A quadratic equation with rational coefficients one of whose roots is $tan ∣ ∣ ∣ \frac{π}{12} ∣ ∣ ∣$ is

x^{2} - 2 x + 1 = 0

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x^{2} - 2 x + 4 = 0

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x^{2} - 4 x + 1 = 0

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x^{2} - 4 x - 1 = 0

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Solution

The correct option is B $x^{2} - 4 x + 1 = 0$
Given one root is $tan (\frac{π}{12}) .$
We have $tan (\frac{π}{12}) = 2 - \sqrt{3}$ .
So, the other root is $2 + \sqrt{3}$ .
Now, sum of roots $= 2 + \sqrt{3} + 2 - \sqrt{3} = 4$ and product of roots $(2 + \sqrt{3}) (2 - \sqrt{3}) = 1$ .
So, the quadratic equation will be $x^{2} - 4 x + 1 = 0$ .

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