Recognise equations that become quadratic after substitution
Introduce a repeated expression as a new variable, solve its quadratic and return to the original domain.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Paper 1 is part of every route. Written practice is self-checked; it does not award official examination marks.
Before you begin
- Solve quadratic equations and apply index laws.
- Know that a real principal square root is non-negative and a denominator cannot be zero.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If u = x², what is x⁴ in terms of u?
What you will learn
- Recognise quadratic structure in powers, square roots and reciprocal expressions.
- Undo a substitution without losing signs or accepting impossible intermediate values.
Look for a quantity and its square
In x⁴ − 13x² + 36 = 0, the repeated quantity is x². Let u = x² to obtain u² − 13u + 36 = 0. The roots u = 4 and 9 are not yet x-values: undo the substitution to obtain x = ±2 and ±3. Since u = x² for real x, an intermediate negative u would have to be rejected.
In x + √x − 6 = 0, use u = √x, with u ≥ 0 and x = u². The quadratic u² + u − 6 = 0 gives u = 2 or −3, but only u = 2 is permitted. Therefore x = 4. Squaring the rejected u to manufacture x = 9 would not satisfy the original equation, because √9 is +3 rather than −3.
Keep restrictions when clearing denominators
For x + 6/x = 5, first record x ≠ 0. Multiplying by x gives x² − 5x + 6 = 0, with roots 2 and 3. Both are non-zero and satisfy the original equation. Multiplying by an expression can introduce candidates at its forbidden zeros, so the original restrictions must remain part of the check.
The same structure can occur in trigonometry: tan² θ − 3 tan θ + 2 = 0 becomes u² − 3u + 2 = 0 with u = tan θ. This gives tan θ = 1 or 2, not θ = 1 or 2. To finish an angle question, use the relevant trig methods, exclude undefined tangent values and find every angle in the specified interval. For 0° ≤ θ < 180°, these give 45° and about 63.4°. More detailed trig teaching follows in section 1.5.
Pause and explain
Let u = √x. If the quadratic gives u = 2 or −3, which value can be used?
Worked example
Solve x⁴ − 13x² + 36 = 0 over real x.
Show the worked solution
- Let u = x² ≥ 0 to get u² − 13u + 36 = 0.
- Factor (u − 4)(u − 9) = 0, so u = 4 or 9.
- Undo x² = 4 and x² = 9, giving both square-root signs; substitute into the original to check.
Answer x = −3, −2, 2 or 3.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve x⁴ − 10x² + 9 = 0 over real x.
Give me a hint
Let u = x², factor the quadratic, then take both square-root signs.
Compare my reasoning
- Use u = x² ≥ 0 to obtain u² − 10u + 9 = 0.
- Factor (u − 1)(u − 9) = 0; both intermediate roots are non-negative.
- Undo the substitution: x² = 1 or 9 gives x = ±1 or ±3.
x = −3, −1, 1 or 3.
Look for these in your work
- Identify the correct quadratic substitution.
- Return to every permitted real x-value.
Solve x + √x − 6 = 0 over real x.
Give me a hint
Use u = √x with u ≥ 0, rather than squaring everything immediately.
Compare my reasoning
- The real domain is x ≥ 0; write x = u² and obtain u² + u − 6 = 0.
- Factor (u + 3)(u − 2) = 0 and reject u = −3 because u is a principal square root.
- For u = 2, x = 4; check 4 + 2 − 6 = 0.
x = 4 only.
Look for these in your work
- State and retain u ≥ 0.
- Check the surviving value in the original equation.
Solve x + 6/x = 5 over real x, explaining the restriction before multiplying.
Give me a hint
The expression is undefined at zero; clearing the denominator gives a quadratic.
Compare my reasoning
- Record x ≠ 0, then multiply the equation by x.
- Rearrange to x² − 5x + 6 = 0 = (x − 2)(x − 3).
- Both x = 2 and 3 are allowed; their original expressions are 2 + 3 and 3 + 2, each equal to 5.
x = 2 or x = 3, with x ≠ 0 retained.
Look for these in your work
- Retain the original denominator restriction.
- Verify both candidates before accepting them.
Common mistakes
- Roots in u are intermediate values; undo the substitution.
- A square-root substitution has u ≥ 0, while x² = a > 0 gives both x signs.
- Keep denominator restrictions and check candidates in the original equation.
What can you explain now?
Solve x⁴ + x² − 2 = 0 over real x.
Compare with the explanation
x = −1 or x = 1.
With u = x² ≥ 0, (u + 2)(u − 1) = 0. Reject u = −2 and solve x² = 1.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Contrast u = x² with u = √x: what restrictions apply, and how do you return to x? Then describe what remains after a quadratic in tan θ is solved.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.