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Since I don't have enough reputation to comment, I am asking this question again.

I cannot understand why we can assume that the quadratic has real roots and then say $D\ge0$. The answer states that $b^2 - 4ac = (b + 2ax)^2$ and since this is a squared term it satisfies for all $x$ belonging to $\mathbb R$. But that is the question we began with, how do we know that $x$ is real. Plugging a non-real value of $x$ does not satisfy the relation of $D$.

N. F. Taussig
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arnav
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    You don’t assume that the quadratic has real roots. You check for which $y$ will the quadratic have the real root. If for some $y$ we have $D\ge 0$ ($D$ depends on $y$), the quadratic has a real root, i.e. there exists a solution $x$ that satisfies the quadratic. But then, the same $x$ satisfies $f(x)=y$ so $y$ is in range. Conversely, if $D<0$ then there is no $x$ to satisfy the quadratic, i.e. there does not exist $x$ such that $f(x)=y$, i.e. $y$ is not in the range. Altogether: $y$ is in range if and only if $D\ge 0$. –  Jun 17 '22 at 19:45
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    (Remember the definition of the range of function: $y$ is in the range of $f$, by definition, if there exists $x$ in the domain of $f$ such that $f(x)=y$.) –  Jun 17 '22 at 19:48
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    "how do we know that $x$ is real" $,$ Granted, the linked question could have been worded better, but the assumption there is that $,f,$ is considered as a real function $,f : \mathbb R \to \mathbb R,$ ("for real values of y, the given equation has real roots"). Therefore, the $,x,$ must be real. The same question for a complex function $,f : \mathbb C \to \mathbb C,$ would require a different solution, and the final answer would be different. – dxiv Jun 17 '22 at 20:05
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    @StinkingBishop Thank you for the explanation! I think I understand, will need some time to grasp this though – arnav Jun 17 '22 at 21:01
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    Thank you !! Ryang – arnav Jun 22 '22 at 04:56

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