The quadratic equation whose one root is 2 - | vedclass

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Question

(B) Given that the first root is $\alpha = 2 - \sqrt{3}$.Since the coefficients are rational,the other root must be the conjugate,$\beta = 2 + \sqrt{3

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given that the first root is $\alpha = 2 - \sqrt{3}$.Since the coefficients are rational,the other root must be the conjugate,$\beta = 2 + \sqrt{3}$.The sum of the roots is $\alpha + \beta = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.The product of the roots is $\alpha \beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.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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