Algebra Know-It-ALL

Solution

Both of these equations are presented as functions of x, so we can mix the right sides directly without hav-
ing to manipulate. When we do that, we get

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

The next several steps are tricky. It’s easy to make mistakes with the signs and the grouping. To minimize
the risk of error, let’s do negative additions, rearrange and regroup the symbols, and change the negative
additions to subtractions after we’ve put everything in the proper order. We can add the negatives of a 2 x^2 ,
b 2 x, and c 2 to each side, getting

a 1 x^2 +b 1 x+c 1 + (−a 2 x^2 )+ (−b 2 x)+ (−c 2 )= 0

Using the commutative law for addition, we can rewrite this as

a 1 x^2 + (−a 2 x^2 )+b 1 x+ (−b 2 x)+c 1 + (−c 2 )= 0

The distributive law of multiplication over addition allows us to group the coefficients and the constants
to get

[a 1 + (−a 2 )]x^2 + [b 1 + (−b 2 )]x+ [c 1 + (−c 2 )]= 0

Changing the negative additions back to subtractions, we obtain

(a 1 −a 2 )x^2 + (b 1 −b 2 )x+ (c 1 −c 2 )= 0

We can solve this equation with the quadratic formula. For reference, here it is, yet one more time, in its
classical form. (Have you memorized it yet?) When we see the quadratic equation

ax^2 +bx+c= 0

the roots are given by

x= [−b± (b^2 − 4 ac)1/2] / (2a)

To solve the equation at hand, we can make these substitutions in the classical version of the quadratic
formula:

• Write (a 1 −a 2 ) in place of a


• Write (b 1 −b 2 ) in place of b


• Write (c 1 −c 2 ) in place of c

When we make these changes in “copy-and-paste” fashion, we get

x= {−(b 1 −b 2 )± [(b 1 −b 2 )^2 − 4(a 1 −a 2 )(c 1 −c 2 )]1/2} / [2(a 1 −a 2 )]

We can simplify this slightly by getting rid of the minus sign in the first expression on the right side of the
equals sign, and then reversing the positions of b 1 and b 2 inside the parentheses. We can also multiply out

Two Quadratics 455