In the general polynomial standard form, we have a= 3, b= 0, and c= 75. The discriminant
is therefore
d=b^2 − 4 ac
= 02 − (4 × 3 × 75)
=− 900
The positive-or-negative square root of the discriminant is
±(d1/2)=±j(|−900|1/2)
=±j(9001/2)
=±j 30
The roots can now be found as
x= [−b±j(|d|1/2)] / (2a)
= (0 ±j30) / (2 × 3)
=±j30/6
=±j 5
Are you astute?
Do you suspect that this particular equation can be solved more easily without resorting to the quadratic
formula? If so, you’re right! If we subtract 75 from each side, we get
3 x^2 =− 75Dividing through by 3 gives us
x^2 =− 25When we take the positive-or-negative square root of both sides, we obtain
x=±[(−25)1/2]
=±j 5
Imaginary roots: the general case
Let’s see what happens with a general quadratic when the coefficient of x is 0. If we write it in
polynomial standard form, we get
ax^2 + 0 x+c= 0Complex Roots by Formula 383