388 Quadratic Equations with Complex Roots
Multiplying through be 3 gives us
3 x^2 + 75 = 0
That’s the original equation.
The general case
Consider a general quadratic in which the coefficient of x^2 is a positive real number a, the
coefficient of x is 0, and the stand-alone constant is a positive real number c. Then we have
ax^2 +c= 0
Subtracting c from each side, we get
ax^2 =−c
Dividing through by a, which we know is not 0 because we’ve stated that it’s positive, we
obtain
x^2 =−c/a
which can be rewritten as
x^2 =−1(c/a)
Because a and c are both positive, we know that the ratio c/a is positive as well. Its positive
and negative square roots are therefore both real numbers. We can take the square root of both
sides of the above equation, getting
x=±[−1(c/a)]1/2
=±(−1)1/2 [±(c/a)1/2]
=±j[(c/a)1/2]
=j[(c/a)1/2] or −j[(c/a)1/2]
Knowing these roots, we can get the binomial factor form of the original quadratic. In
one case,
x=j[(c/a)1/2]
If we subtract j[(c/a)1/2] from each side, we get
x−j[(c/a)1/2]= 0
In the other case,
x=−j[(c/a)1/2]