Teaching unit · Algebra 1 (commonly taken Grades 8 to 10)

Solving quadratics: completing the square and the quadratic formula

Two exact methods that work even when a quadratic will not factor

About four to five lessons of 45 to 60 minutes

Start here · hook

The equation factoring can't crack

Try to solve x² - 6x - 2 = 0 by factoring: you need two integers that multiply to -2 and add to -6. There are none, every integer pair fails. Does that mean the equation has no solution? No, it has two real solutions, they are just not whole numbers.

Completing the square and the quadratic formula solve ANY quadratic equation exactly, whether the roots are whole numbers, fractions, irrational numbers written with a square root, or complex numbers written with i, using the same handful of steps every time. They are not a backup plan for when factoring fails: the quadratic formula is completing the square done once, in general, for every possible a, b and c.

When does a launched rocket hit the ground?its height h = -16t² + 64t + 3 rarely lands on a whole-number time
A rectangle's area equationsolving for an unknown side length often gives an equation that will not factor
The exact width of a border around a gardena real design problem, not a made-up one, usually needs an exact (possibly irrational) answer
Where a parabola crosses the x-axisthe discriminant tells you the answer BEFORE you solve: twice, once, or never

Learning objective

What students will be able to do

Students will transform a quadratic equation into completed-square form (x - p)² = q, solve quadratic equations exactly by completing the square, derive and apply the quadratic formula, and use the discriminant to classify roots and write complex solutions in a ± bi form.

Success criteria

I can rewrite x² + bx + c = 0 as (x - p)² = q by completing the square, stating p and q.
I can solve x² = k or (x - p)² = q by taking square roots, simplifying the radical exactly.
I can solve a quadratic equation by completing the square, including when the answer is irrational.
I can apply the quadratic formula, x = (-b ± $\sqrt{b² - 4ac}$ ) / 2a, to solve ax² + bx + c = 0.
I can compute the discriminant b² - 4ac and use its sign to say whether an equation has two, one, or no real solutions.
When the discriminant is negative, I can write the two complex solutions in a ± bi form.

Curriculum anchor

Standards this unit teaches

HSA-REI.B.4.aCommon Core (US)
Complete the square to derive the quadratic formula
Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)² = q that has the same solutions. Derive the quadratic formula from this form.
HSA-REI.B.4.bCommon Core (US)
Solve quadratic equations by the most appropriate method
Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.

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

Quadratic equation: an equation that can be written ax² + bx + c = 0, with a not equal to 0
Completing the square: rewriting x² + bx + c as (x - p)² + a constant, by adding and subtracting (b/2)²
Quadratic formula: x = (-b ± $\sqrt{b² - 4ac}$ ) / 2a, a formula that solves any quadratic equation ax² + bx + c = 0
Discriminant: the value b² - 4ac; its sign tells you the number of real solutions before you solve
Radical: an expression written with a root symbol, such as $\sqrt{11}$ , left exact rather than rounded to a decimal
Irrational number: a number, like $\sqrt{11}$ , that cannot be written as an exact fraction; its decimal never terminates or repeats
Imaginary unit: the number i defined by i² = -1, so $\sqrt{-k}$ = i√k for positive k

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. Taking square roots: the simplest case

Concrete

When a quadratic equation has no x term on its own, only x² (or a squared bracket), you can solve it directly by taking the square root of both sides. This is the same idea as 'undoing' squaring anywhere else in maths, just applied to an equation.

x² = k has two solutions whenever k is positive: x = √k and x = -√k, written together as x = ±√k. If k is a perfect square (like 36 or 49), the answer is a whole number. If it is not, the answer is irrational, and you simplify the radical instead of rounding it: find the largest perfect-square factor of k, pull its square root out the front, and leave the rest under the root sign.

Worked example

Solve x² - 20 = 0.

Isolate x²: x² = 20.
Take the square root of both sides: x = ± $\sqrt{20}$ .
Simplify the radical: 20 = 4 × 5, and 4 is a perfect square, so $\sqrt{20}$ = $\sqrt{4}$ × $\sqrt{5}$ = 2 $\sqrt{5}$ .
x = ±2 $\sqrt{5}$ (≈ ±4.47), left exact, not rounded.

The two exact solutions of x² = 20, plotted at their (approximate) positions on the number line. Neither is a whole number, but both are exact.

Answer: x = 2 $\sqrt{5}$ or x = -2 $\sqrt{5}$ .

Check for understanding, ask

Why does x² = 20 have two solutions, not one?
Would x² = -20 have a real solution? Why or why not?
How do you decide whether √k simplifies, or is already in simplest form?

2. Completing the square: turning any quadratic into (x - p)² = q

Pictorial

Most quadratics have an x term, so you cannot take a square root right away. Completing the square rewrites x² + bx + c = 0 as an equivalent equation (x - p)² = q, which CAN be solved by the square-root method from section 1, for any b and c at all.

The rule: p is always half of the x-coefficient's opposite, p = -b/2, and q = p² - c. Once you can find p and q, you already know how to finish: take the square root of both sides, exactly like section 1. This also reveals something about the graph: the parabola y = x² + bx + c is symmetric about the vertical line x = p, the same p from the completed square.

y = x² - 6x - 2. The parabola is symmetric about x = 3, exactly the p found when completing the square below (b = -6, so p = -(-6)/2 = 3).

Worked example

Rewrite x² - 6x - 2 = 0 as (x - p)² = q, then solve it.

Move the constant across: x² - 6x = 2.
Half of -6 is -3; square it: (-3)² = 9. Add 9 to both sides: x² - 6x + 9 = 2 + 9 = 11.
The left side is now a perfect square: (x - 3)² = 11. So p = 3, q = 11.
Take square roots: x - 3 = ± $\sqrt{11}$ (11 has no perfect-square factor, so it does not simplify further).
x = 3 ± $\sqrt{11}$ (≈ 6.32 or ≈ -0.32).

Answer: x = 3 + $\sqrt{11}$ or x = 3 - $\sqrt{11}$ .

Check for understanding, ask

What number do you always add to both sides when completing the square, in terms of b?
Why is (x - 3)² = 11 easier to solve than x² - 6x - 2 = 0 directly?
How does p in (x - p)² = q relate to the graph of y = x² + bx + c?

3. Deriving and using the quadratic formula

Abstract

Completing the square works for any equation, but redoing all those steps every time is slow. Do it ONCE for the general equation ax² + bx + c = 0 (dividing by a first, then completing the square exactly as in section 2), and the result is the quadratic formula, a shortcut that skips straight to the answer.

Derive the formula from ax² + bx + c = 0: divide by a to get x² + (b/a)x = -c/a. Add (b/2a)² to both sides, giving (x + b/2a)² = (b² - 4ac)/4a², or equivalently (2ax + b)² = b² - 4ac. Take both square roots, 2ax + b = ± $\sqrt{b² - 4ac}$ , then isolate x. The result is x = (-b ± $\sqrt{b² - 4ac}$ )/2a.

Plug in a, b and c, compute the discriminant b² - 4ac first (it decides whether the roots are rational, irrational radicals, or complex), then finish the exact arithmetic.

Worked example

Use the quadratic formula to solve 2x² - 3x - 5 = 0.

Identify a = 2, b = -3, c = -5.
Discriminant: b² - 4ac = (-3)² - 4(2)(-5) = 9 + 40 = 49.
$\sqrt{49}$ = 7 (a perfect square, so both roots will be rational).
x = (-b ± $\sqrt{b² - 4ac}$ ) / 2a = (3 ± 7) / 4.
x = (3 + 7)/4 = 10/4 = 5/2, or x = (3 - 7)/4 = -4/4 = -1.

The two solutions of 2x² - 3x - 5 = 0, found with the quadratic formula.

Answer: x = 5/2 or x = -1.

Check for understanding, ask

Why must a not equal 0 in the quadratic formula?
What happens in the formula if the discriminant is a perfect square, versus when it is not?
Could you solve 2x² - 3x - 5 = 0 by factoring instead? Try it and compare.

4. The discriminant: predicting and writing solutions

Abstract

The discriminant, b² - 4ac, is the part of the quadratic formula under the square root sign. Its SIGN alone tells you how many real solutions the equation has, before doing any more work, because it is exactly what decides whether you end up taking the square root of a positive number, zero, or a negative number.

If b² - 4ac > 0, there are two real solutions (the parabola crosses the x-axis twice). If b² - 4ac = 0, there is exactly one real solution, a repeated root (the parabola touches the x-axis once, at its vertex). If b² - 4ac < 0, there are no real solutions (the parabola never reaches the x-axis), but there are two complex solutions. Use i² = -1, so $\sqrt{-k}$ = i√k, then simplify the quadratic formula into a ± bi form.

Worked example

Classify and solve 3x² + 2x + 5 = 0.

Identify a = 3, b = 2, c = 5.
Discriminant: b² - 4ac = 2² - 4(3)(5) = 4 - 60 = -56.
Since -56 < 0, the equation has no real solutions.
Use $\sqrt{-56}$ = i $\sqrt{56}$ = 2i $\sqrt{14}$ in the quadratic formula: x = (-2 ± 2i $\sqrt{14}$ )/6.
Reduce the common factor of 2: x = (-1 ± i $\sqrt{14}$ )/3.

Answer: No real solutions; the complex solutions are x = (-1 + i $\sqrt{14}$ )/3 and x = (-1 - i $\sqrt{14}$ )/3.

Check for understanding, ask

If the discriminant is 0, how many DIFFERENT x-values solve the equation?
Can you tell the discriminant's sign just by looking at a, b and c, without fully computing it? Try 5x² + x + 5 = 0.
Why does $\sqrt{-56}$ become 2i $\sqrt{14}$ ?

Watch for

Common misconceptions and how to address them

MisconceptionCompleting the square and the quadratic formula are two unrelated tricks to memorise separately.

Why it happens: Each is usually taught as its own recipe of steps, so the connection between them is easy to miss.

How to address it: The quadratic formula IS completing the square, done once for the general ax² + bx + c = 0 instead of a specific equation. Deriving it live (HSA-REI.B.4a) makes this obvious rather than asking students to trust two disconnected procedures.

Misconceptionx² = 20 has only one solution, because you 'just take the square root'.

Why it happens: Students carry over the single-answer habit from evaluating $\sqrt{20}$ on a calculator, forgetting the equation itself has two solutions.

How to address it: Every positive k gives x² = k TWO solutions, +√k and -√k, since both squares equal k. Always write the ± explicitly, and check both answers by substitution.

MisconceptionWhen completing the square, you add (b/2)² to only one side of the equation.

Why it happens: Students focus on 'making the left side a perfect square' and forget the equation must stay balanced.

How to address it: Whatever is added to complete the square on the left must also be added to the right side, keeping the equation true. Say it every time: 'add the same number to both sides.'

MisconceptionA negative discriminant means you made an arithmetic mistake.

Why it happens: Every earlier quadratic example in this and the prior AU unit had a real answer, so a negative result feels like an error rather than a valid outcome.

How to address it: A negative discriminant is valid: it means there are no REAL solutions (the graph never crosses the x-axis). Continue with $\sqrt{-k}$ = i√k to write the two complex solutions in a ± bi form.

Misconception $\sqrt{11}$ should be rounded to a decimal before it counts as 'the answer'.

Why it happens: Calculators make decimals feel more like 'real answers' than exact radical expressions.

How to address it: This course always keeps radicals exact (2 $\sqrt{5}$ , 3 ± $\sqrt{11}$ ) as the actual answer; a decimal is only a rounded approximation shown for intuition, never a substitute for the exact value in the answer key.

Do it together

Guided practice (with answers)

1. Solve x² = 36 by taking square roots.
Answer: x = ±6, because $\sqrt{36}$ = 6 exactly (36 is a perfect square).
2. Solve x² = 12 by taking square roots, simplifying the radical.
Answer: x = ±2 $\sqrt{3}$ , because 12 = 4 × 3, and $\sqrt{12}$ = $\sqrt{4}$ × $\sqrt{3}$ = 2 $\sqrt{3}$ .
3. Rewrite x² + 8x + 3 = 0 in the form (x - p)² = q.
Answer: (x + 4)² = 13, so p = -4 and q = 13, because half of 8 is 4 (p = -4), and q = p² - c = 16 - 3 = 13.
4. Solve x² - 4x - 1 = 0 by completing the square.
Answer: x = 2 + $\sqrt{5}$ or x = 2 - $\sqrt{5}$ , because p = 2 (half of -4, negated), q = 2² - (-1) = 5, so (x - 2)² = 5.
5. Use the quadratic formula to solve 3x² + 2x - 8 = 0.
Answer: x = 4/3 or x = -2, because the discriminant is 2² - 4(3)(-8) = 100, $\sqrt{100}$ = 10, and x = (-2 ± 10)/6.
6. Find the discriminant of 5x² - 4x + 1 = 0, classify the roots and write the complex solutions.
Answer: Discriminant = (-4)² - 4(5)(1) = 16 - 20 = -4, so there are no real solutions. Using $\sqrt{-4}$ = 2i, x = (4 ± 2i)/10 = (2 ± i)/5.

On their own

Independent practice worksheets

Practise taking square roots, completing the square, the quadratic formula and the discriminant, with computed, never-wrong answer keys and exact (unrounded) radicals.

Algebra 1: taking square rootsthe core skill from section 1, freshly generated each time Algebra 1: completing the square (transform)practise finding p and q from section 2, before solving Algebra 1: completing the square (solve)full solves, including exact irrational answers Algebra 1: the quadratic formulathe shortcut from section 3, with the discriminant shown in the working Algebra 1: the discriminantclassify the number of real solutions from section 4, without solving Year 9 quadratics (factoring) worksheetrevisit the factoring method this unit builds on

Reach every student

Differentiation

Support

Start with perfect-square cases only (x² = k where k is already a perfect square), so no radical simplification is needed while the square-root method itself is new.
Provide a reference table of squares (1² = 1 through 15² = 225) as scaffolding while simplifying radicals.
Use only even middle coefficients at first when completing the square, so p = -b/2 is always a whole number, not a fraction.
Give the quadratic formula with a, b and c already substituted in, so the arithmetic is the only new step for the first few problems.

Extension

Ask students to solve the same equation by two different methods (e.g. completing the square and the quadratic formula) and confirm the answers match exactly.
Have students derive the quadratic formula themselves from ax² + bx + c = 0 with a general a, b and c, matching HSA-REI.B.4a directly rather than being shown the derivation.
Give several negative-discriminant equations and ask students both to explain why their graphs have no real x-intercepts and to write their complex solutions in a ± bi form.
Explore how changing only c shifts the parabola up or down, and connect that to how many times it crosses the x-axis, without changing a or b.

Check it stuck

Assessment: exit ticket

A short exit ticket sampling all four skills: square roots, completing the square, the quadratic formula and the discriminant.

1. Solve x² = 27 by taking square roots, simplifying the radical.
Answer: x = ±3 $\sqrt{3}$ , because 27 = 9 × 3, and $\sqrt{27}$ = $\sqrt{9}$ × $\sqrt{3}$ = 3 $\sqrt{3}$ .
2. Solve x² + 6x - 3 = 0 by completing the square.
Answer: x = -3 + 2 $\sqrt{3}$ or x = -3 - 2 $\sqrt{3}$ , because p = -3, q = 9 - (-3) = 12, and $\sqrt{12}$ = 2 $\sqrt{3}$ .
3. Find the discriminant of 4x² - 4x + 1 = 0, state how many real solutions it has, and solve if possible.
Answer: Discriminant = (-4)² - 4(4)(1) = 0, one repeated real solution, x = 1/2 (check: 4(1/2)² - 4(1/2) + 1 = 1 - 2 + 1 = 0).
4. Solve x² + 4x + 8 = 0, writing the complex solutions in a ± bi form.
Answer: Discriminant = 4² - 4(1)(8) = -16. Since $\sqrt{-16}$ = 4i, x = (-4 ± 4i)/2 = -2 ± 2i.

For the teacher

Teacher notes and timings

Rough timing across four to five lessons: Lesson 1 taking square roots (section 1), Lesson 2 completing the square (section 2), Lesson 3 the quadratic formula (section 3), Lesson 4 the discriminant (section 4), Lesson 5 mixed practice plus the exit ticket.
Prior knowledge: the factoring-based quadratics unit (year-9-quadratics-expand-factorise-solve, linked above) explicitly scopes itself to 'monic quadratics with integer roots' and says non-integer-root cases 'need the quadratic formula, met later'. This unit is that later: it deliberately does not repeat expanding, factorising or integer-root solving, which are already covered there.
Language to repeat: completing the square and the quadratic formula are not two competing techniques, the formula IS completing the square done once in general, for any a, b and c (HSA-REI.B.4a). Showing the derivation live connects the two rather than presenting them as separate recipes.
Real/complex boundary: the generated worksheet classifies negative-discriminant cases as having no real roots, while section 4 explicitly carries those cases through to complex roots in a ± bi form, as HSA-REI.B.4b requires. Broader complex-number arithmetic remains an Algebra 2 topic.
Radicals are always left in exact simplified form (e.g. 2 $\sqrt{5}$ , 3 ± $\sqrt{11}$ ) in the answer keys, never rounded; a decimal approximation is shown alongside purely for number sense, never as the graded answer.
Curriculum note: Common Core HSA-REI.B.4a is the transform-and-derive skill (sections 2 and 3); HSA-REI.B.4b is the solve-by-any-appropriate-method skill (sections 1 to 3, plus the method-choice conversation in the hook) and the complex-root skill (section 4). Verified live at thecorestandards.org/Math/Content/HSA/REI/B/4/, /a/ and /b/ on 2026-07-14.

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