Algebra Know-It-ALL

4. Here are the roots again, for reference:

x=j7 or x=−j 3

Both of these statements are equations. In the first one, we can subtract j7 from each side.

In the second one, we can add j3 to each side. This gives us

x−j 7 = 0

and

x+j 3 = 0

The binomial factor form of a quadratic with the above mentioned roots is therefore

(x−j7)(x+j3)= 0

5. To find the polynomial standard form of the equation we just found, we must multiply
   out the left side using the product of sums rule. Let’s convert the first binomial to a sum
   to avoid sign confusion. Then we can take it from there:

[x+ (−j7)](x+j3)= 0

x^2 +xj 3 + (−j 7 x)+ (−j7)(j3)= 0

x^2 +j 3 x+ (−j 7 x)+ (−j×j)× 7 × 3 = 0

x^2 −j 4 x+ 21 = 0

This is a twist we haven’t seen yet! One of the coefficients is imaginary.

6. Once again, for reference, let’s state the general polynomial standard form of a quadratic
   equation:

ax^2 +bx+c= 0

In the equation we derived in solution to Prob. 5, we have a= 1, b=−j4, and c= 21. Plug-

ging these values into the quadratic formula and then working out the arithmetic, we get

x= [−b± (b^2 − 4 ac)1/2] / (2a)

= {−(−j4)± [(−j4)^2 − 4 × 1 × 21]1/2} / (2 × 1)

Now let’s be careful with (−j4)^2. This is the sort of expression that can easily cause us to make

a mistake. We can break it down like this, and then solve, being careful with the signs:

(−j4)^2 = (− 1 ×j× 4)^2

= (−1)^2 ×j^2 × 42

= 1 × (−1)× 16

=− 16

Chapter 23 677