I have a question here from a textbook.
Show that $x^2 +2kx +9 \ge 0$ for all real values of $x$ if $k^2 \le 9$
Here's my proof:
I found values of $k$ to be between $ -3\le k\le3 $
For all real values of $x$, $b^2 -4ac \le 0 \Rightarrow $ no real roots.
I feel there is a relationship between the value of y and $b^2-4ac$
Any help appreciated.
Note that the correct statement is $b^2 -4ac \le 0 \Rightarrow$ no real distinct roots.
From this condition we obtain
$$4k^2-36\le 0\iff-3\le k\le3$$
Proof:
If the quadratic
$f(x) = x^2+ 2kx +9$ has only complex roots then
$f(x) \ge 0.$
Let $z_1=z$ , $z_2 = \overline z$ be the complex roots.
Then for real $x$:
$f(x) =(x-z)(x-\overline z)=(x-z)(\overline{x-z})\ge 0,$
since $c\overline c =|c|^2 \ge 0$, where $c \in \mathbb{C}.$
Now follow zam's and gimusi's arguments.
$$0\le x^2+2kx+9=(x+k)^2+9-k^2$$
Clearly, $(x+k)^2\ge0$ for real $x+k$ So, we need $$9-k^2\ge0$$
Your intuition is right. For $k^2=9$ we have the only root $x_1=x_2=-k$ and for $k^2<9$ we have two distinct and conjugate complex roots e.g. for $k=1$ we have $$x^2+2x+9=(x+1)^2+8>0$$
