Let $a+b+c=0$. Does $1^2a+1b+c=0 \implies b^2 \geq 4ac \ $ ? Obviously we know the inequality holds in general, but does considering a "quadratic in one" imply this?
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Depends on the values of $a , b $ and $c$ . – Spectre Aug 22 '20 at 15:41
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$\because a + b + c = 0$, any of the scenarios given may come : $ a + c = -b, b + c = -a, a + b = -c $. – Spectre Aug 22 '20 at 15:42
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And probably you are right, because I said it out of thought and not out of written work. – Spectre Aug 22 '20 at 15:43
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1This only implies that a function of $f(x)=ax^2+bx+c$ has a root of $1$. which obviously implies the discriminant is positive. – Popular Power Aug 22 '20 at 15:48
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Yes, you're right (assuming $a$, $b$, $c$ are real). You can prove it in another way:
$$b=-a-c$$ $$b^2=a^2+c^2+2ac$$
Now, by the A.M.-G.M. inequality, $$\frac{a^2+c^2}{2}\ge ac$$
Thus, $$b^2\ge4ac$$
Using the quadratic method, you'd have to assume a quadratic like this: $$ax^2+bx+c=0$$
Clearly, $1$ is a root. The other root must be real as well if $a$, $b$ and $c$ are real. So - $$b^2\ge4ac$$
Nikhil Anand
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2That was my reasoning as well. The $1$ is just a(an?) $x$ in disguise. – Popular Power Aug 22 '20 at 15:50
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'an' $x$ when said , so in written language too that is the right one, I suppose.... – Spectre Aug 22 '20 at 15:58
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$1^2a+1b+c=0$ means that $x=1$ is a solution of $ax^2+bx+c=0$
In general when $a\not=0$ the solutions are $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ and since $x=1$ is real you are correct that this directly implies $b^2\ge 4ac$
If $a=0$ then $4ac =0$ while $b^2 \ge 0$ so that is true too
Henry
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