Nature of roots of a quadratic equation with irrational co-efficients

Question

What would be the nature of the roots of the equation $$2x^2 - 2\sqrt{6} x + 3 = 0$$

My book says that as the discriminant is 0 so the roots are rational and equal. But discriminant can be used for determining the nature of roots only when the roots are rational numbers. Is the answer in the book wrong because actually the nature of roots should be irrational?

Answer 1

Rationality and multiplicity of roots are two separate questions. The discriminant tells you whether there are repeated roots. It offers no opinion on whether those repeated roots are rational.

Of course, for a polynomial whose roots are all rational the coefficients (divided by the leading coefficient) would all be rational as well.

Answer 2

The discriminant can always be used. Also, your suspicion that the roots are irrational does not conflict with the claim that they are real. Note that $$ 2x^2-2\sqrt 6x+3=(\sqrt 2 x-\sqrt 3)^2$$

Answer 3

If the discriminant of $ax^2+bx+c$ is $0$, there is a double root, i. e. the quadratic can be written as $\,a(x-\xi)^2\,$ for some real number $\xi$ (if the coefficients $a,b,c$ are real, of course). Furthermore, this double root is equal to: $$\xi=-\frac b{2a}.$$ So here the double root is actually irrational, equal to $\,\dfrac{\sqrt6}2$.