Algebra Know-It-ALL

528 Review Questions and Answers

When we subtract the quantity (a^2 x^2 +b^2 ) from each side, we get

2 abx=− 2 abx

We can add 2abx to each side and then divide through by 4ab (which is “legal” because we’ve

been told that a≠ 0 and b≠ 0, so we can be sure that 4ab≠ 0). When we do that, we obtain

x= 0 as the only root of the “mixed quadratic.” We can substitute this value for x into the

original equations, obtaining y=b^2 in both cases. Therefore, the system has the single solu-

tion (0,b^2 ).

Question 27-8

Imagine that we’re trying to solve a general two-by-two system of equations, and we mix them

to get a single equation in one variable. When we solve that “mixed” equation, we get two dif-

ferent roots, one of which has multiplicity 1, and the other of which has multiplicity 2. What

does this say about the solutions of the original system?

Answer 27-8

The solutions of the original two-by-two system have the same multiplicity pattern as the

roots of the “mixed” equation. In this case, that means there is one solution with multiplicity

1, and another solution with multiplicity 2.

Question 27-9

Consider the following pair of equations:

y= (x+ 1)^3

and

y=x^3 + 2 x^2 +x

How can we find the real solutions to this two-by-two system?

Answer 27-9

Let’s multiply out the first equation to get it into polynomial standard form. When we do

that, the system becomes

y=x^3 + 3 x^2 + 3 x+ 1

and

y=x^3 + 2 x^2 +x

We can mix the right sides of these equations, getting

x^3 + 3 x^2 + 3 x+ 1 =x^3 + 2 x^2 +x