Learning Objectives
Select a nonlinear-algebra method from the form of an expression or equation.
Connect factors, roots, graph features, and model parameters.
Keep sign and domain conditions in a complete solution.
Prerequisites: the four nonlinear algebra lessons in this section. Use exact values unless rounding is requested. The questions are original skill practice rather than official exam items.
1. Match Structure with a Method
Structure | Method |
|---|---|
Powers with the same base | Apply the relevant exponent rule |
Radicals with square factors | Extract squares, then combine like radicals |
A polynomial sum | Check common factors and standard patterns |
A quadratic equation | Factor, isolate a square, or use the formula |
A constant percentage change | Use a repeated multiplier |
A rational expression | Find excluded inputs before simplifying |
A method can change how an expression looks without changing its values on the original domain. Keep that domain attached. Solving an equation, in contrast, asks which values satisfy a condition and may produce several answers.
2. Connect a Factorization with a Graph
Worked Example: Read Several Features
Consider f(x) = x² − 2x − 8. Factoring gives (x − 4)(x + 2), so its x-intercepts are (4, 0) and (−2, 0).
Completing the square gives f(x) = (x − 1)² − 9. The vertex is (1, −9), and the axis of symmetry is x = 1.
The roots' average is (4 + (−2))/2 = 1, confirming the symmetry line. The y-intercept is (0, −8), obtained by setting x = 0.
Each form exposes a different feature while representing the same function on all real inputs.
3. Preserve Conditions while Simplifying
For (x² − 4)/(x − 2), factoring and cancellation produce x + 2 for x ≠ 2. If the expression is set equal to 4, solving the simplified equation would suggest x = 2, but that input is excluded. The original equation therefore has no solution.
Common Mistakes
Cancellation can expose an apparent solution that the original denominator excludes.
A principal square root is nonnegative; a squared equation can have two signed solutions.
A percentage multiplier applies to the current amount at each stage.
4. Mixed Practice
1. Simplify (2x³)²/x² for x ≠ 0.
2. Simplify √75 + √12.
3. Factor x² − x − 20 and use the result to solve x² − x − 20 = 0.
4. Solve 3x² − 12 = 0.
5. Find the vertex and range of f(x) = 2(x − 4)² + 1 on all real inputs.
6. A quantity starts at 50 and doubles each stage. Give a model with initial stage 0 and find its value at stage 4.
7. Solve (x² − 9)/(x − 3) = 8, respecting the domain.
8. Solve |2x − 1| = 7.
Worked Solutions
1. Square the coefficient and multiply the exponent: (2x³)² = 4x⁶. Dividing by x² gives 4x⁴, with x ≠ 0 retained.
2. √75 = 5√3 and √12 = 2√3. Their sum is 7√3.
3. The factors are (x − 5)(x + 4), since −5 and 4 multiply to −20 and add to −1. The equation has solutions x = 5 and x = −4.
4. Divide by 3 to get x² = 4. Both x = 2 and x = −2 satisfy the equation.
5. The vertex is (4, 1). Since the square term is nonnegative and its coefficient is positive, the range is [1, ∞).
6. Q(t) = 50·2ᵗ, so Q(4) = 50(16) = 800.
7. Exclude x = 3, then simplify to x + 3 = 8. The candidate x = 5 is allowed. In the original expression, (25 − 9)/(5 − 3) = 8, so it checks.
8. Solve 2x − 1 = 7 or 2x − 1 = −7. The solutions are x = 4 and x = −3.
5. Summary
- Choose a method from the expression's structure.
- Connect algebraic forms with graph features and model meanings.
- Keep excluded inputs and sign conditions visible.
- Check solutions in the original equation rather than only a simplified version.