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Sign of Quadratic Expression

If y=x2+kx+1 ...

Question

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

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Solution

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

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