Solve this problem please
1 answer:
You might be interested in
This is a polynomial of degree 2, and remember that complex roots always come in pairs; therefore, in a second degree polynomial you can only have 0 real roots and 2 complex roots, or 2 real roots and 0 complex roots. There is no way you can have 1 real root and 1 complex root because, as stated before, complex (non-real) roots always come in pairs; so the only way of having 1 real root is dumping the polynomial of degree 2 (
) by changing the 14 to 0. That way we'll left with 
which only has one real root:
