What form does completing the square produce?

Completing the square rewrites y = ax² + bx + c in the form y = a(x + p)² + q. When a = 1, p is half of b and q = c − p². When a ≠ 1, factor out a first. This form reveals the turning point (−p, q) and is useful for solving equations, sketching graphs, and proving results using (x + p)² ≥ 0.

1. What is Completing the Square?

Completing the square is a method to rewrite a quadratic in a different form:

$y = ax^2 + bx + c \quad \Rightarrow \quad y = a(x + p)^2 + q$

This form is useful for:

Solving quadratic equations
Finding the turning point (vertex) of the graph
Creating the equation when given the turning point
Proving results (since ((x + p)^2 \geq 0) for all (x))

2. When a = 1

For (y = x^2 + bx + c):

Step 1: (p) is half of the coefficient of (x): (p = \frac{b}{2})

Step 2: (q = c - p^2)

Result: (y = (x + p)^2 + q)

The turning point is ((-p, q)).

Example: (y = x^2 + 6x + 5)

(p = \frac{6}{2} = 3), (q = 5 - 9 = -4)

(y = (x + 3)^2 - 4) → Turning point at ((-3, -4))

3. When a ≠ 1

4. Worked Example: a = 1

5. Worked Example: a ≠ 1

6. Using (x + p)² ≥ 0 for Proofs

7. Examiner Tips

Completing the square

Summary

1. What is Completing the Square?

Completing the square is a method to rewrite a quadratic in a different form:

$y = ax^2 + bx + c \quad \Rightarrow \quad y = a(x + p)^2 + q$

This form is useful for:

Solving quadratic equations
Finding the turning point (vertex) of the graph
Creating the equation when given the turning point
Proving results (since ((x + p)^2 \geq 0) for all (x))

2. When a = 1

For (y = x^2 + bx + c):

Step 1: (p) is half of the coefficient of (x): (p = \frac{b}{2})

Step 2: (q = c - p^2)

Result: (y = (x + p)^2 + q)

The turning point is ((-p, q)).

Example: (y = x^2 + 6x + 5)

(p = \frac{6}{2} = 3), (q = 5 - 9 = -4)

(y = (x + 3)^2 - 4) → Turning point at ((-3, -4))

3. When a ≠ 1

For (y = ax^2 + bx + c) with (a \neq 1):

Step 1: Factor (a) out of the (x^2) and (x) terms:

$y = a\left(x^2 + \frac{b}{a}x\right) + c$

Step 2: Complete the square inside the bracket. Here (p = \frac{b}{2a}) (half of (b/a)):

$y = a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c$

Step 3: Simplify to get (y = a(x + p)^2 + q).

The turning point is ((-p, q)).

4. Worked Example: a = 1

Write (y = x^2 - 4x + 7) in the form ((x + p)^2 + q).

(p = \frac{-4}{2} = -2), so we need ((x - 2)^2)

(x^2 - 4x = (x - 2)^2 - 4)

(y = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3)

Turning point: (2, 3)

5. Worked Example: a ≠ 1

Write (y = 2x^2 + 12x + 5) in the form (a(x + p)^2 + q).

Factor out 2: (y = 2(x^2 + 6x) + 5)

Complete square: (x^2 + 6x = (x + 3)^2 - 9)

(y = 2[(x + 3)^2 - 9] + 5 = 2(x + 3)^2 - 18 + 5 = 2(x + 3)^2 - 13)

Turning point: (−3, −13)

6. Using (x + p)² ≥ 0 for Proofs

A squared term is always (\geq 0). So:

(a(x + p)^2 + q) has minimum (q) when (a > 0) (at (x = -p))
(a(x + p)^2 + q) has maximum (q) when (a < 0) (at (x = -p))

This can be used to prove minimum/maximum values and inequalities.

7. Examiner Tips

The question may explicitly ask you to complete the square, or remind you of the form
Sometimes you must spot that completing the square is needed (e.g. for turning points)
When (a \neq 1), always factor (a) out of the (x^2) and (x) terms first

For (y = ax^2 + bx + c) with (a \neq 1):

Step 1: Factor (a) out of the (x^2) and (x) terms:

$y = a\left(x^2 + \frac{b}{a}x\right) + c$

Step 2: Complete the square inside the bracket. Here (p = \frac{b}{2a}) (half of (b/a)):

$y = a\left(x + \frac{b}{2a}\right)^2 - a\left(\frac{b}{2a}\right)^2 + c$

Step 3: Simplify to get (y = a(x + p)^2 + q).

The turning point is ((-p, q)).