Algebra Know-It-ALL

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

Whenx=−1/2, we have

y= 2 x^3

= 2 × (−1/2)^3

= 2 × (−1/8)

=−1/4

The solutions to the system are (x,y)= (0,0) and (x,y)= (−1/2,−1/4). The solution (0,0)

has multiplicity 2. Consider again the cubic that we got by mixing:

2 x^3 +x^2 = 0

This factors into

(x)(x)(2x+ 1) = 0

The root x= 0 occurs once for each factor of x, or twice in total. That means the corre-

sponding solution (0,0) has multiplicity 2 in the two-by-two system.

8. First, let’s check (0,0) in the original two-variable quadratic equation:

x^2 +y= 0

02 + 0 = 0

0 + 0 = 0

0 = 0

Next, let’s check (−1/2,−1/4) in that equation:

x^2 +y= 0

(−1/2)^2 + (−1/4)= 0

1/4+ (−1/4)= 0

0 = 0

Next, let’s check (0,0) in the original two-variable cubic equation:

2 x^3 −y= 0

2 × 03 − 0 = 0

2 × 0 − 0 = 0

0 − 0 = 0

0 = 0