Word problems with quadratics · Grade 9
In Word problems with quadratics (Grade 9) you practise 4 skills: Problems about lengths and areas · Problems about numbers · Problems about speed, distance and time · 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
For example: A rectangle is 8 cm longer than it is wide, and its area is 33 cm². How wide is it?
Answer and solution
The answer: 3 cm
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
For example: Two consecutive positive whole numbers multiply to 30. What is the smaller one?
Answer and solution
The answer: 5
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 speed, distance and time
For example: A car travelled 200 km in 4 hours, at a steady speed. What was its speed?
Answer and solution
The answer: 50 km/h
Speed is distance divided by time: 200 ÷ 4 = 50
| Speed (km/h) | Time (hours) | Distance (km) | |
|---|---|---|---|
| The car | 50 | 4 | 200 |
Problems about percentages
For example: In a sale with 10% off, a bike costs 747 shekels. What is the price of the bike without the discount?
Answer and solution
The answer: 830 shekels
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 shekels.
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Parts of this topic that are not built yet: Integrated literacy tasks · Problems about working together.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education's grade 9 maths curriculum (2026–27).