Next, let’s check [j,(1−j3)] in the original quadratic equation:
2 x^2 − 3 x−y+ 3 = 0
2 j^2 − 3 j− (1 −j3)+ 3 = 0
2 × (−1)+ (−j3)+ [−(1−j3)]+ 3 = 0
− 2 + (−j3)+ (−1)+j 3 + 3 = 0
−j 3 +j 3 + (−2)+ (−1)+ 3 = 0
0 = 0
Finally, let’s check [−j,(1+j3)] in that equation:
2 x^2 − 3 x−y+ 3 = 0
2(−j)^2 − 3(−j)− (1 +j3)+ 3 = 0
2 × (−1)+j 3 + [−(1+j3)]+ 3 = 0
− 2 +j 3 + (−1)+ (−j3)+ 3 = 0
j 3 + (−j3)+ (−2)+ (−1)+ 3 = 0
0 = 0
- Here are the two equations in their original forms:
x^2 +x−y=− 1
and
x^2 − 2 x−y= 2
We can morph these into functions of x, obtaining
y=x^2 +x+ 1
and
y=x^2 − 2 x− 2
When we mix the right sides of these equations, we obtain
x^2 +x+ 1 =x^2 − 2 x− 2
Adding the quantity (−x^2 + 2 x+ 2) to each side gives us
3 x+ 3 = 0
which resolves to x=−1. That’s the only root of the equation we got by morphing and
mixing. To obtain the y-value for the solution to the two-by-two system, we can plug the
Chapter 27 703
