Answer:
To get 2 imaginary solutions, c must be less than -2
Step-by-step explanation:
The general form of the quadratic equation is:
[tex]ax^2+bx+c=0[/tex]
Solve the quadratic equation by using the formula:
[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The equation to solve is:
[tex]-2x^2+4x+c=0[/tex]
In our equation: a=-2, b=4, c=unknown
For the roots to be imaginary, the argument of the square root must be negative, that is:
[tex]b^2-4ac<0[/tex]
Substituting the known values:
[tex]4^2-4(-2)c<0[/tex]
[tex]16+8c<0[/tex]
Subtracting 16:
[tex]8c<-16[/tex]
Solving:
[tex]c<-2[/tex]
Thus, to get 2 imaginary solutions, c must be less than -2
