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From what I understand:

$D > 0$ and a perfect square $\Longrightarrow$ Real and Rational Roots

$D > 0$ but not a perfect square $\Longrightarrow$ Real and Irrational Roots

$D = 0$ $\Longrightarrow$ Double Root

$D < 0$ $\Longrightarrow$ Complex Roots

My question is that the discriminant of $\sqrt{3}x^2-4\sqrt{3}$ $\Longrightarrow$ 48

48 is not a perfect square yet the roots for $\sqrt{3}x^2-4\sqrt{3}=0$ are real and rational.

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  • $\begingroup$ Not sure about the "perfect square" and rational claim. By the quadratic formula (here with $b=0$): $x = \dfrac{\pm\sqrt{-4ac}}{2a}$, and both the numerator and denominator are irrational. Then the roots happen to be rational this time, but no guarantee. $\endgroup$
    – peterwhy
    Feb 3 at 2:17
  • $\begingroup$ @peterwhy My teacher just said the rule is if the discriminant is a perfect square then the roots are real and rational. $\endgroup$
    – Ox Ph
    Feb 3 at 2:21
  • $\begingroup$ And you found a counterexample. $\endgroup$
    – peterwhy
    Feb 3 at 2:22
  • $\begingroup$ @peterwhy So it's not a universal rule? $\endgroup$
    – Ox Ph
    Feb 3 at 2:22
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    $\begingroup$ Was your teacher perhaps talking about $ax^2+bx+c=0$ when $a=1$? $\endgroup$ Feb 3 at 2:31

4 Answers 4

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Notice how $$\sqrt{3}x^2- 4\sqrt{3}=\sqrt{3} (x^2-4)$$ As such the roots of the polynomial (i.e. the solutions to $\sqrt{3}x^2- 4\sqrt{3}=0$) have to be the same as $x^2-4=0$

If you want to find a rule for the solution to be rational given a quadratic equation $$ ax^2+bx+c=0 $$ is that $ \frac{D}{4a^2}$ is a perfect square, $\frac{b}{2a}$ is rational and $D\ge0$. Indeed the roots of a second degree polynomials are $$ x_{1,2}=-\frac{b}{2a} \pm \sqrt{\frac{D}{4a^2}} $$ As you can prove from the usual formula

I want to clarify that this is a sufficient condition, but is not necessary, as e.g $$x^2+(1+\sqrt{2}) x + \frac{\sqrt{2}}{2}+\frac{1}{4}=0$$ does not satisfy this condition (you can check that $D=2$), but still has a rational root.

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    $\begingroup$ That factoring of that radical is really smart for the explanation $\endgroup$
    – imranfat
    Feb 3 at 4:11
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"My teacher just said the rule is if the discriminant is a perfect square then the roots are real and rational"

That rule is for polynomials with rational coefficients.

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You really cant comment on roots being rational or irrational for D>0 .

roots are given by -b +- sqrt(b^2-4ac) /2a

Even if the D is perf square roots being rational/irrational depend on b and a

Like in your question the a=sqrt3 so the sqrt3 in D is cancelled by the a=sqrt3.

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Let $f(x) = ax^{2} + bx + c$ whose coefficients are real numbers, $x\in\mathbb{R}$ and $a\neq 0$.

Then we can rewrite it as follows: \begin{align*} f(x) & = ax^{2} + bx + c\\\\ & = a\left(x^{2} + \frac{bx}{a} + \frac{c}{a}\right)\\\\ & = a\left[\left(x^{2} + \frac{bx}{a} + \frac{b^{2}}{4a^{2}}\right) - \left(\frac{b^{2} - 4ac}{4a^{2}}\right)\right]\\\\ & = a\left(x + \frac{b}{2a}\right)^{2} - \frac{\Delta}{4a} \end{align*}

Hence we split the behavior of $f$ into three cases:

  • If $\Delta > 0$, then $f$ has two (distinct) real roots.
  • If $\Delta = 0$, then $f$ has two (equal) real roots.
  • If $\Delta < 0$, then $f$ has no real roots.

As you have noticed, $\Delta = 48$ when $f(x) = \sqrt{3}x^{2} - 4\sqrt{3}$.

Then we may conclude that $f$ has two distinct real roots.

Can you take it from here?

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