Question

Can someone help me know how to do this problem In finding what I can put in for (c) in order to get 2 imaginary solutions

Answer 1

Answer:

To get 2 imaginary solutions, c must be less than -2

Step-by-step explanation:

The general form of the quadratic equation is:

$ax^2+bx+c=0$

Solve the quadratic equation by using the formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The equation to solve is:

$-2x^2+4x+c=0$

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:

$b^2-4ac$

Substituting the known values:

$4^2-4(-2)c$

$16+8c$

Subtracting 16:

$8c$

Solving:

$c$

Thus, to get 2 imaginary solutions, c must be less than -2