Algebra Know-It-ALL

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

Let a be the coefficient of x^2 , let b be the coefficient of x, and let c be the stand-alone

constant. Then a= 2, b= 0, and c= 2. The quadratic formula tells us that

x= [−b± (b^2 − 4 ac)1/2] / (2a)

= [0 ± (0^2 − 4 × 2 × 2)1/2] / (2 × 2)

=± (−16)1/2 / 4

=±j4 / 4

=±j

The roots are x=j or x=−j. We can plug these into either of the original functions to get

they-values. This time, let’s use the quadratic function. When x=j, we have

y= 2 j^2 − 3 j+ 3

= 2 × (−1)−j 3 + 3

=− 2 −j 3 + 3

= 1 −j 3

Whenx=−j, we have

y= 2(−j)^2 − 3(−j)+ 3

= 2 × (−1)+j 3 + 3

=− 2 +j 3 + 3

= 1 +j 3

The solutions to the system are (x,y)= [j,(1−j3)] and (x,y)= [−j,(1+j3)].

4. After we plug in the values, let’s convert subtractions to negative additions to be sure we
   keep the signs in order. First, let’s check [j,(1−j3)] in the original linear equation:

3 x+y− 1 = 0

3 j+ (1 −j3)− 1 = 0

j 3 + 1 + (−j3)+ (−1)= 0

j 3 + (−j3)+ 1 + (−1)= 0

0 = 0

Next, let’s check [−j,(1+j3)] in that equation:

3 x+y− 1 = 0

3(−j)+ (1 +j3)− 1 = 0

−j 3 + 1 +j 3 + (−1)= 0

−j 3 +j 3 + 1 + (−1)= 0

0 = 0