1

Let $P(x)=ax^2+bx+c$ be a polynomial with $a≥0$ and $b≥\frac{1}{8a}$, and $Δ=b^2-4ac$. Show that $P(Δ)≥0$.

My answer: We have $Δ=b^2-4ac$. If$Δ≤0$, then $P(Δ)≥0$. If $Δ>0$, there are two cases:

If $c≥0$, $P(Δ)=aΔ^2+bΔ+c≥0$. If $c<0,Δ=b^2-4ac$ then $Δ>b^2$ and $Δ>-4ac$, then $$P(Δ)=aΔ^2+bΔ+c>a×(-4ac)^2+b^3+c=16a^3c^2+c+b^3$$ is of degree two of variable $c$. We have $Δ_1=1-4×16a^3b^3=1-(4ab)^3$, which is negative if $b≥\frac{1}{4a}$ or we have $b≥\frac{1}{8a}$.

I want the correct solution of the exercise.

Ѕᴀᴀᴅ
  • 34,263
connaissant
  • 71
  • 1
    What is your question? Is this a solution verification? – Bumblebee Nov 22 '21 at 01:21
  • 2
    Also, is it $b\geq \frac{1}{8a}$ or $b\geq \frac{1}{8}\cdot a$? – Alan Abraham Nov 22 '21 at 01:22
  • yes Is this a solution verification, but my answer is not correct – connaissant Nov 22 '21 at 01:26
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 22 '21 at 01:31
  • I want the solution of the exercise – connaissant Nov 22 '21 at 01:33
  • You still need to prove that $P(\Delta) \geq 0$ if $\Delta \leq 0$. i.e. You need to show that $a\Delta^{2} \geq b \Delta + c$ – Kendall Nov 22 '21 at 01:33
  • if $\Delta \leq 0$ a≥0 so the sign of P is the sign of a which is positive – connaissant Nov 22 '21 at 01:37
  • 1
    @connaissant the statement you just made of "if $\Delta \leq 0$ $a \geq 0$ so the sign of $P$ is the sign of $a$ which is positive" is not true in general. You have to prove it! – Kendall Nov 22 '21 at 01:45
  • @Kendall If $,\Delta \lt 0,$ the quadratic has no real roots, so it must have the same sign for all $,x \in \mathbb R,$, and that must be the sign of $,a,$. – dxiv Nov 22 '21 at 03:31
  • @dxiv I see I misread part of the question that lead me to an incorrect conclusion. You are indeed correct. – Kendall Nov 22 '21 at 09:36

1 Answers1

2

Completing the square differently, this is non-negative when $\,a \gt 0\,$, $\,\Delta \ge 0\,$, $\,2b - \dfrac{1}{4a} \ge 0\,$:

$$ \begin{align} \frac{1}{a} P(\Delta) &= \Delta^2+\frac{b}{a}\Delta \color{red}{-2 \frac{b}{a}\Delta + \frac{b^2}{4a^2} + 2 \frac{b}{a}\Delta - \frac{b^2}{4a^2}}+\frac{c}{a} \\ &= \Delta^2 - 2 \frac{b}{2a}\Delta + \frac{b^2}{4a^2}+\frac{2b}{a}\Delta-\frac{b^2-4ac}{4a^2} \\ &= \left(\Delta - \frac{b}{2a}\right)^2 + \left(2 b - \frac{1}{4a}\right)\frac{\Delta}{a} \end{align} $$

dxiv
  • 76,497