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# Polynomial Tactics

Prove that if the discriminant of a quadratic equation is less than zero, and the coefficient of $$x^2$$ is positive then $$P(x)>0$$ for all rea values of $$x$$.

In short, for polynomial $$P(x) = ax^2+bx+c$$, if $$D<0$$ and $$a>0$$, then $$P(x)>0$$ for all real values of $$x$$.

Also, if $$D<0$$ and $$a<0$$, then, $$P(x) < 0$$.

Note by Naitik Sanghavi
8 months ago

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$$P(x) = ax^{2} + bx + c$$
$$P(x) = a(x^{2} + \dfrac{b}{a}x + \dfrac{c}{a})$$
$$P(x) = a\left( \left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a^{2}}\right)$$
$$P(x) = a\left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a}$$

Now for $$a > 0$$
$$a\left(x+\dfrac{b}{2a}\right)^{2} \ge 0$$
$$a\left(x+\dfrac{b}{2a}\right)^{2} + \dfrac{4ac-b^{2}}{4a} \ge \dfrac{4ac-b^{2}}{4a}$$
$$P(x) \ge \dfrac{4ac-b^{2}}{4a}$$
$$D < 0 \rightarrow 4ac - b^{2} > 0$$
$$P(x) \ge \dfrac{4ac-b^{2}}{4a} > 0$$ for all x. Similar proof for $$a < 0$$ · 8 months ago