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mathmath333
 one year ago
Find the range of \(x\)
mathmath333
 one year ago
Find the range of \(x\)

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} x^2+x+1>0,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ \end{align}}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2This quadratic expression has no roots, and it points upwards. It means that it must lie above the xaxis always, thus meaning that the inequality holds for all \(x\).

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0it has imaginary roots

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Well, that's obvious. All polynomials have roots. It comes without saying that I mean real roots.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Anyway, do you see why that holds for all \(x\)?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0cuz \(x\in \mathbb{R},x\cancel{\in} \mathbb{I}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2In inequalities, we don't talk about complex numbers anyway. Do you see why \(x^2 + x + 1\) is positive for all \(x\)?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Another way to think is that the minimum value of the function is \(3/4\), so the function will always be greater than 0.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[x^2 + x + 1 = \left(x + \frac{1}{2}\right) ^2 + \frac{3}{4}\]
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