Question 237493
equation is 2x^2 - 3x + k = 0


standard form of equation is ax^2 + bx + c = 0


in this equation:


a = 2
b = -3
c = k


discriminant is b^2 - 4ac


if the discriminant is negative, then the roots will be imaginary.


b^2 - 4ac becomes:


(-3)^2 - 4*2*k


this equals 9 - 8*k


solve for k to get:


k = 9/8


if k is greater than 9/8, then the discriminant will be negative and the roots will be imaginary.


example:


let k = 9/8


equation becomes:


2x^2 - 3x + 9/8 = 0


solve for the roots to get:


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


a = 2
b = -3
c = 9/8


formula becomes:


x = -(9/8) +/- sqrt(9 - 4*2*9/8) / 4


this becomes:


x = (-9/8) +/- sqrt(9-9) / 4 = real roots.


now let c = 10/8


equation becomes:


x = (-10/8) +/- sqrt (9 - 4*2*10/8) / 4


this becomes:


x = (-10/8) +/- sqrt (9 - 10) / 4


sqrt (9-10) is negative causing the roots to be imaginary.


answer is that k > 9/8 makes the roots imaginary.