Algebra Know-It-ALL

678 Worked-Out Solutions to Exercises: Chapters 21 to 29

Now let’s substitute back in where we left off. That gives us an expression that’s still tricky.

But it can be simplified like this:

x= [−(−j4)± (− 16 − 4 × 1 × 21)1/2] / (2 × 1)

= [j 4 ± (−100)1/2] / 2

= (j 4 ±j10) / 2

This breaks down to

x= (j 4 +j10)/2 or x= (j 4 −j10)/2

which simplifies to

x=j14/2 or x=j(−6)/2

and further to

x=j7 or x=−j 3

These are the roots we chose in Prob. 4 to “manufacture” the quadratic.

7. We’ve been told to find the roots of this quadratic:

(x+ 2 +j3)(x− 2 −j3)= 0

We can convert this pair of trinomial factors to a pair of binomial factors by changing the

first subtraction in the second factor to addition, and also by grouping the terms within

the factors, as follows:

[x+ (2 +j3)][x+ (−1)(2+j3)]= 0

which can be rewritten as

[x+ (2 +j3)][x− (2 +j3)]= 0

Now we have an equation in binomial factor form, where the factors both consist of the

variable x plus or minus a numerical constant. The roots can therefore be found by solv-

ing the following two first-degree equations:

x+ (2 +j3)= 0

and

x− (2 +j3)= 0

In the top equation, we can subtract the quantity (2 +j3) from each side, getting

x=−(2+j3)

=− 2 −j 3