Algebra Know-It-ALL

4 × 9 + 18 + (−62)+ 8 = 0

36 + 18 + (−62)+ 8 = 0

54 + (−62)+ 8 = 0

− 8 + 8 = 0

0 = 0

Next, we check (−5,−39) in the second original equation:

3 x^2 +y+ 5 x− 11 = 0

3 × (−5)^2 + (−39)+ 5 × (−5)− 11 = 0

3 × 25 + (−39)+ (−25)− 11 = 0

75 + (−39)+ (−25)− 11 = 0

36 + (− 25) − 11 = 0

11 − 11 = 0

0 = 0

Completing the job, we check (3, −31) in that equation:

3 x^2 +y+ 5 x− 11 = 0

3 × 32 + (−31)+ 5 × 3 − 11 = 0

3 × 9 + (−31)+ 15 − 11 = 0

27 + (−31)+ 15 − 11 = 0

− 4 + 15 − 11 = 0

11 − 11 = 0

0 = 0

Are you confused?

You might ask, “Is it possible for a two-by-two system, in which one or both of the equations is quadratic,

to have imaginary or complex solutions?” The answer is “Yes. In that case, when we use the quadratic

formula, those solutions will appear.” You’ll see this happen in later in this chapter.

Here’s a challenge!

Suppose that a 1 ,a 2 ,b 1 ,b 2 ,c 1 , and c 2 are real numbers, and neither a 1 nor a 2 is equal to 0. Consider these

two functions of x:

y=a 1 x^2 +b 1 x+c 1

and

y=a 2 x^2 +b 2 x+c 2

Derive a general formula for solving this system.

454 More Two-by-Two Systems