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(B) Given that the first root is $\alpha = 2 - \sqrt{3}$.Since the coefficients of the quadratic equation are assumed to be rational,the irrational ro

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$x^2 - 4x + 1 = 0$

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(B) Given that the first root is $\alpha = 2 - \sqrt{3}$.Since the coefficients of the quadratic equation are assumed to be rational,the irrational roots must occur in conjugate pairs.Therefore,the second root is $\beta = 2 + \sqrt{3}$.The sum of the roots is $\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.The product of the roots is $\alpha \cdot \beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.The standard form of a quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - 4x + 1 = 0$.

The quadratic equation whose one root is $2 - \sqrt{3}$ will be

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