PSAT/NMSQT Math · Lesson 2 of 4
Nonlinear equations and equivalent expressions
Recognize when a relationship is quadratic, exponential, or rational, and simplify without changing the domain.
Choosing the structure
Constant additive change suggests a linear model. Constant multiplicative change suggests an exponential model. A product of two linear factors suggests a quadratic, whose graph is a parabola and whose zeros are the roots of those factors.
Equivalent expressions agree on their common domain. (x² − 1)/(x − 1) equals x + 1 only when x ≠ 1. Reporting the simplified polynomial without that restriction describes a different function at x = 1.
Solving nonlinear equations
If an equation contains a square root, isolating the radical and squaring both sides can introduce an extra solution. Substitute every candidate into the original equation and reject the ones that fail.
For a quadratic set equal to zero, the solutions are the inputs that make a factor zero. If the question asks for the vertex instead, a root is not the requested value.
Worked example
For which x is (x² − 1)/(x − 1) equal to 5, if x ≠ 1?
- For x ≠ 1 the expression equals x + 1.
- Set x + 1 = 5, so x = 4.
- x = 4 is allowed. Check: (16 − 1)/(4 − 1) = 15/3 = 5.
Why this works. Simplify within the original domain, then solve, then confirm the result was not excluded.
Check your understanding
- Decide whether a table grows by adding or by multiplying.
- Reject extraneous solutions after squaring.
- Keep denominator restrictions after canceling.