Algebra Know-It-ALL

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

Plugging in the first root and converting the subtractions to additions, we can proceed

like this, refining the equation step-by-step:

[− 2 + (−j3)]^2 + [(5 + (−j12)]= 0

[− 2 + (−j3)][− 2 + (−j3)]+ [(5 + (−j12)]= 0

4 +j 6 +j 6 + (−j)^29 + 5 + (−j12)= 0

Keeping in mind that (−j)^2 =−1, we can simplify to

4 +j 6 +j 6 + (−9)+ 5 + (−j12)= 0

When we add all the terms on the left side, we get 0 = 0. The first root checks! Now we’ll

plug in the second root and convert the subtraction to negative addition. We can proceed

like this, step-by-step:

(2 +j3)^2 + [5 + (−j12)]= 0

(2+j3)(2+j3)+ [5 + (−j12)]= 0

4 +j 6 +j 6 +j^29 + 5 + (−j12)= 0

4 +j 6 +j 6 + (−9)+ 5 + (−j12)= 0

That’s the same equation we got when we plugged in the other root. When we add all the

terms on the left side, we get 0 = 0. The second root checks!

10. We’ve been told to convert the following equation, which consists of two trinomial
    factors, into the polynomial standard form for a quadratic:

(j 2 x+ 2 +j3)(−j 5 x+ 4 −j5)= 0

This might look daunting at first, but our assigned task is merely a matter of multiplying

things out, taking care with the signs, and not rushing it! Let’s begin by converting the

subtraction in the second factor to addition. That gives us

(j 2 x+ 2 +j3)[(−j 5 x+ 4 + (−j5)]= 0

We can use the product of sums rule, remembering that −j×j= 1. It goes like this:

(j 2 x)(−j 5 x)+j 8 x+ (j 2 x)(−j5)

+ (−j 10 x)+ 8 + (−j10)

+ (j3)(−j 5 x)+j 12 + (j3)(−j5)

= 0

Simplifying the individual terms, we get

10 x^2 +j 8 x+ 10 x

+ (−j 10 x)+ 8 + (−j10)

+ 15 x+j 12 + 15

= 0