Quadratic Equations – SS1 Mathematics Lesson Note (First Term)

A quadratic equation is an equation in which the highest power of the unknown is 2.

Behavioural objectives

By the end of the lesson, students should be able to:

1. identify a quadratic equation and write it in the form ax² + bx + c = 0;
2. state the rule that if a × b = 0, then a = 0 or b = 0 or both are 0;
3. solve equations that are already in factor form;
4. solve quadratic equations by factorisation and check the answers by substitution;
5. solve equations such as m² = 16 by using the difference of two squares.

What is a quadratic equation?

A quadratic equation has an equal sign and an unknown whose highest power is 2.

• Standard form: ax² + bx + c = 0, where a, b and c are numbers and a is not 0.
• a: the number multiplying x² (the coefficient of x²).
• b: the number multiplying x (the coefficient of x).
• c: the constant term, with no unknown.
• Roots: the values of the unknown that make the equation true. They are also called solutions.
• Number of roots: a quadratic equation has at most two different roots.

What are examples of quadratic equations?

Each equation below is quadratic, even when it is not yet written in standard form.

• a) 2x² – 5x – 3 = 0
• b) n² + 50 = 27n
• c) 0 = (4a – 9)(2a + 1)
• d) 49 = k²

What is the zero product rule?

The zero product rule says that if the product of two numbers is 0, then at least one of the numbers is 0.

• Examples: 3 × 0 = 0, 0 × 5 = 0 and 0 × 0 = 0.
• In general: if a × b = 0, then a = 0, or b = 0, or both a and b are 0.
• Use: set each factor equal to 0 and solve each small equation.
• Warning: the rule works only when the product equals 0. It does not work for (x – 2)(x + 7) = 5.

How do you solve an equation that is already in factor form?

Set each factor equal to 0, then solve each equation.

1. Check that one side of the equation is 0.
2. Write each bracket or factor equal to 0.
3. Solve each simple equation.
4. State all the solutions.

Example 1

Solve (x – 2)(x + 7) = 0.

1. Either x – 2 = 0 or x + 7 = 0.
2. x = 2 or x = -7.

The equation has two solutions: 2 and -7.

Example 2

Solve d(d – 4)(d + 6)² = 0.

1. The left side has four factors: d, (d – 4), (d + 6) and (d + 6).
2. Any one of them may be 0: d = 0, or d – 4 = 0, or d + 6 = 0 (twice).
3. d = 0, 4, or -6 twice.

The root -6 is called a repeated root because its factor appears twice.

How do you solve a quadratic equation by factorisation?

To solve by factorisation, write the equation as an expression equal to 0, factorise it, and apply the zero product rule.

1. Rearrange the equation so that one side is 0.
2. Factorise the other side.
3. Set each factor equal to 0.
4. Solve each equation to get the roots.
5. Check each root by substitution into the original equation.

Example 1: when a = 1

Solve m² – 5m – 14 = 0.

1. Find two numbers with product -14 and sum -5. They are -7 and 2.
2. Factorise: (m + 2)(m – 7) = 0.
3. Either m + 2 = 0 or m – 7 = 0.
4. m = -2 or m = 7.

Check m = 7: 49 – 35 – 14 = 0.

Example 2: when a is not 1

Solve 4y² + 5y – 21 = 0.

1. Multiply a and c: 4 × (-21) = -84.
2. Find two numbers with product -84 and sum 5. They are 12 and -7.
3. Split the middle term: 4y² + 12y – 7y – 21 = 0.
4. Group: 4y(y + 3) – 7(y + 3) = 0.
5. Factorise: (y + 3)(4y – 7) = 0.
6. Either y + 3 = 0 or 4y – 7 = 0.
7. y = -3 or 4y = 7, so y = 7/4.

Check by substitution:

• If y = -3: 4(9) + 5(-3) – 21 = 36 – 15 – 21 = 0.
• If y = 7/4: 4 × 49/16 + 35/4 – 21 = 49/4 + 35/4 – 21 = 84/4 – 21 = 0.

How do you solve an equation like m² = 16?

Move all terms to one side, then factorise using the difference of two squares: a² – b² = (a – b)(a + b).

Example

Solve m² = 16.

1. Rearrange: m² – 16 = 0.
2. Factorise: (m – 4)(m + 4) = 0.
3. Either m – 4 = 0 or m + 4 = 0.
4. m = 4 or m = -4.

In the same way, 49 = k² gives k² – 49 = 0, so (k – 7)(k + 7) = 0 and k = 7 or k = -7.

What are the special cases and common mistakes?

Some quadratic equations have a missing term or a repeated root, and some mistakes cost marks.

Special cases

• No constant term: take out the common factor. For x² – 6x = 0, x(x – 6) = 0, so x = 0 or x = 6.
• Perfect square: the roots are equal. For x² + 6x + 9 = 0, (x + 3)² = 0, so x = -3 twice.

Common mistakes

• Not rearranging: always make one side 0 before factorising.
• Dividing by the unknown: dividing x² = 6x by x loses the root x = 0.
• Losing the negative root: m² = 16 has two roots, +4 and -4.
• Skipping the check: substitute each root to catch sign errors.

Evaluation

1. State the rule for two numbers whose product is 0.
2. Solve (x – 2)(x + 7) = 0 and say how many solutions it has.
3. Solve m² – 5m – 14 = 0, given that it factorises to (m + 2)(m – 7) = 0.
4. Solve d(d – 4)(d + 6)² = 0.
5. Solve x² – 49 = 0 by using the difference of two squares.

Assignment

Solve the following equations. Check each result by substitution.

1. (4b – 12)(b – 5) = 0
2. (11 – 4x)² = 0
3. (d – 5)(3d – 2) = 0
4. u² – 8u – 9 = 0
5. c² = 25
6. 2x² = 3x + 5
7. a² – 3a = 0
8. p² + 7p + 12 = 0