Practice Maths

Solving Quadratic Equations

Key Terms

A quadratic equation has the form ax² + bx + c = 0. It can have 0, 1, or 2 real solutions.
Method 1 — Null factor law
Factorise the left side, set each factor = 0. Works when the quadratic factors nicely.
Method 2 — Completing the square
Rewrite in the form (x + p)² = q, then take the square root. Gives exact surd answers.
Method 3 — Quadratic formula
x = [−b ± √(b² − 4ac)] / (2a). Always works — the most powerful method.
The discriminant Δ = b² − 4ac: Δ > 0 (two real roots), Δ = 0 (one repeated root), Δ < 0 (no real roots).
When solutions involve surds, leave them in exact form (e.g. x = 3 + √5) unless a decimal is specifically requested.
MethodWhen to use itKey step
Null factor lawEquation factorises easily(x − p)(x − q) = 0 → x = p or x = q
Completing the squareNeed exact surd answers; a = 1(x + b/2)² = (b/2)² − c
Quadratic formulaAlways; especially a ≠ 1x = [−b ± √(b²−4ac)] / 2a
Discriminant checkBefore solving, to count rootsΔ = b² − 4ac
Hot Tip Always rearrange to ax² + bx + c = 0 before solving. Check your answers by substituting back. If a question says “exact values”, leave surds unsimplified — do NOT round to decimals.

Worked Example 1 — Factorising (Null Factor Law)

Question: Solve x² − 5x + 6 = 0.

Step 1: Factorise: (x − 2)(x − 3) = 0.

Step 2 (null factor law): x − 2 = 0 or x − 3 = 0.

Answer: x = 2 or x = 3.

Worked Example 2 — Quadratic Formula with Surd Solutions

Question: Solve 2x² − 4x − 1 = 0. Give exact values.

Step 1 (discriminant): Δ = (−4)² − 4(2)(−1) = 16 + 8 = 24 > 0. Two real solutions.

Step 2 (formula): x = [4 ± √24] / (2×2) = [4 ± 2√6] / 4 = (2 ± √6) / 2.

Answer: x = (2 + √6)/2 or x = (2 − √6)/2.

Worked Example 3 — Completing the Square

Question: Solve x² + 6x + 2 = 0 by completing the square.

Step 1: x² + 6x = −2.

Step 2: Add (6/2)² = 9 to both sides: (x + 3)² = 7.

Step 3: x + 3 = ±√7.

Answer: x = −3 + √7 or x = −3 − √7.

Three Methods and When to Use Each

There is no single “best” method for solving quadratics. The right choice depends on the equation’s structure, but knowing all three methods and their trade-offs is essential.

Factorisation is fastest when integer roots exist, but requires pattern recognition. It works beautifully for equations like x² − 5x + 6 = 0 but is hopeless for x² − 3x − 1 = 0.

Completing the square always works and reveals the vertex form of the parabola simultaneously. It is the method of choice when you need the turning point or are deriving the quadratic formula itself. It requires dividing by the leading coefficient first if a ≠ 1.

The quadratic formula always works, requires no creativity, and handles messy coefficients effortlessly. It is the fallback when other methods fail, and it is derived from completing the square applied to ax² + bx + c = 0 in general.

The Null Factor Law: The WHY Behind Factorisation

Why does setting each factor equal to zero work? The fundamental property being used is: if a product of real numbers equals zero, at least one of the factors must be zero. In symbols: if AB = 0, then A = 0 or B = 0.

This is not obvious — it fails for other values. If AB = 6, you cannot conclude A = 6 or B = 6. Zero is the unique value where this works, because zero is the only number that “absorbs” multiplication: anything times zero is zero.

This is why you must rearrange to = 0 before factorising. The equation (x − 2)(x + 3) = 4 cannot be solved by setting x − 2 = 4 or x + 3 = 4. First expand, rearrange to = 0, then factorise the new expression.

Completing the Square: Step-by-Step

The goal is to rewrite ax² + bx + c = 0 in the form a(x + p)² = q, from which you can take the square root directly.

For x² + 6x + 2 = 0 (a = 1):

  1. Move the constant: x² + 6x = −2.
  2. Add (half the x-coefficient)² to both sides. Half of 6 is 3, and 3² = 9. So: x² + 6x + 9 = 7.
  3. The left side is a perfect square: (x + 3)² = 7.
  4. Take the square root: x + 3 = ±√7.
  5. Solve: x = −3 ± √7.

Why does this work? Because (x + 3)² = x² + 6x + 9 — the constant 9 we added is exactly what “completes” the square. The key insight is (b/2)² is always the completing value when the coefficient of x² is 1.

The Discriminant: Counting Roots Geometrically

The discriminant Δ = b² − 4ac determines the nature of roots without solving. Its geometric meaning comes from the parabola y = ax² + bx + c and its relationship with the x-axis.

When Δ > 0: the parabola crosses the x-axis twice — two distinct real roots. When Δ = 0: the vertex of the parabola sits exactly on the x-axis — one repeated root (the x-intercept and vertex coincide). When Δ < 0: the parabola does not reach the x-axis at all — no real roots, meaning the quadratic is always positive (if a > 0) or always negative (if a < 0).

This geometric interpretation is powerful: if you sketch the parabola and can see it does not cross the x-axis, you know immediately the equation has no real solutions, without any calculation.

Real-World Quadratics

Quadratic equations arise naturally whenever a quantity is proportional to the square of another. Projectile motion is the classic example: if an object is launched upward at v m/s from height h metres, its height at time t is given by s = −½gt² + vt + h (where g ≈ 9.8 m/s²). Setting s = 0 gives a quadratic in t, whose solutions are when the object is at ground level.

Area problems also generate quadratics: if a rectangle has perimeter 20 m and area 24 m², letting the length be x gives x(10 − x) = 24, which rearranges to x² − 10x + 24 = 0, solved by factorisation as (x − 4)(x − 6) = 0.

Always check that solutions make physical sense: negative lengths or times before launch are mathematically valid roots but physically invalid answers.

Exam Tip: The quadratic formula works for every quadratic equation, including ones that factorise. If you cannot spot the factorisation quickly, go straight to the formula — do not waste time hunting for factors. The formula is always reliable. Make sure you write the full formula, substitute carefully, and simplify the surd in your answer.
Exam Tip: When Δ < 0, there are no real solutions. This makes geometric sense — the parabola does not cross the x-axis. Do not say “no solutions” without explaining why — write “no real solutions since Δ < 0”. In complex number work (Year 12 Specialist), Δ < 0 gives complex roots, but in Methods we only deal with real solutions.

Mastery Practice

See Answers ➔

  1. Solve by factorising (null factor law). Fluency

    1. (a) x² − 7x + 12 = 0
    2. (b) x² + x − 6 = 0
    3. (c) 2x² + 5x − 3 = 0
    4. (d) 3x² − 12x = 0
    5. (e) x² − 9 = 0
    6. (f) 4x² − 4x + 1 = 0

  2. Solve by completing the square. Give exact answers. Fluency

    1. (a) x² + 4x − 3 = 0
    2. (b) x² − 6x + 1 = 0
    3. (c) x² + 2x − 5 = 0
    4. (d) x² − 8x + 10 = 0

  3. Use the quadratic formula to solve each equation. Give exact values. Fluency

    1. (a) x² − 4x + 1 = 0
    2. (b) 2x² + 3x − 2 = 0
    3. (c) 3x² − 5x + 1 = 0
    4. (d) x² + x + 3 = 0 (comment on the result)

  4. Without solving, use the discriminant to determine the number and type of solutions. Fluency

    1. (a) x² − 6x + 9 = 0
    2. (b) 2x² + x + 4 = 0
    3. (c) x² − 3x − 5 = 0
    4. (d) 4x² − 12x + 9 = 0

  5. Rearrange into standard form, then solve by the most efficient method. Fluency

    1. (a) x(x + 5) = 14
    2. (b) (x + 1)(x − 3) = 5
    3. (c) x² = 3x + 10
    4. (d) 2x(x − 1) = x + 3

  6. Exact surd solutions. Understanding

    Leave all answers in exact form.
    1. (a) Solve x² − 2x − 4 = 0. Simplify the surd in your answer fully.
    2. (b) Solve 2x² + 2x − 3 = 0. Give your two solutions and rationalise any surd denominators.
    3. (c) Show that the solutions of x² + bx + c = 0 can be written as x = −b/2 ± √(b²/4 − c). (Hint: complete the square.)

  7. Equations with a parameter. Understanding

    Finding conditions on a constant.
    1. (a) For the equation x² + mx + 9 = 0 to have exactly one solution, find the possible value(s) of m.
    2. (b) For the equation x² − 4x + k = 0 to have two distinct real solutions, find the range of values of k.
    3. (c) The equation kx² + 2x + k = 0 has a repeated root. Find all possible values of k.

  8. Quadratics in geometry. Understanding

    Setting up and solving quadratic equations from geometric contexts.
    1. (a) A rectangle has a perimeter of 24 cm and an area of 32 cm². Let the length be x cm. Form and solve a quadratic to find the dimensions.
    2. (b) The hypotenuse of a right-angled triangle is 10 cm. One leg is 2 cm longer than the other. Find the exact lengths of both legs.

  9. Vieta’s formulas (sum and product of roots). Problem Solving

    Key result: If α and β are roots of ax² + bx + c = 0, then α + β = −b/a and αβ = c/a.
    1. (a) Prove that if α and β are roots of x² + bx + c = 0 then α + β = −b and αβ = c. (Hint: expand (x − α)(x − β) and compare coefficients.)
    2. (b) Without solving, find the sum and product of the roots of 3x² − 5x − 2 = 0.
    3. (c) The roots of x² − px + q = 0 satisfy α + β = 5 and αβ = 3. Find the values of p and q, then find the roots exactly.
    4. (d) One root of x² − 7x + k = 0 is twice the other. Use Vieta’s formulas to find the roots and k.

  10. Equations reducible to quadratics. Problem Solving

    Substitution technique. Use the substitution u = x² (or u = x − 1/x) to reduce higher-degree equations to quadratics.
    1. (a) Solve x&sup4; − 5x² + 4 = 0. (Let u = x².)
    2. (b) Solve x&sup4; − 13x² + 36 = 0.
    3. (c) Solve (x² − 2x)² − 11(x² − 2x) + 24 = 0. (Let u = x² − 2x.)
    4. (d) Solve x + 4/x = 5 for x ≠ 0.