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#.