Nonlinear equations and mixed systems
Solve quadratic, polynomial, absolute-value, radical and rational equations, then verify candidates.
What you will tackle
- Solve quadratics by factoring, completing the square and the quadratic formula
- Use the discriminant to classify real roots
- Solve factored polynomial, absolute-value, radical and rational equations with restrictions
- Solve mixed linear–nonlinear systems and interpret intersections
- Rearrange multivariable formulas
Different structures call for different methods
For ax² + bx + c = 0, x = (−b ± √(b² − 4ac))/(2a), with a ≠ 0. A positive, zero or negative discriminant gives two, one repeated or no real roots. In a factored polynomial, set each factor to zero. An equation |x − a| = b with b > 0 has the two cases x − a = b and x − a = −b; b < 0 gives no real solution.
Transformed equations need a final check
A rational denominator cannot be zero. Squaring a radical equation may introduce an extraneous candidate, so check the original. To combine y = x + 2 and y = x², substitute to solve x² = x + 2; each root gives a graph intersection. Rearranging a literal formula follows the same inverse operations, retaining restrictions on denominators and roots.
Build your notes, then practise
Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.
Guided practice
Guided 1
Solve x² − 2x − 1 = 0, |x − 4| = 3, and y = x² with y = x + 2.
Show a starting hint
An absolute value is a distance and can produce two positions.
Compare with the explained answer
- The quadratic formula gives (2 ± √8)/2 = 1 ± √2.
- The absolute-value cases give x = 7 or x = 1.
- x² − x − 2 = (x − 2)(x + 1), so intersections are (2,4) and (−1,1).
Answer / response criteria: 1 ± √2; 1 or 7; intersections (2,4) and (−1,1).
Guided 2
Make h the subject of V = πr²h with r > 0. Then solve the intersections of y = x² and y = 2x + 3.
Compare with the explained answer
- h = V/(πr²), with nonnegative V for the physical model.
- Set x² = 2x + 3 and factor (x − 3)(x + 1) = 0.
- At x = 3 and −1 the y values are 9 and 1.
Answer / response criteria: h = V/(πr²); intersections (3,9) and (−1,1).
Independent practice
Independent 1
Solve (x − 1)(x + 2)(x − 4) = 0; solve √(x + 6) = x; classify x² − 6x + k = 0 using k.
Compare with the explained answer
- The product is zero at x = 1, −2, 4.
- Squaring gives x² − x − 6 = 0, with candidates 3 and −2; only 3 satisfies the original.
- The discriminant is 36 − 4k: two real roots for k < 9, one for k = 9, none for k > 9.
Answer / response criteria: 1, −2, 4; x = 3 only; root counts depend on k as stated.
Independent 2
Make r the subject of A = 4πr² for a physical sphere. Solve y = x² − 2 with y = −x.
Compare with the explained answer
- For A ≥ 0, the physical radius is r = √(A/(4π)).
- Equating gives x² + x − 2 = 0 = (x + 2)(x − 1).
- The intersections are (−2,2) and (1,−1).
Answer / response criteria: r = √(A/(4π)); intersections (−2,2),(1,−1).
Application practice
Application 1
For a cylinder V = πr²h, make r the subject and find it when V = 80π and h = 5. Separately solve 6/(x − 1) = 3 and record the excluded input.
Compare with the explained answer
- Divide by πh to get r² = V/(πh); a physical radius is the nonnegative square root.
- r = √(80π/(5π)) = 4.
- For the rational equation x ≠ 1; multiplying gives 6 = 3(x − 1), so x = 3.
Answer / response criteria: r = √(V/(πh)), giving 4; rational solution x = 3, with x = 1 excluded.
Application 2
A path is modelled by height h = t² and a sensor line by h = 3t + 4. Find intersections and retain nonnegative time. For a separate cylinder with V = 80π and h = 5 > 0, recover r and state the square-root conditions.
Compare with the explained answer
- t² − 3t − 4 = (t − 4)(t + 1), giving t = 4 or −1.
- The positive-time intersection is (4,16).
- For V ≥ 0 and h > 0, r = √(V/(πh)) = 4; the nonnegative root is the physical radius.
Answer / response criteria: The physical path intersection is t = 4, h = 16; cylinder radius 4.
What can you explain now?
A path is modelled by height h = t² and a sensor line by h = 3t + 4. Find intersections and retain nonnegative time. For a separate cylinder with V = 80π and h = 5 > 0, recover r and state the square-root conditions.
Compare with the explanation
The physical path intersection is t = 4, h = 16; cylinder radius 4.
Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.
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Source and review status
Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.
Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.
Official source specification / guide