$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$

Question

If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.

My attempt is as follows:

$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0$$

First condition: If $\forall$ $x>0$, inequality is satisfied, it means $x=0$ must have been the root of the equation $\quad \{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}=0\quad$ at which given inequality would not be satisfied.

Second condition: If one root is real, then other root must be real as complex roots occur as conjugates if all quadratic equation coefficients are real. So $D>=0$

Third condition: $a>0$ as $\forall$ $x>0$, $\quad\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0\quad$

First condition: $$x=0$$ $$a+\sqrt{2}=0$$ $$a=-\sqrt{2}$$

Second condition: $$D>=0$$

$$4(a^2-2)^2-4(a-\sqrt{2})(a^2+a-3)(a+\sqrt{2})>=0$$ $$(a-\sqrt{2})(a+\sqrt{2})(a^2-2-a^2-a+3)>=0$$ $$(a-\sqrt{2})(a+\sqrt{2})(a-1)<=0$$ $$a\in \left(-\infty,-\sqrt{2}\right]\quad \cup \quad[1,\sqrt{2}]$$

Third condition:

$$a>0$$ $$(a-\sqrt{2})(a^2+a-3)>0$$ $$(a-\sqrt{2})\left(a-\left(\frac{-1+\sqrt{13}}{2}\right)\right)\left(a-\left(\frac{-1-\sqrt{13}}{2}\right)\right)>0$$

$$a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$

Taking intersection of all three conditions would give

$$a\in \{-\sqrt{2}\}\quad$$

But answer is $a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$

What am I missing here, I tried to think of it a lot but didn't any breakthroughs. Please help me in this.

Answer 1

For the first condition ($x = 0$), you want $a+\sqrt{2} > 0$ so $a > -\sqrt{2}$.

Answer 2

Why are you taking the intersection of those three conditions? It doesn't make much sense to me.

For example, the second condition $\Delta\geq 0$ is not necessary. If $a>\sqrt{2}$ then we have $\Delta<0$ and the inequality is satisfied for all real $x$ (not only for $x>0$): $$\underbrace{(a^2+a-3)(a-\sqrt{2})}_{>0}x^2+2(a^2-2)x+(a+\sqrt{2})>0.$$ Also $a=\sqrt{2}$ works because the inequality boils down to $(\sqrt{2}+\sqrt{2})>0$ which holds.

Hence $[\sqrt{2},+\infty)$ is a subset of the possible values of the parameter $a$.

Answer 3

Well I got the answer, thanks to @Robert Z

I) First lets see what should be the condition when $a>0 \text{\{a here means coefficient of $x^2$\}}$

So $$\text{parameter }a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$

Now there can be two sub-cases:

1) When $D<0$, then at all values of $x$, inequality will hold.

So for $D<0$ and $a>0$, $\text{parameter }a\in \left(-\sqrt{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right)$

2) When $D>=0$, then inequality will hold only when $x=0$ is the root of the given quadratic equation.

So for $a>0$ and $D>=0$ and $\text{$x=0$ is the root}$, we get parameter $a\in \{-\sqrt{2}\}$

II)When $a<0$, as in that case inequality cannot hold till $+\infty$, so no solutions.

III) When $a=0$, then parameter $a$ can be $\{\sqrt{2},\frac{-1+\sqrt{13}}{2},\frac{-1-\sqrt{13}}{2}\}$

If parameter $a=\sqrt{2}$, then $\sqrt{2}+\sqrt{2}>0$ which is always true, so $\sqrt{2}$ is the solution.

If parameter $a=\frac{-1+\sqrt{13}}{2}$, then we get $x<4.5$, so inequality will hold only for $x<4.5 $, but we want inequality to hold $\forall x>0$, so $a=\frac{-1+\sqrt{13}}{2}$ can't be the solution.

If parameter $a=\frac{-1-\sqrt{13}}{2}$, then we get $x>0.133$, so inequality will hold only for $x>0.133$, but we want inequality to hold $\forall x>0$, so $a=\frac{-1-\sqrt{13}}{2}$ can't be the solution.

So taking union of all cases I),II),III), we get $\text {parameter }a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$ which is the correct answer.