Algebra Know-It-ALL

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]