Question 237493: What value of K will the roots of the equation 2x^2-3x+K=0 be imaginary?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
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