# Why does $\sqrt{3}x^2-4\sqrt{3}=0$ not follow discriminant rules?

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.

• 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. Feb 3 at 2:17
• @peterwhy My teacher just said the rule is if the discriminant is a perfect square then the roots are real and rational. Feb 3 at 2:21
• And you found a counterexample. Feb 3 at 2:22
• @peterwhy So it's not a universal rule? Feb 3 at 2:22
• Was your teacher perhaps talking about $ax^2+bx+c=0$ when $a=1$? Feb 3 at 2:31

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.

• That factoring of that radical is really smart for the explanation Feb 3 at 4:11

"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.

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.

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?