Question

Does there exist a quadratic equation whose coefficients are all distinct irrational but both the roots are rational? why?

Answer 1

yes it exists as the rule of quadratic equation does not break. it says that the root should be rational

Answer 2

Answer:

Yes, consider the quadratic with all distinct irrational coefficients

$i.e \: \sqrt{3 {x}^{2} } - 7 \sqrt{3x} + 12 \sqrt{3} = 0.$

The roots of this quadratic equation are 3 and 4, which are rationals.