meteraiqn

2022-02-01

Sign of $P\left(\Delta \right)$ with $P\left(x\right)=ax²+bx+c$ and $\Delta ={b}^{2}-4ac$

Natalia Thomas

Expert

Completing the square differently, this is non-negative when $a>0,,\mathrm{\Delta }\ge 0,2b-\frac{1}{4a}\ge 0$:
$\begin{array}{rl}\frac{1}{a}P\left(\mathrm{\Delta }\right)& ={\mathrm{\Delta }}^{2}+\frac{b}{a}\mathrm{\Delta }-2\frac{b}{a}\mathrm{\Delta }+\frac{{b}^{2}}{4{a}^{2}}+2\frac{b}{a}\mathrm{\Delta }-\frac{{b}^{2}}{4{a}^{2}}+\frac{c}{a}\\ & ={\mathrm{\Delta }}^{2}-2\frac{b}{2a}\mathrm{\Delta }+\frac{{b}^{2}}{4{a}^{2}}+\frac{2b}{a}\mathrm{\Delta }-\frac{{b}^{2}-4ac}{4{a}^{2}}\\ & =\left(\mathrm{\Delta }-\frac{b}{2a}{\right)}^{2}+\left(2b-\frac{1}{4a}\right)\frac{\mathrm{\Delta }}{a}\end{array}$$\begin{array}{rl}\frac{1}{a}P\left(\mathrm{\Delta }\right)& ={\mathrm{\Delta }}^{2}+\frac{b}{a}\mathrm{\Delta }-2\frac{b}{a}\mathrm{\Delta }+\frac{{b}^{2}}{4{a}^{2}}+2\frac{b}{a}\mathrm{\Delta }-\frac{{b}^{2}}{4{a}^{2}}+\frac{c}{a}\\ & ={\mathrm{\Delta }}^{2}-2\frac{b}{2a}\mathrm{\Delta }+\frac{{b}^{2}}{4{a}^{2}}+\frac{2b}{a}\mathrm{\Delta }-\frac{{b}^{2}-4ac}{4{a}^{2}}\\ & =\left(\mathrm{\Delta }-\frac{b}{2a}{\right)}^{2}+\left(2b-\frac{1}{4a}\right)\frac{\mathrm{\Delta }}{a}\end{array}$

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