Question

Quadratic equation problem.. please help...

Answer 1

\[ax^2+bx+c=0 \\ \text{ means } x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[\text{ if } b^2-4ac=0 \text{ then there is one solution } \]
\[\text{example of one real solution: } \\ x^2+6x+9=0 \\ x=\frac{-6 \pm \sqrt{6^2-4(1)(9)}}{2(1)}=\frac{-6 \pm \sqrt{36-36}}{2}=\frac{-6 \pm 0}{2}=\frac{-6}{2}=-3 \]
\[ \text{example of one imaginary solution: } \\ x^2+6i x-9=0 \\ x=\frac{-6i \pm \sqrt{(6i)^2-4(1)(-9)}}{2(1)}=\frac{-6i \pm \sqrt{36i^2+36}}{2} =\frac{-6i}{2}=-3i\]
\[\text{ example of two real solutions: } \\ x^2+5x+6=0 \\ x=-2 \text{ or } x=-3\] \[\text{ example of two imaginary solutions: } \\ x^2+9=0 \\ x=3i \text{ or } x=-3i \]

Answer 2

so all of them are possible

but i bet you have some restrictions on the values of a,b, and c

Answer 3

and that hasn't been shared