Part Three 509In this case, the coefficient of x is imaginary. The coefficient of x^2 , as well as the stand-alone
constant, are real. The complete polynomial quadratic equation is
x^2 +j 5 x− 6 = 0Question 23-9
Is it possible for one root of a quadratic equation to be pure real and the other pure imaginary? If so,
provide an example of such an equation in polynomial standard form. If not, explain why not.
Answer 23-9
Yes, this is possible. Here is an example of such an equation in binomial factor form:
(x+ 2)(x+j3)= 0The roots of this equation are x=−2 or x=−j3, as we can verify by plugging them in. To
convert this to polynomial standard form, we multiply the product of binomials out:
(x+ 2)(x+j3)=x^2 +j 3 x+ 2 x+j 6
=x^2 + (2 +j3)x+j 6
Here, the coefficient of x is complex but not pure imaginary, and the stand-alone constant is
pure imaginary. The coefficient of x^2 is real. The complete polynomial quadratic equation is
x^2 + (2 +j3)x+j 6 = 0Question 23-10
Is it possible for a quadratic equation to have two nonconjugate complex roots, neither or
which is pure imaginary? If so, provide an example of such an equation in polynomial stan-
dard form. If this sort of situation is impossible, explain why.
Answer 23-10
This, too, is possible! Suppose the roots are 1 +j and 2 +j. These are non-conjugate complex
numbers, and neither of them is pure imaginary. We can construct a binomial factor quadratic
with these numbers as roots by subtracting the roots from x, like this:
[x− (1 +j)][x− (2 +j)]= 0When we multiply the left side of this equation out to obtain a polynomial, taking extra pre-
cautions to be sure that we don’t mess up with the signs, we obtain
[x− (1 +j)][x− (2 +j)]= [x+ (−1)+ (−j)][x+ (−2)+ (−j)]
=x^2 + (− 2 x)+ (−jx)+ (−x)+ 2 +j+ (−jx)+j 2 + (−1)
=x^2 + (− 3 x)+ (−j 2 x)+j 3 + 1
=x^2 + (− 3 −j2)x+ (1 +j3)