How many solutions can a quadratic equation have and why does it vary? how would the corresponding graphs look like?

Hello from MrBillDoesMath!

0, 1, or 2  

Discussion:

Examples:

1) x^2 + 9 = 0 has 0 solutions over the real numbers

2) x^2 + 9 = 0 has 2 solutions over the complex number ( 3i, -3i)

3) the quadratic (x-3)^2 = 0 has a single solution, x = 3, of multiplicity two. (which some regard as two solutions).

The bottom line is the number of roots is given by the quadratic solution

( -b +\- sqrt( b^2 - 4ac) ) /2a

In particular if the discriminant  (b^2 -4ac) = 0, the quadratic has one solution. If the discriminant is >0 there are two (real) solutions, and if the discriminant is <0 there are 2 imaginary (complex) solutions.

Thank you,

MrB

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14