Unreal root of quadratic equation

Question

Set a positive real number such that

$$a^3=6(a+1)$$

Prove that the equation

$$x^2 + ax + a^2 -6 = 0$$ there is no real solution

Solution attempt:

condition : $$a^2 - 4a^2 + 24 <0$$ $$a^2>8$$ $$a>2\sqrt{2}$$ or $$a<-2\sqrt{2}$$

By inspection

(a = 1) -> 1<12

(a = 2) -> 8<18

(a = 3) -> 27 > 24

So, 2 < a < 3

for a = 2sqrt2 , have 16sqrt2 < 6(2sqrt2 + 1)

The question would be resolved if the previous inequality were contradicted, but it didn't work. How can I proceed now?

Remembering that it is a matter of high school, more specifically of a Balkan olympics / 2007

Answer 1

We can use here the Cardano's formula ($\Delta=1$).

Show that $a=\sqrt[3]2+\sqrt[3]4$ and $\left(\sqrt[3]2+\sqrt[3]4\right)^2>8.$

Actually, we can use the following identity: $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$

Answer 2

I think you have already solved it, it just remains to reason the result. Let $f(a)=a^3-6(a+1)$. Since $f(2\sqrt{2})<0$ and $f(3)>0$, it means that $a>2\sqrt{2}$, which implies, as you have shown, the discriminant $a^2-4a+24<0$ so the equation has no real solution.

Answer 3

The discriminant $\Delta$ of the polynomial $x^2+ax+(a^2-6)$ is $24-3a^2$. If $\Delta<0$, then the polynomial does not have any real roots. So we need to show that $a>2\sqrt{2}$ or $a<-2\sqrt{2}$. Since $a^3=6(a+1)$, we have $a^3-6a-6=0$. This cubic equation has a single positive real root (exercise). Moreover, $(2\sqrt{2})^3-6(2\sqrt{2})-6<0$, and so $a$ must satisfy $a>2\sqrt{2}$. Therefore, $\Delta<0$, and the result follows.

Answer 4

Just to give a different approach, note that $a^3=6(a+1)$ implies $a^2=6(1+{1\over a})$, which turns $x^2+ax+a^2-6$ into $x^2+ax+{6\over a}$, hence

$$\begin{align} x^2+ax+a^2-6\text{ has no real solution}&\iff x^2+ax+{6\over a}\text{ has no real solution}\\ &\iff a^2-{24\over a}\lt0\\ &\iff a^3\lt24\\ &\iff6(a+1)\lt24\\ &\iff a\lt3 \end{align}$$

But it's easy to see that if $a\ge3$ then we have a contradiction:

$$0=a^3-6a-6=a(a^2-6)-6\ge3(9-6)-6=3\gt0$$

Thus the inequality $a\lt3$ holds, which gives us the desired result.

The virtue here (if any) is that we can avoid fussing around with any roots of $2$.