Does there exist a quadratic equation whose c | vedclass

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(C) Yes,such a quadratic equation exists.Consider the quadratic equation $\sqrt{3}x^2 - 7\sqrt{3}x + 12\sqrt{3} = 0$.Here,the coefficients are $a = \s

Vedclass · Accepted Answer

Yes,for example,$\sqrt{3}x^2 - 7\sqrt{3}x + 12\sqrt{3} = 0$.

Vedclass · Answer

(C) Yes,such a quadratic equation exists.Consider the quadratic equation $\sqrt{3}x^2 - 7\sqrt{3}x + 12\sqrt{3} = 0$.Here,the coefficients are $a = \sqrt{3}$,$b = -7\sqrt{3}$,and $c = 12\sqrt{3}$,which are all distinct irrational numbers.Dividing the entire equation by $\sqrt{3}$,we get $x^2 - 7x + 12 = 0$.Factoring the quadratic,we get $(x - 3)(x - 4) = 0$.Thus,the roots are $x = 3$ and $x = 4$,which are both rational numbers.

Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

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