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
Core coverage follows the detailed SAT skills in Appendix B of the September 2026 framework. The conditional School Day Essay is separate. The framework is maintained rather than issued as an annual subject syllabus.
Original teaching and explained self-check practice are available. Independent subject-specialist review and difficulty calibration remain pending; this is not an exam-board endorsement or a mastery certificate.
Official source specification / guide