I've to solve the following polynomial inequality
$$x^2 - 6x + 11 > 0$$ By using quadratic formula, I got the value of $x$ as below
$$ \frac{6\pm \sqrt{-8}}{2}$$
These are imaginary roots and the graph will never touch $x$-axis. So, I'm not sure what would be the solution set for $x$? Is it something like $$\{x | −\infty < x < +\infty \}$$
$\endgroup$ 12 Answers
$\begingroup$$$x^2-6x+11>0$$ $$x^2-6x+9+2>0$$ $$(x-3)^2+2>0$$ because $$(x-3)^2\geq 0 ,\space\forall x\in\mathbb R$$
$\endgroup$ 2 $\begingroup$When solving equations you need to state the domain of $x$. If your problem is over the complex numbers, then the solution set is as you have given: $$x = 3 \pm i \sqrt{2}.$$
If your problem requires $x$ to be a real number then you have shown that there are no such solutions to this equation. In this case the solution set for $x$ is the empty set $\{\emptyset \}$.
One would say "there are no real solutions to this equation." You might like to look up the fundamental theorem of algebra for polynomial equations of order $N$ over the complex numbers...
Anyway, since you see that the graph of your polynomial is always above the x axis, as we've shown that there are no real solutions, then any value of x will satisfy the equality. Thus the solution is $x \in \mathbb{R}$.
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