Question

If $y=x^2+kx+1$ intersects the $x$-axis at two different points, then the minimum positive integral value of $k$ is

Answer 1

$y=x^2+kx+1$
This intersects the $x$-axis at two points, so it has two real and distinct roots.
$D>0$
$\Rightarrow k^2-4>0$
$\Rightarrow k^2>4$
$\therefore k<-2$ or $k>2$
Hence, the minimum positive integral value of $k$ is $3$.