Algebra Know-It-ALL

700 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= 3, and c= 0. The quadratic formula tells us that

x= [−b± (b^2 − 4 ac)1/2] / (2a)

= [− 3 ± (3^2 − 4 × 2 × 0)1/2] / (2 × 2)

= (− 3 ± 3) / 4

= 0/4 or −6/4

= 0 or −3/2

The roots are x= 0 or x=−3/2. We can plug these into either of the original functions to

get the y-values. The linear function is easier. When x= 0, we have

y=− 3 x+ 1

=− 3 × 0 + 1

= 0 + 1

= 1

Whenx=−3/2, we have

y=− 3 x+ 1

=− 3 × (−3/2)+ 1

= 9/2 + 1

= 9/2 + 2/2

= 11/2

The solutions to the system are (x,y)= (0,1) and (x,y)= (−3/2,11/2).

2. First, let’s check (0,1) in the original linear equation:

3 x+y− 1 = 0

3 × 0 + 1 − 1 = 0

0 + 1 − 1 = 0

1 − 1 = 0

0 = 0

Next, let’s check (−3/2,11/2) in that same equation:

3 x+y− 1 = 0

3 × (−3/2)+ 11/2 − 1 = 0

−9/2+ 11/2 − 1 = 0

2/2− 1 = 0

1 − 1 = 0

0 = 0