If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?

Question

If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?

This is what I have done:
Let $y = 3^x$
$$9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$$ $$\implies9y^2 + (t^2 - 4t - 2)y + 1 > 0$$
For the LHS to be greater than zero, $b^2 - 4ac$ has to be $\lt 0$, since coefficient of $y^2$ is greater than $0$(which will give us an upward opening parabola).
$$(t^2 - 4t - 2)^2 - 36< 0$$
The answer that I get is different from the correct answer. The correct answer to this is $t \in \mathbb{R} - \{2\}$. What did I do wrong?

Answer 1

With the substitution $y= 3^x$, as you mention, we get the equivalent inequality: $$9y^2+(t^2-4t-2)y>0$$ but with the added condition $y> 0$ as $3^x > 0$.

Hence we see that if $b=t^2-4t-2\ge 0$, the LHS is positive for all allowable $y$, so we need to worry only about $b< 0$. Here using the discriminant condition, it is sufficient to show that $$b^2< 4ac \iff (t^2-4t-2)^2< 36 \iff -6 < t^2-4t - 2 < 6$$ As we need only consider $b< 0$, only the left inequality matters. So we have $t^2-4t+4 = (t-2)^2 > 0 \implies t \neq 2$.

Answer 2

You need $b^2 - 4ac$ not $b - 4ac$.

Answer 3

Y needs to be positive, and this will give you another inequality for t

Answer 4

Alternately, by positivity of $3^x$ and AM-GM: $$(t^2-4t-2)3^x+3^{2x+2}+1\ge (t^2-4t-2)3^x+2\sqrt{3^{2x+2}} = (t^2-4t+4) \cdot 3^x$$ with equality possible when $x=-1$. So we need $t^2-4t+4 = (t-2)^2 > 0$ which means $t \neq 2$.

Answer 5

Given $$9^{x+1}+(t^2-4t-2)3^x+1>0$$

We can write it as $$3^x\bigg[9\cdot 3^x+(t^2-4t-2)+\frac{1}{3^x}\bigg]>0$$

So $$\underbrace{3^x}_{>0} \bigg[\bigg(3\cdot 3^{\frac{x}{2}}-\frac{1}{3^{\frac{x}{2}}}\bigg)^2+(t-2)^2\bigg]>0$$

So for Above statement is True $\forall \ x \; \mathbb{R},$ If $t\in \mathbb{R}-\{2\}$