How do you know if a polynomial of degree 2 has an x-intercept?

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Answer 1 · 2016-03-06T00:55:47

Once you have it in standard $ax^2+bx+c$ form, check the discriminant $Delta = b^2-4ac$ is non-negative.

Any polynomial of degree $2$ in $x$ can be expressed in the form:

$ax^2+bx+c$

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac$

If $Delta >= 0$ then $ax^2+bx+c = 0$ has at least one Real root.

Roots of $ax^2+bx+c = 0$ are $x$ intercepts.

The roots of $ax^2+bx+c = 0$ are given by the formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)$

Notice that the discriminant is the expression under the square root, so the square root takes Real values if and only if $Delta >= 0$ .