Algebra Know-It-ALL

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

This equation has the single root x= 1, which occurs with multiplicity 3. We can plug

x= 1 into either of the original functions to get the y-value. Let’s use the first function.

That gives us

y= 2 x^3 +x^2 +x− 4

= 2 × 13 + 12 + 1 − 4

= 2 × 1 + 1 + 1 − 4

= 2 + 1 + 1 − 4

= 0

The solution to the system is therefore (x,y)= (1,0), with multiplicity 3.

10. First, we check (1,0) in the first original two-variable cubic:

4 x^3 + 2 x^2 + 2 x− 2 y− 8 = 0

4 × 13 + 2 × 12 + 2 × 1 − 2 × 0 − 8 = 0

4 × 1 + 2 × 1 + 2 − 0 − 8 = 0

4 + 2 + 2 − 8 = 0

8 − 8 = 0

0 = 0

Next, we check (1,0) in the second original cubic:

3 x^3 − 2 x^2 + 4 x−y− 5 = 0

3 × 13 − 2 × 12 + 4 × 1 − 0 − 5 = 0

3 × 1 − 2 × 1 + 4 − 5 = 0

3 − 2 + 4 − 5 = 0

1 + 4 − 5 = 0

5 − 5 = 0

0 = 0

Chapter 28

1. See Table C-2, which shows selected values for the functions

y=− 3 x+ 1

and

y= 2 x^2 + 1

Bold numerals indicate the real solutions we found when we worked out Prob. 1 in Chap. 27.

The fractional values are also shown in decimal form so they’ll be easier to graph.

2. See Fig. C-4. Each horizontal increment is 1/2 unit. Each vertical increment is 3 units.