Factorization of Quadratic Equations II: – Word Problems – SS1 Mathematics Lesson Note (First Term)

A word problem leading to a quadratic equation is a problem whose facts, when written in symbols, form an equation such as ax² + bx + c = 0. To solve it, let the unknown be x, form the equation, factorise it, solve for x, and reject any answer that does not make sense, such as a negative age.

Behavioural objectives

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

1. Represent the unknown quantities in a word problem with letters.
2. Form a quadratic equation from the information in a word problem.
3. Solve the equation by factorisation.
4. Decide which answers are acceptable, for example by rejecting a negative age.
5. Check each answer against the facts in the question.

What is a quadratic equation?

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

• General form: ax² + bx + c = 0, where a is not 0.
• Roots: the values of x that make the equation true. A quadratic equation has up to two roots.
• Factorisation: writing the expression as a product of two brackets, such as (x – 3)(x + 2).
• Zero product rule: if (x – p)(x – q) = 0, then x – p = 0 or x – q = 0.

What are the steps for solving a word problem on quadratic equations?

There are six steps: read, assign letters, form the equation, factorise, solve and check.

1. Read: read the question slowly and list the facts given.
2. Assign: let one unknown quantity be x and write the others in terms of x.
3. Form: use the facts to write an equation, then rearrange it to ax² + bx + c = 0.
4. Factorise: write the left side as two brackets.
5. Solve: set each bracket equal to 0 to find the values of x.
6. Check: reject any value that does not fit the problem, then test the answer in the question.

How do you write words as algebra?

Each common phrase in a word problem has a standard algebraic form.

What types of word problems lead to quadratic equations?

Four types are common at this level: numbers, ages, consecutive integers and area.

• Numbers: two numbers with a given difference, sum or product.
• Ages: ages now, in the future or in the past, with a given product.
• Consecutive integers: numbers such as x and x + 1 with a given product.
• Area: a rectangle whose sides are related and whose area is given.

How do you solve a problem on two numbers?

Example 1: two numbers with a given difference and product

Question: Find two numbers whose difference is 5 and whose product is 266.

1. Let the smaller number be x.
2. Then the larger number is x + 5.
3. Their product is x(x + 5) = 266.
4. x² + 5x – 266 = 0
5. (x – 14)(x + 19) = 0
6. x = 14 or x = -19
7. If x = 14, the other number is 14 + 5 = 19.
8. If x = -19, the other number is -19 + 5 = -14.

Answer: the two numbers are 14 and 19, or -19 and -14.

• Why both pairs are accepted: the question does not say the numbers must be positive.
• Check: 19 – 14 = 5 and 14 × 19 = 266.

How do you solve a problem on ages?

Example 2: ages of a mother and her daughter

Question: Tina is 3 times as old as her daughter. In four years, the product of their ages will be 1536. How old are they now?

1. Let the daughter’s age be x years.
2. Tina’s age is 3x years.
3. In four years, the daughter’s age is (x + 4) years.
4. In four years, Tina’s age is (3x + 4) years.
5. The product of their ages: (x + 4)(3x + 4) = 1536.
6. 3x² + 16x + 16 = 1536, so 3x² + 16x – 1520 = 0.
7. (x – 20)(3x + 76) = 0
8. x = 20 or x = -76/3 (about -25.3)
9. Age cannot be negative, so x = 20.

Answer: the daughter is 20 years old and Tina is 3 × 20 = 60 years old.

• Check: in four years they will be 24 and 64, and 24 × 64 = 1536.

How do you solve a problem on consecutive integers?

Example 3: two consecutive positive integers

Question: The product of two consecutive positive integers is 156. Find them.

1. Let the smaller integer be x.
2. The next integer is x + 1.
3. x(x + 1) = 156
4. x² + x – 156 = 0
5. (x – 12)(x + 13) = 0
6. x = 12 or x = -13
7. The question says positive, so reject x = -13.

Answer: the integers are 12 and 13. Check: 12 × 13 = 156.

How do you solve a problem on the area of a rectangle?

Example 4: sides of a rectangle

Question: The length of a rectangle is 3 cm more than its width. Its area is 40 cm². Find its length and width.

1. Let the width be x cm.
2. The length is (x + 3) cm.
3. Area = length × width, so x(x + 3) = 40.
4. x² + 3x – 40 = 0
5. (x – 5)(x + 8) = 0
6. x = 5 or x = -8
7. A length cannot be negative, so reject x = -8.

Answer: the width is 5 cm and the length is 8 cm. Check: 5 × 8 = 40.

What mistakes should students avoid?

• Forgetting to rearrange: the equation must equal 0 before you factorise.
• Using only one root: find both values of x, then decide which one fits.
• Wrong future age: add the same number of years to every person’s age.
• Not checking: always test the answer against the original facts.
• Keeping impossible answers: lengths, areas and ages cannot be negative.

Evaluation

1. In Example 1, why is the larger number written as x + 5?
2. Check that 14 and 19 have a difference of 5 and a product of 266.
3. In Example 2, check that ages of 24 and 64 in four years give a product of 1536.
4. Why was x = -19 accepted in Example 1, but a negative value of x rejected in Example 2?
5. The product of two consecutive positive integers is 132. Find the integers.

Assignment

Solve all three questions by forming and factorising a quadratic equation.

1. The area of a rectangle is 60 cm². The length is 11 cm more than the width. Find the width.
2. A man is 37 years old and his child is 8. How many years ago was the product of their ages 96?
3. Let α and β be the roots of 2x² – 9x + 4 = 0, with α greater than β. Factorise to find the roots, then find: (a) α + β, (b) αβ, (c) α – β, (d) αβ ÷ (α + β).