"Sign of P(Δ) with ²P(x)=ax²+bx+c and Δ=b2−4ac" solved and explained

High School
Algebra
Algebra I
Quadratic function and equation

meteraiqn

Answered question

2022-02-01

Sign of

P (Δ)

with

P (x) = a x ² + b x + c

and

Δ = b^{2} - 4 a c

Answer & Explanation

Natalia Thomas

Beginner2022-02-02Added 7 answers

Completing the square differently, this is non-negative when $a > 0,, Δ \geq 0, 2 b - \frac{1}{4 a} \geq 0$ :
$\begin{aligned} \frac{1}{a} P (Δ) & = Δ^{2} + \frac{b}{a} Δ - 2 \frac{b}{a} Δ + \frac{b^{2}}{4 a^{2}} + 2 \frac{b}{a} Δ - \frac{b^{2}}{4 a^{2}} + \frac{c}{a} \\ = Δ^{2} - 2 \frac{b}{2 a} Δ + \frac{b^{2}}{4 a^{2}} + \frac{2 b}{a} Δ - \frac{b^{2} - 4 a c}{4 a^{2}} \\ = (Δ - \frac{b}{2 a})^{2} + (2 b - \frac{1}{4 a}) \frac{Δ}{a} \end{aligned}$ $\begin{aligned} \frac{1}{a} P (Δ) & = Δ^{2} + \frac{b}{a} Δ - 2 \frac{b}{a} Δ + \frac{b^{2}}{4 a^{2}} + 2 \frac{b}{a} Δ - \frac{b^{2}}{4 a^{2}} + \frac{c}{a} \\ = Δ^{2} - 2 \frac{b}{2 a} Δ + \frac{b^{2}}{4 a^{2}} + \frac{2 b}{a} Δ - \frac{b^{2} - 4 a c}{4 a^{2}} \\ = (Δ - \frac{b}{2 a})^{2} + (2 b - \frac{1}{4 a}) \frac{Δ}{a} \end{aligned}$

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