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?

Why do you think the discriminant can only be used for rational roots? — Daniel Fischer, Jun 15 '15 at 20:06
@DanielFischer I edited it.I had written real instead of rational — user456, Jun 15 '15 at 20:09
You still have the sentence "But discriminant can be used only when the roots are rational numbers." in your question. That is wrong, and I wonder what made you think that. — Daniel Fischer, Jun 15 '15 at 20:11
The discriminant doesn't generally tell you whether the roots are rational or irrational, like Robert Israel wrote in his answer. — Daniel Fischer, Jun 15 '15 at 20:14
Does your book definitely say "the roots are rational and equal"? If so, it is wrong. The discriminant detects equal roots whatever their nature. — Mark Bennet, Jun 15 '15 at 20:50

score 4 · Answer 1 · answered Jun 15 '15 at 20:05

4

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 $$ 2x^2-2\sqrt 6x+3=(\sqrt 2 x-\sqrt 3)^2$$

answered Jun 15 '15 at 20:05

Hagen von Eitzen

374,180

I edited the question Sir – user456 Jun 15 '15 at 20:10

score 3 · Accepted Answer · answered Jun 15 '15 at 20:11

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.

score 1 · Answer 3 · answered Jun 15 '15 at 20:21

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 $\,\dfrac{\sqrt6}2$.

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

3 Answers3