \( 1 - \frac{\sqrt{3}}{2} \): \[ 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \] Now, substituting these back into the expression for \( x \): \[ x = \sqrt{\frac{2 + \sqrt{3}}{2}} - \sqrt{\frac{2 - \sqrt{3}}{2}} \] ### Step 2: Factor out the square root of \( \frac{1}{2} \) We can factor out \( \sqrt{\frac{1}{2}} \) from both terms: \[ x = \sqrt{\frac{1}{2}} \left( \sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}} \right) \] ### Step 3: Calculate \( \sqrt{2 + \sqrt{3}} \) and \( \sqrt{2 - \sqrt{3}} \) To find \( \sqrt{2 + \sqrt{3}} \) and \( \sqrt{2 - \sqrt{3}} \), we can approximate their values: 1. \( \sqrt{2 + \sqrt{3}} \approx \sqrt{2 + 1.732} \approx \sqrt{3.732} \approx 1.93 \) 2. \( \sqrt{2 - \sqrt{3}} \approx \sqrt{2 - 1.732} \approx \sqrt{0.268} \approx 0.52 \) ### Step 4: Substitute back to find \( x \) Now substituting these approximations back into the expression for \( x \): \[ x \approx \sqrt{\frac{1}{2}} (1.93 - 0.52) = \sqrt{\frac{1}{2}} (1.41) \approx 1.41 \times 0.707 \approx 1.00 \] ### Step 5: Calculate \( \frac{\sqrt{2} - x}{\sqrt{2} + x} \) Now we need to find the value of: \[ \frac{\sqrt{2} - x}{\sqrt{2} + x} \] Using \( \sqrt{2} \approx 1.41 \) and \( x \approx 1.00 \): \[ \frac{1.41 - 1.00}{1.41 + 1.00} = \frac{0.41}{2.41} \approx 0.17 \] ### Final Answer The value of \( \frac{\sqrt{2} - x}{\sqrt{2} + x} \) is approximately \( 0.17 \).
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