Loav · VCE Mathematical Methods · Units 3 & 4
F06 · 2026 curriculum · Original teaching guide
1. Solving is a chain of logically justified steps
An equation asks which allowed inputs make two expressions equal. An inequality asks where one expression is larger or smaller than another. In both cases, the goal is a solution set, not merely a line of rearranged algebra. Begin by identifying the domain and the requested kind of answer: exact values, approximations, intervals, parameter conditions or a contextual interpretation.
Some operations preserve equivalence automatically, such as adding the same expression to both sides. Others need conditions. Dividing by an expression assumes it is non-zero; taking a real logarithm requires positivity; squaring can turn opposite-sign quantities into the same value. Distinguish a step that gives an equivalent equation from a step that only produces candidates. The final substitution check is essential whenever a non-reversible operation has been used.
Before expanding, inspect structure. A zero product suggests factorisation. A repeated exponential or circular expression may suggest a substitution. A quadratic suggests the discriminant or completed square. A pair of equations suggests elimination or substitution. A graph can reveal likely intersections, but the algebra explains whether all intersections have been found and whether they satisfy the original restrictions.
2. Squaring requires a sign check
The principal square root is always non-negative. Thus an equation √A = B requires both A ≥ 0 and B ≥ 0 before squaring. The squared equation A = B² alone does not enforce B ≥ 0. It can therefore have extra solutions. State the sign condition first, solve the squared equation, and reject candidates that violate the condition or fail the original equation.
Worked example 1: reject the extra square-root candidate
Solve √(2x + 5) = x − 1. The radicand requires x ≥ −5/2, while the non-negative left side requires x − 1 ≥ 0. Together these give x ≥ 1. Squaring yields 2x + 5 = (x − 1)², hence x² − 4x − 4 = 0.
The quadratic formula gives x = 2 ± 2√2. The smaller candidate is below 1 and is rejected. For x = 2 + 2√2, the right side is 1 + 2√2, which is positive, and its square is 9 + 4√2 = 2x + 5. Thus the original square-root equation holds. The only solution is 2 + 2√2.
The same caution applies when clearing a denominator. Record every excluded input first, multiply through, solve the resulting equation, then retain the restrictions. Multiplying by a denominator can be safe for an equality at permitted inputs, but its sign creates an additional issue for inequalities. Avoid multiplying an inequality by an expression of unknown sign unless you split into the necessary cases.
3. Inequalities need intervals, not just boundary roots
To solve a polynomial inequality, move everything to one side and factor where possible. The zeros divide the number line into intervals on which the sign cannot change without passing through another zero. Determine the sign in each interval using test values or factor signs. Then include or exclude the boundary roots according to whether the inequality allows equality.
Multiplying or dividing an inequality by a negative number reverses its direction. Multiplying by a positive number preserves it. If the multiplier contains the unknown, you may not know its sign in advance. A sign diagram avoids that ambiguity by examining the original expression on each interval. Do not “cancel” a variable factor in an inequality and assume the solution set stays unchanged.
Worked example 2: a quadratic inequality with boundary inclusion
Solve x² − x − 6 ≥ 0. Factor to obtain (x − 3)(x + 2) ≥ 0. The critical inputs are −2 and 3. For x < −2, both factors are negative, so their product is positive. For −2 < x < 3, the factors have opposite signs, so their product is negative. For x > 3, both are positive.
The equality is allowed, so include both roots. The solution is (−∞, −2] ∪ [3, ∞). A graph of the upward-opening parabola agrees: the graph is on or above the x-axis outside the roots. If the original domain were [0, 5], the final answer would instead be [3, 5], obtained by intersecting the algebraic solution with that domain.
4. A parameter represents a family of problems
A parameter is held fixed while the variable changes, but different parameter values define different members of a family. A question such as “for which k are there two solutions?” asks you to classify that family. Do not choose one convenient numerical value of k and treat its graph as proof for every value. Instead, identify the algebraic feature that controls the change.
For a genuine quadratic ax² + bx + c = 0, with a ≠ 0, the discriminant Δ = b² − 4ac determines the number of distinct real roots: two when Δ > 0, one repeated root when Δ = 0, and none when Δ < 0. This classification assumes the leading coefficient is non-zero. If a depends on a parameter, test the values that make a = 0 separately before using quadratic reasoning.
A discriminant counts unrestricted real roots. A function domain or context can remove some of them. Two real roots need not both be positive, lie in a given interval, or satisfy logarithm conditions. For a quadratic with real roots r and s, the relationships r + s = −b/a and rs = c/a can help determine their signs. Use those relationships together with the discriminant rather than treating either test as sufficient alone.
Worked example 3: two intersections in a restricted domain
For which real k do y = x² and y = 2kx − 3 have two distinct intersections with x > 0? Equating gives x² − 2kx + 3 = 0. Its discriminant is 4k² − 12, so two distinct real roots require k < −√3 or k > √3.
The root product is 3, so the two real roots have the same sign. Their sum is 2k. For k < −√3 the sum is negative, so both roots are negative and neither is allowed. For k > √3 the sum is positive, so both are positive. Therefore the required condition is k > √3. At k = √3 there is only one distinct intersection, so the boundary is excluded.
5. Degenerate cases are often the point of the question
Consider (k − 1)x = k + 2. If k ≠ 1, division gives x = (k + 2)/(k − 1). But at k = 1 the equation becomes 0x = 3, which has no solution. Writing the fraction without its condition hides an entire case. If both sides had become zero, every real x would satisfy the equation instead.
For simultaneous linear equations, elimination can produce the same three possibilities: a unique value, a contradiction, or an identity. A contradiction such as 0 = 5 means no common solution. An identity such as 0 = 0 means the elimination step has not singled out one point; inspect the remaining equation to describe the full solution set. When two equations describe the same line, there are infinitely many common points.
For example, the equations y = 2x + 1 and y = kx + 1 give (2 − k)x = 0. If k ≠ 2, the unique intersection is (0, 1). If k = 2, the two equations coincide, giving every point on y = 2x + 1. Ordinary elimination and a separate parameter case explain the geometry.
6. Exact, approximate and contextual answers serve different purposes
An exact answer retains integers, fractions, radicals, logarithms or exact angles. An approximate answer uses a requested numerical precision. Keep the exact expression through intermediate steps, then round once. If a context asks for a whole number of objects or a time satisfying a threshold, ordinary rounding may not be appropriate: interpret which side of the threshold is required.
For a technology-assisted equation, state the equation being solved and the interval or restrictions supplied. Use a graph to locate possible solutions, then a numerical solver to refine them where exact algebra is impractical. Check the residual by substituting the approximation back into the original equation. A small residual supports numerical accuracy, but it does not by itself prove that no other solutions exist elsewhere.
Practise, then check
- Solve √(x + 6) = x exactly, showing which candidate is rejected.
- Solve (x + 4)(x − 1) < 0 and then restrict your answer to x ≥ 0.
- Solve (k − 2)x = k² − 4 for every real parameter k.
- Classify the number of distinct real roots of x² − 2kx + 4 = 0 as k varies.
- Find all k for which (k − 1)x² + 2x + 1 = 0 has exactly one distinct real solution. Remember the non-quadratic case.
- Solve the simultaneous equations y = 3x + 2 and y = kx − 1, separating the exceptional parameter value.
- Find all real solutions of e²ˣ − eˣ − 6 = 0 and explain why one quadratic candidate is inadmissible.
Full reasoned answers
- The right side must be non-negative, so x ≥ 0. Squaring gives x + 6 = x², or (x − 3)(x + 2) = 0. Reject x = −2 because it violates x ≥ 0 and gives √4 = 2 rather than −2. The solution x = 3 satisfies √9 = 3.
- The roots are −4 and 1. The product is negative between them, so the unrestricted answer is (−4, 1). Intersecting with x ≥ 0 gives [0, 1). Include zero because the product there is −4; exclude 1 because the inequality is strict.
- Factor the right side as (k − 2)(k + 2). For k ≠ 2, divide by k − 2 to obtain x = k + 2. For k = 2, the original equation is 0 = 0, so every real x is a solution. Cancellation without the separate case would miss this result.
- The discriminant is 4k² − 16 = 4(k² − 4). There are two distinct real roots for k < −2 or k > 2, one repeated root for k = −2 or 2, and no real roots for −2 < k < 2. The leading coefficient is always 1, so no degree-degeneration case is needed.
- If k ≠ 1, the equation is quadratic and Δ = 4 − 4(k − 1) = 8 − 4k. Exactly one distinct real root occurs at k = 2, giving (x + 1)² = 0. Separately, at k = 1 the equation is 2x + 1 = 0, with the single solution −1/2. Thus k ∈ {1, 2}.
- Equating gives (3 − k)x = −3. For k ≠ 3, x = 3/(k − 3), and y = 2 + 9/(k − 3). For k = 3, the equations are parallel lines with different intercepts, or algebraically 0 = −3, so there is no solution.
- Let u = eˣ, so u > 0. Then u² − u − 6 = (u − 3)(u + 2) = 0. Reject u = −2 because an exponential is positive. The remaining equation eˣ = 3 gives x = ln 3, which satisfies the original equation.
Exam habits and technology checks
Organise a parameter solution into clearly labelled cases and state what each case implies. Use interval notation for inequalities and a set of values for isolated roots. Check exceptional values in the original equation, not in a formula obtained by dividing by something that becomes zero there. End with a sentence that answers the actual question, especially when it asks for a number of solutions or a restricted parameter range.
Curriculum and sources
Scope: VCE Mathematical Methods Units 3 and 4, current 2023–2027 study design, Functions, relations and graphs and Algebra, number and structure. See the official VCAA Mathematical Methods curriculum and VCAA Mathematical Methods FAQs, updated February 2026. This guide covers equations, inequalities, simultaneous equations and parameter cases.
Further reading: Michael Evans, Kay Lipson and David Greenwood, Cambridge Senior Mathematics VCE Mathematical Methods Units 3 & 4, second edition (2023), ISBN 9781009110495. Chapter 2, sections 2B and 2E, pp. 66–67 and 76–78; Chapter 4, sections 4A and 4H, pp. 138–147 and 183–187. Explanations, diagrams, worked examples and practice questions in this guide are original. This is an independent Loav learning resource, not a VCAA publication.
