Complete the square and interpret a quadratic model
Rewrite a quadratic in completed-square form · Solve using the resulting square equation · Identify an extremum and select feasible contextual roots
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Binomial squares
- Quadratic equations
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Rewrite a quadratic in completed-square form
- Solve using the resulting square equation
- Identify an extremum and select feasible contextual roots
Add and subtract the same square
For x²+bx+c, take half of b and square it: x²+bx+c = (x+b/2)²+c−(b/2)². Thus x²+6x+4 = (x+3)²−5. A squared term is non-negative, revealing minimum −5 at x = −3.
For a non-unit leading coefficient, factor it from the quadratic and linear terms first. A negative leading coefficient makes a downward parabola with a maximum. The turning-point input must lie in any restricted domain.
Equation and context
Setting (x+3)²−5 = 0 gives x+3 = ±√5, hence x = −3±√5. In a rectangle problem, if width w and length w+3 have area 40, then w²+3w−40 = 0; factorisation gives 5 or −8, but only the positive dimensions are feasible.
Sources for this lesson
Pause and explain
What is the minimum of (x+3)²−5 on all real x?
Worked example
Complete the square for x²+6x+4, state its minimum and solve it equal to zero.
Show the worked solution
- (x+3)²−9+4 = (x+3)²−5.
- Minimum −5 at −3. Solving gives x+3 = ±√5.
Answer (x+3)²−5; minimum −5; roots −3±√5.
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.
Complete the square for x²−8x+3 and state its minimum.
Give me a hint
Half of −8 is −4.
Compare my reasoning
- x²−8x+3 = (x−4)²−16+3.
- The square is non-negative.
(x−4)²−13; minimum −13 at x = 4.
Look for these in your work
- Keep the added/subtracted square
- Locate the zero-square input
Solve x²−8x+3 = 0 by completing the square.
Give me a hint
Use the form from the previous method.
Compare my reasoning
- (x−4)² = 13.
- x−4 = ±√13.
x = 4±√13.
Look for these in your work
- Use both square roots
- Keep exact values
A rectangular garden has width w, length w+3 and area 40 square units. Find the dimensions and explain the rejected root.
Give me a hint
Create an area equation before factorising.
Compare my reasoning
- w(w+3) = 40 gives (w+8)(w−5) = 0.
- w = 5 gives length 8; w = −8 gives negative width and length and is infeasible.
Width 5 and length 8; reject w = −8.
Look for these in your work
- Construct the quadratic
- Apply physical constraints
Common mistakes
- Subtract the square you added.
- An algebraic negative length is not feasible.
What can you explain now?
What is the minimum of (x+3)²−5 on all real x?
Compare with the explanation
−5, at x = −3.
The square is zero there and non-negative elsewhere.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
What is the minimum of (x+3)²−5 on all real x?
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.