Let $P(x) = x^2 +2bx + c$ be a quadratic form where $b,c$ are real numbers.If $b^2 < c$ , show that $P(x) > 0$ for all $x$ .Is the converse also true?
The value of $x$ after solving the equation will be : $x = -b+(b^2 -c)^{1/2}$. and when we will put it again in the quadritic form after applying the given condition, we will get the answer as $0$ . Can anyone give a proper solution or some hint and point out where i am wrong and does the converse also hold in such condition? why? If not, why?Thanks a lot for the help.
$$P(x)=(x+b)^2+c-b^2$$
Now, $(x+b)^2\ge0$ for all real $x,b\implies P(x)\ge c-b^2$
So, $P(x)$ will be $>0$ for all real $x$ if $c-b^2>0\iff c>b^2$
Conversely, if $c>b^2, P(x)=(x+b)^2+c-b^2>0$ for all real $x$