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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.

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Keep in mind that the quadratic formula is locating the x-intercepts of the parabola described by $P(x)$. Since $a = 1 > 0$ in this quadratic function (a "monic polynomial"), the parabola "opens upward". If $b^2 - c < 0$, there are no x-intercepts, so the parabola must be entirely "above" the x-axis (this reasoning dates back to shortly after the introduction of analytic geometry). Because of the connection between the character of the polynomial and the geometry of the parabola, the converse holds as well. –  RecklessReckoner May 10 '13 at 6:32
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1 Answer 1

up vote 5 down vote accepted


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 $x$

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thanks a lot for the help... :) –  under-root May 10 '13 at 4:52
@sunnyverma, you are welcome –  lab bhattacharjee May 10 '13 at 5:10
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