Question

Find the range of \(x\)

Answer 1

\(\large \color{black}{\begin{align} x^2+x+1>0,\quad x\in \mathbb{R} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

\[x \in R\]

Answer 3

This quadratic expression has no roots, and it points upwards. It means that it must lie above the x-axis always, thus meaning that the inequality holds for all \(x\).

Answer 4

Are you in 11th?

Answer 5

it has imaginary roots

Answer 6

Well, that's obvious. All polynomials have roots. It comes without saying that I mean real roots.

Answer 7

Anyway, do you see why that holds for all \(x\)?

Answer 8

cuz \(x\in \mathbb{R},x\cancel{\in} \mathbb{I}\)

Answer 9

In inequalities, we don't talk about complex numbers anyway.

Do you see why \(x^2 + x + 1\) is positive for all \(x\)?

Answer 10

Another way to think is that the minimum value of the function is \(3/4\), so the function will always be greater than 0.

Answer 11

how u got 3/4

Answer 12

\[x^2 + x + 1 = \left(x + \frac{1}{2}\right) ^2 + \frac{3}{4}\]