1.1 Quadratics

2026 Syllabus Objectives

By the end of this topic, you should be able to:

  1. Carry out the process of completing the square for a quadratic polynomial ax² + bx + c and use a completed square form (e.g., to locate the vertex of the graph of y = ax² + bx + c or to sketch the graph)

  2. Find the discriminant of a quadratic polynomial ax² + bx + c and use the discriminant (e.g., to determine the number of real roots of the equation ax² + bx + c = 0). Knowledge of the term 'repeated root' is included.

  3. Solve quadratic equations and quadratic inequalities in one unknown (by factorising, completing the square, and using the formula)

  4. Solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic (e.g., x + y + 1 = 0 and x² + y² = 25)

  5. Recognise and solve equations in x which are quadratic in some function of x (e.g., x⁴ – 5x² + 4 = 0, 6x + √x – 1 = 0)

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1. Completing the Square

What is completing the square?

Completing the square is a method of rewriting a quadratic expression (one that contains x²) into a special form that makes it easier to work with. Instead of writing the quadratic as ax² + bx + c, we write it as a(x - h)² + k.

This special form is called completed square form and it's very useful because:

  • The point (h, k) tells us the turning point (also called the vertex) of the graph
  • It makes sketching the graph much easier
  • It helps us find the inverse function of a quadratic (we'll see this later)

The Standard Forms:

  • Standard form: f(x) = ax² + bx + c
  • Completed square form: f(x) = a(x - h)² + k

where (h, k) represents the coordinates of the turning point.

Step-by-Step Method:

When the coefficient of x² is not 1, follow these steps:

Step 1: Take out the coefficient of x² as a factor from the first two terms

Step 2: Inside the bracket, halve the coefficient of x, then square it. Add and subtract this value inside the bracket.

Step 3: Expand and simplify, keeping one squared term and combining the constant terms.

Worked Example:

Express f(x) = 2x² - 12x + 10 in the form a(x - b)² + c

Solution:

Step 1: Take out the 2 from the first two terms: f(x) = 2[x² - 6x] + 10

Step 2: Half of -6 is -3. Square it to get 9. Add and subtract 9: f(x) = 2[x² - 6x + 9 - 9] + 10

Step 3: Rewrite as a perfect square and simplify: f(x) = 2[(x - 3)² - 9] + 10 f(x) = 2(x - 3)² - 18 + 10 f(x) = 2(x - 3)² - 8

Therefore: a = 2, b = 3, c = -8

The turning point is at (3, -8).

Using Completed Square Form to Sketch Graphs:

Once you have the completed square form a(x - h)² + k, you can quickly sketch the graph:

  1. Turning point: The coordinates are (h, k)
  2. Shape: If a > 0, the parabola opens upwards (U-shape). If a < 0, it opens downwards (n-shape)
  3. y-intercept: Substitute x = 0 into the original equation to find where the graph crosses the y-axis

For our example f(x) = 2(x - 3)² - 8:

  • Turning point: (3, -8)
  • Shape: U (because a = 2 > 0)
  • y-intercept: f(0) = 2(0)² - 12(0) + 10 = 10

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