2 - x = x² + mx + m

x² + mx + x + m - 2 = 0

x² + (m + 1)x + (m - 2) = 0

Now it's in the form ax² + bx + c = 0.

Next step is to find the discriminant.(b²-4ac)

Discriminant

= (m+1)² -4(1)(m-2)

= (m² + 2m + 1) - (4m - 8)

= m² - 2m + 9

= m² - 2m + 1 + 8

= (m - 1)² + 8

(m - 1)² is always ≥ 0 for all real values of m.

Because m is a real number, m - 1 is a real number. Any square of a real number is always 0 or positive.

So, (m-1)² + 8 ≥ 8 for all real values of m.


This tells you that (m - 1)² + 8 > 0 for all real values of m


Since the discriminant is > 0 for all real values of m, there are real and unequal roots. So the line will always cut the curve at 2 distinct points

Alternatively after obtaining the discriminant m2 - 2m + 9, we can also talk about ITS discriminant which is 4 - 36 < 0, and along with the fact that coefficient of m2 is 1 which is positive, m2 - 2m + 9 > 0 for all m.

So the original discriminant > 0 for all m.

Why not you include that in the answer you posted to him as well, such that it's clearer for him