# Word problems with quadratics · Year 10

> Word problems with quadratics, Year 10: Problems about numbers and more. Explanation, worked examples and free practice for the national curriculum in England.

In Word problems with quadratics (Year 10) you practise 3 skills: Problems about lengths and areas · Problems about numbers · Problems about percentages.

In a word problem, call the unknown x and turn each fact into an expression: 3 longer is x plus 3, and the next whole number is x plus 1. The link in the story, such as an area, a product or a time, gives the equation.

When the equation is quadratic, move every term to one side and factorise. There are usually two solutions, so go back to the story: a length, a speed or a positive whole number cannot be negative, and sometimes both solutions fit.

In motion problems the time is the distance divided by the speed, so x can end up in a denominator. Multiply both sides by the product of the denominators to get a quadratic equation.

**Tip:** At the end, put the answer back into the story itself, not just the equation. That shows whether it makes sense.

## What you learn

### Problems about lengths and areas

- Example: A rectangle is 8 cm longer than it is wide, and its area is 33 cm². How wide is it?
- Answer: 3 cm
- Solution: Call the width x. The length is x + 8 , and the area is the length times the width: x(x + 8) = 33 Expand the brackets and move every term to one side: x² + 8x − 33 = 0 Factorise: (x − 3)(x + 11) = 0 Set each factor to 0: x = 3 or x = −11 . A length cannot be negative, so: x = 3

### Problems about numbers

- Example: Two consecutive positive whole numbers multiply to 30. What is the smaller one?
- Answer: 5
- Solution: Call the smaller number x. The next one is x + 1 . The product: x(x + 1) = 30 Expand the brackets and move every term to one side: x² + x − 30 = 0 Factorise: (x − 5)(x + 6) = 0 Set each factor to 0: x = 5 or x = −6 . A positive whole number cannot be negative, so: x = 5

### Problems about percentages

- Example: In a sale with 10% off, a bike costs £747. What is the price of the bike without the discount?
- Answer: £830
- Solution: A discount of 10% means multiplying by 0.9. Call the old price x and write an equation: 0.9x = 747 Divide both sides by 0.9: x = 747 ÷ 0.9 = 830 The old price was £830.

The exercises are made by rules and the answers are worked out in code. The topic follows the national curriculum in England: mathematics (key stages 3 and 4).

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