Teaching unit · UK Year 11 (Key Stage 4 / GCSE HIGHER tier, ages 15 to 16)

The quadratic formula and completing the square

Solving any quadratic exactly, reading the discriminant, and finding a turning point from completed-square form

About three lessons of 45 to 60 minutes

Start here · hook

Factorising is a lock-pick that only fits some doors. These two methods open every quadratic.

By Year 11 you can factorise x² - 5x + 6 into (x - 2)(x - 3) in seconds. But try x² - 5x + 3. No pair of whole numbers multiplies to 3 and adds to -5, so factorising simply cannot start. The equation still has two perfectly real solutions; they just happen to be irrational numbers involving $\sqrt{13}$ .

Completing the square rewrites any quadratic around a single squared bracket, which can then be square-rooted directly, and as a bonus the rewritten form hands you the turning point of the curve for free. The quadratic formula, x = (-b ± $\sqrt{b² - 4ac}$ ) / 2a, is completing the square done once, in general, for every quadratic at the same time. Inside it sits the discriminant b² - 4ac, a single number that announces how many real solutions exist before you solve anything.

x² - 5x + 3 = 0 cannot be factorisedthe formula gives x = (5 ± $\sqrt{13}$ )/2 exactly, no rounding, no guessing
y = x² - 6x + 11 rewritten as (x - 3)² + 2the turning point (3, 2) can be read straight off the completed square
b² - 4ac = -8 for some quadraticnegative discriminant: the curve never crosses the x-axis, so there are no real roots to find
A projectile's height is a quadratic in timewhen it lands is a quadratic equation whose coefficients are rarely factorisable

Learning objective

What students will be able to do

Students will solve quadratic equations exactly by completing the square and by the quadratic formula (rearranging into standard form first where necessary), simplify surd answers fully, use the discriminant to state the number of real roots without solving, and find the turning point of a quadratic curve from its completed-square form.

Success criteria

I can write x² + bx + c in the form (x + p)² + q.
I can solve a quadratic exactly by completing the square, leaving surd answers in simplified exact form.
I can rearrange an equation into ax² + bx + c = 0 and solve it with the quadratic formula.
I can evaluate the discriminant b² - 4ac and state whether the equation has two, one or no real roots.
I can find the turning point of y = x² + bx + c by completing the square, and say whether it is a maximum or a minimum.

Curriculum anchor

Standards this unit teaches

GCSE Algebra #18 (Higher tier)UK GCSE Mathematics (DfE, England)
Quadratic equations: completing the square and the quadratic formula
Subject content statement (Department for Education, "GCSE mathematics: subject content and assessment objectives", published 1 November 2013, reference DFE-00233-2013, "Algebra" section, "Solving equations and inequalities", item 18, https://www.gov.uk/government/publications/gcse-mathematics-subject-content-and-assessment-objectives): students should "solve quadratic equations (including those that require rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph". The clauses "(including those that require rearrangement)" and "by completing the square and by using the quadratic formula" are BOLD type in the source: Higher tier only. Factorising with integer roots is the Foundation half, taught earlier.
GCSE Algebra #11 (bold clause, Higher tier)UK GCSE Mathematics (DfE, England)
Turning points by completing the square
Item 11 of the same "Algebra" section reads "identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square", where "and turning points by completing the square" is BOLD type: deriving a turning point algebraically (rather than reading it off a drawn graph) is assessed at Higher tier only. This unit teaches exactly that derivation.

Before you start

Prior knowledge

This unit builds on skills students should already have met. Revisit any that are shaky first.

Key vocabulary

Words to teach and display

Completed-square form: a quadratic written as (x + p)² + q, built around one squared bracket
Quadratic formula: x = (-b ± $\sqrt{b² - 4ac}$ ) / 2a, the exact solutions of ax² + bx + c = 0 for any a, b, c
Discriminant: the value b² - 4ac inside the formula's square root; its sign tells you how many real roots exist
Surd: an exact, unrounded root such as $\sqrt{13}$ , kept as a symbol because its decimal never ends
Turning point: the single lowest (or highest) point of a parabola, where the curve changes direction

Teaching sequence

Teach it: concrete, pictorial, abstract

The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.

1. Completing the square, and solving exactly with it

Concrete

Half the x-coefficient names the bracket; squaring the bracket overshoots the constant, so correct it. That two-step routine turns x² + bx + c into (x + b/2)² + (c - (b/2)²), a form that can be square-rooted directly.

Rewrite x² + 6x + 4. Half of 6 is 3, and (x + 3)² = x² + 6x + 9. That 9 is 5 more than the 4 we want, so subtract it: x² + 6x + 4 = (x + 3)² - 5. Nothing has been solved yet; the expression has only been reshaped around one squared bracket.

Now solve (x + 3)² - 5 = 0. Move the 5 across: (x + 3)² = 5. Square-root both sides, remembering both signs: x + 3 = ± $\sqrt{5}$ . So x = -3 + $\sqrt{5}$ or x = -3 - $\sqrt{5}$ , exactly. A calculator decimal like -0.76 is an approximation; the surd form is the true value and is what 'give exact answers' means at GCSE.

Worked example

Solve x² - 8x + 3 = 0 by completing the square, giving exact answers.

Half of -8 is -4: (x - 4)² = x² - 8x + 16.
Correct the constant: 3 - 16 = -13, so x² - 8x + 3 = (x - 4)² - 13.
Set it to zero and move the 13: (x - 4)² = 13.
Square-root both sides with ±: x - 4 = ± $\sqrt{13}$ .
Add 4: x = 4 + $\sqrt{13}$ or x = 4 - $\sqrt{13}$ .

Answer: x = 4 ± $\sqrt{13}$ , exactly (13 has no square factor, so the surd is already fully simplified).

Check for understanding, ask

Why does the corrected constant use (b/2)² and not b²?
Why must square-rooting an equation produce a ± and not just the positive root?

2. Turning points from completed-square form

Pictorial

A square is never negative. That single fact makes the completed-square form a machine for finding turning points: (x + p)² + q is smallest exactly when the square is zero, at x = -p, where y = q.

Take y = x² - 6x + 11 = (x - 3)² + 2. The bracket (x - 3)² is zero when x = 3 and positive everywhere else, so y is smallest at x = 3, where y = 2. The turning point is (3, 2) and it is a minimum, because the positive x² coefficient opens the parabola upward.

Notice the sign flip: (x - 3)² + 2 turns at x = +3, and (x + 5)² - 4 turns at x = -5. The x-coordinate is always the value that KILLS the bracket. This algebraic route is the Higher-tier partner of reading a turning point off a drawn graph: no graph is needed, and the answer is exact.

Worked example

Find the turning point of y = x² + 10x + 18, and state whether it is a maximum or a minimum.

Half of 10 is 5: (x + 5)² = x² + 10x + 25.
Correct the constant: 18 - 25 = -7, so y = (x + 5)² - 7.
The square is zero at x = -5, and there y = -7.
The x² coefficient is positive, so the parabola opens upward.

Answer: Turning point (-5, -7); it is a minimum.

Check for understanding, ask

Why is the y-coordinate of the turning point exactly the constant q in (x + p)² + q?
If y = (x - 2)² + 3, why can y never equal 1?

3. The quadratic formula and the discriminant

Abstract

Completing the square on the general equation ax² + bx + c = 0, once, produces x = (-b ± $\sqrt{b² - 4ac}$ ) / 2a. Learning the formula IS learning the finished general version of section 1. Before substituting, the equation must genuinely be in standard form: rearrange first, then read off a, b and c with their signs.

Solve 2x² = 6x + 5. Rearranged: 2x² - 6x - 5 = 0, so a = 2, b = -6, c = -5. The discriminant is b² - 4ac = 36 - 4(2)(-5) = 36 + 40 = 76. Then x = (6 ± $\sqrt{76}$ )/4. Simplify the surd: 76 = 4 x 19, so $\sqrt{76}$ = 2 $\sqrt{19}$ , and x = (6 ± 2 $\sqrt{19}$ )/4 = (3 ± $\sqrt{19}$ )/2, fully simplified after cancelling the common factor 2.

The discriminant alone answers 'how many real roots?'. Positive: the ± creates two different values, two distinct real roots. Zero: plus-nothing and minus-nothing agree, one repeated root. Negative: no real number squares to a negative, so no real roots; the curve never touches the x-axis. This is why an examiner can ask about the roots of a quadratic without asking you to find them.

Worked example

Use the quadratic formula to solve x² - 5x + 3 = 0 exactly, and confirm first, from the discriminant, that two distinct real roots exist.

a = 1, b = -5, c = 3.
Discriminant: b² - 4ac = 25 - 12 = 13. Positive, so two distinct real roots exist.
x = (5 ± $\sqrt{13}$ ) / 2.
13 is square-free, so $\sqrt{13}$ does not simplify and the answer is already exact and fully simplified.

Answer: x = (5 + $\sqrt{13}$ )/2 or x = (5 - $\sqrt{13}$ )/2.

Check for understanding, ask

Why does the formula's denominator use 2a rather than 2?
Without solving, how many real roots does x² + 4x + 9 = 0 have, and how do you know?

Watch for

Common misconceptions and how to address them

MisconceptionIn completed-square form (x - 3)² + 2, the turning point is at x = -3.

Why it happens: Students read the -3 inside the bracket directly, without asking which x makes the bracket zero.

How to address it: Always ask 'what value of x kills the bracket?'. (x - 3)² dies at x = +3, so the turning point is (3, 2). Substituting the candidate x back into the bracket takes two seconds and catches every sign slip.

MisconceptionSubstituting a negative b into the formula as if it were positive, e.g. using b = 5 for x² - 5x + 3.

Why it happens: The minus sign gets absorbed into the printed equation and forgotten at substitution time, flipping -b and corrupting b² - 4ac when c is also negative.

How to address it: Write the line 'a = 1, b = -5, c = 3' BEFORE touching the formula, brackets included when squaring: b² = (-5)² = 25. The single written line prevents almost every formula error.

MisconceptionA surd answer like (6 ± 2 $\sqrt{19}$ )/4 is finished.

Why it happens: Students stop as soon as the formula produces numbers, without checking the surd or the fraction for common factors.

How to address it: Two simplification checks every time: does the number under the root have a square factor (76 = 4 x 19), and do all terms of the fraction share a factor ((6 ± 2 $\sqrt{19}$ )/4 = (3 ± $\sqrt{19}$ )/2)? GCSE mark schemes reserve the final mark for the fully simplified exact form.

Do it together

Guided practice (with answers)

1. Write x² + 4x + 1 in the form (x + p)² + q.
Answer: (x + 2)² - 3, because half of 4 is 2 and 1 - 4 = -3.
2. Solve (x + 2)² - 3 = 0 exactly.
Answer: x = -2 ± $\sqrt{3}$ , by moving the 3 across and square-rooting with ±.
3. Find the turning point of y = x² - 2x + 5.
Answer: (1, 4), a minimum: y = (x - 1)² + 4.
4. Find the discriminant of 3x² + 2x + 4 = 0 and interpret it.
Answer: b² - 4ac = 4 - 48 = -44: negative, so no real roots.
5. Solve x² + 3x - 2 = 0 exactly with the formula.
Answer: x = (-3 ± $\sqrt{17}$ )/2, since b² - 4ac = 9 + 8 = 17.

On their own

Independent practice worksheets

Practise both methods, the discriminant, and completed-square turning points, with computed, never-wrong answer keys.

The Quadratic Formula (Year 11)exact surd solutions, rearrange-first equations, and discriminant reasoning Completing the Square (Year 11)rewrite, solve exactly, and find turning points by completing the square The Quadratic Formula (Year 10)the same skill, freshly generated, for a Year 10 class meeting it early Surds (Year 11)the surd-simplification skills the exact answers here depend on

Reach every student

Differentiation

Support

Run the two-step chant for every completing-the-square: 'half the x-coefficient names the bracket; fix the constant'. Keep it visible until it is automatic.
Have students write the 'a = ..., b = ..., c = ...' line, signs included, before every formula substitution, and tick each value off as it enters the formula.
Start the discriminant work with sorting: given six ready-made discriminant values, sort the equations into two/one/no real roots before computing any discriminants from scratch.

Extension

Complete the square on the general ax² + bx + c (dividing through by a first) and derive the quadratic formula itself, proving the two methods are one.
Find the turning point of y = 2x² - 12x + 7 by first taking out the factor of 2, previewing non-monic completed-square form.
Use the discriminant to find the value of k for which x² + kx + 25 = 0 has exactly one repeated root.

Check it stuck

Assessment: exit ticket

A three-question exit ticket sampling completed-square solving, a turning point, and the formula with a simplifying surd.

1. Solve x² + 6x + 2 = 0 by completing the square, exactly.
Answer: x = -3 ± $\sqrt{7}$ , from (x + 3)² = 7.
2. Find the turning point of y = x² - 4x + 9.
Answer: (2, 5), a minimum: y = (x - 2)² + 5.
3. Solve x² - 4x - 8 = 0 with the quadratic formula, simplifying the surd.
Answer: x = 2 ± 2 $\sqrt{3}$ , since the discriminant is 16 + 32 = 48 and $\sqrt{48}$ = 4 $\sqrt{3}$ , so x = (4 ± 4 $\sqrt{3}$ )/2 = 2 ± 2 $\sqrt{3}$ .

For the teacher

Teacher notes and timings

Rough timing: Lesson 1 completing the square and exact solving (section 1), Lesson 2 turning points (section 2), Lesson 3 the formula and the discriminant plus the exit ticket (section 3).
Tier note: this is a HIGHER TIER unit. In DfE reference DFE-00233-2013, Algebra item 18's clauses '(including those that require rearrangement)' and 'by completing the square and by using the quadratic formula', and Algebra item 11's clause 'and turning points by completing the square', are all bold type, which the document's type key defines as content assessed only at the higher tier. Foundation students factorise integer-root quadratics and read turning points off drawn graphs instead.
This unit deliberately keeps completing the square monic (x² coefficient 1), the standard GCSE Higher scope; the extension path previews the non-monic form.
No figure accompanies this unit: its objects are equations and exact surd values, which the site renders inline (KaTeX via MathText) rather than as diagrams. The matching quadratic-graph reading skills (Foundation half of Algebra #11) are a separate, graphical unit planned with Foundation batch 5.

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