If $a^{3}=6(a+1)$ then $a>2.83$. $x^{2}+ax+a^{2}-6=0$ has no roots if $a>\sqrt{8}$.

The determinant of the quadratic is $a^{2}-4(a^{2}-6)=3(8-a^{2})$ which must be $\ge 0$ if the solutions are real. but then we must have $a\le 2\sqrt 2$. so we have $6=a(a^{2}-6)\le 2\sqrt 2(8-6)=4\sqrt 2$, contradiction.

We can rewrite the given quadratic equation as $x^{2}+ax+\frac{6}{a}=0$. $D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}$.

kunny wrote: We can rewrite the given quadratic equation as $x^{2}+ax+\frac{6}{a}=0$. $D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}$. Why are we done here?

We can also rewrite the given equation as $ax(x+a)=-6.$ This doesn't 0 or positive roots, now if this has negative roots, .....

kunny wrote: We can also rewrite the given equation as $ax(x+a)=-6.$ This doesn't 0 or positive roots, now if this has negative roots, ..... Shouldn't it be $ax(a+x)=-1$? And I still don't see your point...

No, you are wrong.

behemont wrote: kunny wrote: We can rewrite the given quadratic equation as $x^{2}+ax+\frac{6}{a}=0$. $D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}$. Why are we done here? well, from what i understood, since $D\ge 0$, we have $a\ge 3$. but this implies $a^{3}>6(a+1)$.

kunny wrote: No, you are wrong. God bless my multiplying.. $6*1=1$

Everyone make a mistake.

Suppose that $\Delta=a^{2}-4(a^{2}-6)=24-a^{2}\geq0$. Then $8\geq a^{2}$. If $a=0$ then $0=6$- false. If $a>0$ then $a^{2}=6+\frac{6}{a}\leq 8$ $\Rightarrow \frac{6}{a}\leq 2$ $\Rightarrow$ $3\leq a$ - false. If $a<0$ then $a=-x$. $\Rightarrow$ $x>0$ and $x^{2}\leq 8$. So: $0<x<2\sqrt{2}$ and $6x=x^{3}+6$. $x=\frac{x^{3}}{6}+1=\frac{x^{3}}{6}+\frac{1}{6}+\frac{1}{6}+\frac{4}{6}\geq \frac{3x}{6}+\frac{2}{3}$ $\Rightarrow$ $x\geq \frac{4}{3}$ $6x^{2}=x^{4}+6x$ $\Rightarrow$ $9-6x=(x^{2}-3)^{2}\geq0$ $\Rightarrow$ $x\leq \frac{3}{2}$ $x=1+\frac{x^{3}}{6}\leq \frac{3}{2}$ $\Rightarrow$ $x\leq\sqrt[3]{3}$ $x=1+\frac{x^{3}}{6}\geq 1+\frac{(\frac{4}{3})^{3}}{6}=\frac{113}{81}$ $x=1+\frac{x^{3}}{6}\geq 1+\frac{(\frac{113}{81})^{3}}{6}=\frac{4561543}{3188646}>1,45$ But $1,45>\sqrt [3]{3}$. So there isn't any x real. Excuse me if I'm wrong.

Note that the equation $a^{3}-6a-6=0$ has root the $a_{1}=\sqrt[3]{2}+\sqrt[3]{4}$

I just saw the topic. My solution: If there were real roots, then the discriminant should be positive or equal to 0. That is, according to the equation given in the problem: $ a^{2}-4(a^{2}-6)\geq0\implies-3a^{2}+24\geq0\implies a^{2}\leq8\implies a\leq2\sqrt{2}\ (1)$ (since a is positive) Since a is positive, we can multiply both sides of the inequality $ a^{2}\leq 8$, without changing anything. As a result, we would have: $ a^{2}\leq 8\implies a^{3}\leq 8a\stackrel{hyp}{\implies}6(a+1)\leq 8a\implies 6\leq 2a\implies a\geq 3\ (2)$ From (1), (2) we have: $ 3\leq a\leq 2\sqrt{2}$ But $ 2\sqrt{2}<3$, so we are forced to conclude that there are no real roots of the above equation and we are done. Excuse me if I have missed something.

My solution for problem 1 : We have : α3=6(α+1)α3-9α+3α-9+3=0 α(α-3)(α+3)+3(α-3)+3=0 (α-3)(α2+3α+3)=-3 and we have α-3<0. And α3=6(α+1) α3-8α=-2α+6α(8-α2)=2(α-3) then 8-α2<0 and the equation has no real solutions .

Let, the equality x^2+ax+a^2-6=0 has real root. Then, D=3(8-a^2)>=0.So,a^2<=8,but a is positive integer So,a=1 or a=2. If a=1, a^3=6(a+1) doesn't take place. If a=2, a^2+6(a+1) isn't correct. So,the roots of equality x^2+ax+a^2-6=0, aren't real.

lasha , your solution is very simple if a is positive integer. But a is a real positive number.

Proof I (generally). Show easily that the function $ \{\begin{array}{c}f\ : \ (0,\infty )\rightarrow\mathcal R\\\ f(x)=x^{2}(x+1)\end{array}$ is strict increasing. Therefore, the equation $ f(\frac{1}{a})=\frac{1}{6}$, i.e. the equation $ a^{3}=6(a+1)$ has at most one positive root. Observe that for the function $ \{\begin{array}{c}g\ : \ (0,\infty )\rightarrow \mathcal R\\\ g(a)=a^{3}-6(a+1)\end{array}$ there is the relation $ g(2\sqrt 2)=2(2\sqrt 2-3)<0<3=g(3)$, i.e. $ g(2\sqrt 2)<0<g(3)$. In conclusion, the single root $ a\in (2\sqrt 2, 3)$. The equation $ x^{2}+ax+a^{2}-6=0$ has no real roots iff $ a^{2}> 8$, what is truly. Proof II (similarly to Mostel). $ f(a)=a^{3}-6(a+1)$ $ \implies$ $ f(0)\cdot f(3)<0$ $ \implies$ $ (\exists)\ a>0$ so that $ a^{3}=6(a+1)$. But $ 0<a\le 2\sqrt 2$ $ \implies$ $ a^{2}\le 8\implies a^{3}\le 8a$ $ \implies 6(a+1)\le 8a\implies 3\le a$, what is falsely. Thus $ a>2\sqrt 2$. $ a\ge 3\implies$ $ a^{3}\ge 3a^{2}\implies$ $ 6(a+1)\ge 3a^{2}\implies$ $ 3a\le a^{2}\le 2(a+1)\implies$ $ a\le 2$, what is falsely. Thus $ a<3$. In conclusion, $ a\in (2\sqrt 2,3)$ and the equation $ x^{2}+ax+a^{2}-6=0$ has no real roots iff $ a^{2}> 8$, what is truly. Remark. $ a^{3}-8=6a+6-8=2(3a-1)\implies (a-2\sqrt 2)(3a-1)>0$ $ \implies a\in (0,\frac{1}{3})\cup (2\sqrt 2,\infty )$. If $ a\in (0,\frac{1}{3})$ then $ 6<6(a+1)=a^{3}<\frac{1}{27}$ $ \implies$ $ 6<\frac{1}{27}$, what is falsely. In conclusion, $ a>2\sqrt 2$ a.s.o.

pohoatza wrote: Let $ a$ be positive real number such that $ a^{3}= 6(a+1)$. Prove that the equation $ x^{2}+ax+a^{2}-6 = 0$ has no real solution. D=$ 24-3a^{2}$ If we suppose that $ 24-3a^{2}\geq 0$ we have a^2=<8. But from $ a^{3}= 6(a+1)$ we have $ a^{3}-6a=6$, so must $ a^{2}\geq 6$. Now we have contradictions in $ a(a^{2}-6)=6$, because no have solutions this equation if $ a$ is positive number and $ 6=<a^{2}=<8$! (is very easy to proof)

$ a^{3}=6(a+1)$ gives: $ a^{2}-6=\frac{6}{a}$ $ (1)$. Let's prove, that $ D<0$. $ D=a^{2}-4(a^{2}-6)=\frac{a^{3}-24}{a}$. As $ a>0$, left to prove that $ a^{3}<24$. $ \frac{a^{2}}{a+1}>a-1$, but $ \frac{a^{2}}{a+1}=\frac{6}{a}$. So, $ \frac{6}{a}>a-1$, which follows $ a<3$. $ a^{3}=6(a+1)<6(3+1)=24$, Q.E.D.

$ \delta = 24-3a^{2}$. We have $ a^{3}= 6(a+1)$ Let $ f(x) = a^{3}-6a-6$. $ f'(x) = 3x^{2}-6$. $ f'(x) =\pm\sqrt{2}$ then $ f(x) < 0,\forall x\in (-\infty,\sqrt{2})$ and $ ,f(x)\nearrow ,\forall x\in [\sqrt{2},+\infty)$ But $ f(2\sqrt{2}) < 0$ then if $ f(x) = 0$ so $ x > 2\sqrt{2}$ Hence $ a > 2\sqrt{2}$ so $ \delta < 0$ ,we're done

pohoatza wrote: Let $a$ be positive real number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+ax+a^{2}-6=0$ has no real solution. Sorry for taking back this topic Here is my solution from this equation $a^{3}=6(a+1)$ from kardanos formula we have $a=2^\frac{1}{3} +4^\frac{1}{3} $ and $a^{2}-6=\frac{6}{a}$ and we have the equation $ax^{2}+a^{2}x+6=0$ $D=a^{4}-24a\ge0$ ==> $a<0$ or $a>2\cdot3^\frac{1}{3}$ but from this inequality $(n-1)^{\frac{1}{3}}+(n+1)^{\frac{1}{3}}<2(n)^\frac{1}{3}$ from this we get this equatin has no real solution

.Are JBMO problems supposed to be easy?

For $a>0$ obtain that $a^3=6(a+1)$ $\implies$ $a\left(a-2\sqrt 2\right)\left(a+2\sqrt 2\right)=2(3-a)$ $\implies$ $\left(a-2\sqrt 2\right)(3-a)>0$ $\implies$ $a\in\left(2\sqrt 2,3\right)$ . Observe that $\Delta\equiv a^2-4\left(a^2-6\right)=-3\left(a^2-8\right)<0$ . In conclusion, the mentioned equation $x^{2}+ax+a^{2}-6=0$ has no real solution.

OK, i will try with elementary solution Assume that, $x^{2}+ax+a^{2}-6=0 $ have real solution. So, $a^2 - 4(1)(a^2 - 6) \ge 0$ $a^2 \le 8$ So, $8a \ge a^3 = 6(a+1) \implies a \ge 3 \implies a^2 \ge 9$. Contradiction with $a^2 \le 8$. CMIIW

youarebad wrote: $a^2 \le 8\implies 8a \ge a^3$ Only for $a\ge 0$ ....

pohoatza wrote: Let $a$ be positive real number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+ax+a^{2}-6=0$ has no real solution. Proof 1 : $a\geq 3\Longrightarrow a^3-6a-6=(a-3)(a^2+3a+3)+3>0\ \therefore 0<a<3$. $x^2+ax+a^2-6=0\Longleftrightarrow (a+1)x^2+a(a+1)x+(a+1)(a^2-6)=0$ $\Longleftrightarrow (a+1)x^2+a(a+1)x+a^3-6(a+1)+a^2=0$ $\Longleftrightarrow (a+1)x^2+a(a+1)x+a^2=0$ $\therefore D=a^2(a+1)^2-4a^2(a+1)=a^2(a+1)(a-3)<0\ Q.E.D.$ Proof 2 : $a\geq 3\Longrightarrow a^3-6a-6=(a-3)(a^2+3a+3)+3>0\ \therefore 0<a<3$. $x^2+ax+a^2-6=0\Longleftrightarrow x^{2}+ax+\frac{6}{a}=0$. $\therefore D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}<0\ Q.E.D.$ Proof 3 : $0<a\leq 2\sqrt{2}\Longrightarrow a^3-6a-6=(a-2\sqrt{2})(a^2+2\sqrt{a}+2)+2(2\sqrt{2}-3)<0$ $\therefore a>2\sqrt{2}$. $\therefore D=3(8-a^2)=3(2\sqrt{2}+a)(2\sqrt{2}-a)<0\ Q.E.D.$ By the way, for what grades ,or ages, is the problem posed in JBMO?

behemont wrote: kunny wrote: We can rewrite the given quadratic equation as $x^{2}+ax+\frac{6}{a}=0$. $D=a^{2}-\frac{24}{a}=\frac{a^{3}-24}{a}=\frac{6(a-3)}{a}$. Why are we done here? because we can see that to have a real solution , we just prove that if D > 0 no solutions exist

Is this right? Please check cause I have found holes in my proofs every now and then.

Lots of people here

Solving the given equality by Tartaglia, we get "a" about 2.847, but the equation only has solution if a <= 2.sqrt(2) <= 2.83, so the equation hasn't solution. Q.E.D.

Devastator wrote:

Is this right? Please check cause I have found holes in my proofs every now and then. Nice proof, yes i think that's right.

In order to have no real solution, we must have $a>2\sqrt2$. By AM-GM, \[a^3=3a+3a+6\geq 9\sqrt[3]{2a^2}\implies a\geq \sqrt[7]{9^3\cdot 3}\]so it is suffices to check that \[\sqrt[7]{9^3\cdot 3}> 2\sqrt{2} \iff 3^{12}>2^{19}\iff 531441>524288\]

First we claim that there are no negative solutions to $a^3 = 6a+6$. If $-1<a<0$, then $a^3$ is negative and $6a+6$ is positive, so this is not possible. If $a$ is $-1$, then the LHS is greater than the RHS, and increases much faster. Now note that for the equal to have no real solution, we need $a^2-4(a^2-6) \leq 0$, or $a^2 \geq 8$. This means it suffices to show that $a \geq 2\sqrt2$. We can see that starting at $0$, the RHS is greater than the LHS, and it will continue to be greater until both sides are equal, creating the first root. We simply have to check whether $2\sqrt2$ makes the RHS greater than the LHS. It does, so the value of $a$ is greater than $2\sqrt2$, as desired.

Because the determinant of the given equation is $$a^2 - 4 \cdot 1 \cdot (a^2 - 6) = 24 - 3a^2$$it suffices to show $a^2 > 8$. Let $x \ge y \ge z$ denote the roots of $a^3 - 6a - 6 = 0$, i.e. the solutions to $a^3 = 6(a + 1)$. Vieta's yields $x + y + z = 0$ and $xyz = 6$. Inspection gives $x, y, z \ne 0$, so the sum condition implies either $$x \ge y > 0 > z$$or $$x > 0 > y \ge z$$is true. But $xyz$ is positive, so we conclude that the second condition must hold. Hence, if $k$ is a positive real between $0$ and $x$, we have $$6 = xyz = x \cdot (-y) \cdot (-z) = x \cdot k \cdot (x - k) \le x \cdot \left(\frac{x}{2} \right)^2 = \frac{x^3}{4}$$which implies $$x^2 \ge 4 \sqrt[3]{9} > 4 \sqrt[3]{8} = 8.$$Because $a$ is positive, we know it cannot equal $y$ or $z$, so we're done. $\blacksquare$

Let me supply a proof which covers the scenario when $a$ is a (not necessarily positive) real number with $a^3 = 6(a+1)$. It suffices to show $D = 24 - 3a^2 < 0$, i.e. $a^2 > 8$. If we suppose $a^2 \leq 8$, then since $a^3 - 6a \leq 4\sqrt{2}$ (equivalent to $(a-2\sqrt{2})(a+\sqrt{2})^2 \leq 0$, which works for $a^2 \leq 8$) and $4\sqrt{2} = \sqrt{32} < 6$, we have obtained a contradiction.