Hello, and welcome to today's mathematics lesson. In this session, we will explore how to solve word problems using quadratic equations — a powerful tool that transforms real-life situations into mathematical form and back again. We will work through problems involving numbers, fractions, ages, speeds, work, and geometry. By the end, you will see how a single quadratic equation can unlock the answer to seemingly complex puzzles.
Let us begin with the fundamental approach. When you encounter any word problem based on quadratic equations, follow three clear steps. First, represent the unknown quantity by a variable, typically x. Second, translate the given conditions into an equation in terms of x. Third, solve this equation using factorisation, completing the square, or the quadratic formula. Always remember to check whether your solutions make sense in the original problem context.
Consider problems based on numbers. Suppose you need to find two natural numbers that differ by 3, where the sum of their squares equals 117. Let the smaller number be x, so the larger becomes x + 3. The condition gives us x² + (x + 3)² = 117. Expanding, we get 2x² + 6x + 9 = 117, which simplifies to x² + 3x − 54 = 0. Factorising yields (x + 9)(x − 6) = 0. Since we need natural numbers, we reject x = −9 and accept x = 6. Thus, the numbers are 6 and 9.
Another common pattern involves reciprocals. Imagine two natural numbers whose sum is 8, and the difference of their reciprocals is 2/15. Let one number be x, so the other is 8 − x. The reciprocal condition becomes 1/x − 1/(8−x) = 2/15. Combining the left side gives (8 − 2x)/[x(8−x)] = 2/15. Cross-multiplying and simplifying leads to x² − 23x + 60 = 0. This factors as (x − 20)(x − 3) = 0. Since the sum must be 8, we reject 20 and obtain the numbers 3 and 5.
Now let us turn to problems involving fractions. Suppose the denominator of a positive fraction exceeds twice the numerator by 1, and the sum of the fraction with its reciprocal equals 2.9. Let the numerator be x, so the denominator is 2x + 1. The condition states x/(2x+1) + (2x+1)/x = 29/10. Combining the left side and cross-multiplying eventually yields 8x² − 11x − 10 = 0. Solving gives x = 2 or x = −5/8. Since the fraction must be positive, we take x = 2, giving the fraction 2/5.
Two-digit numbers present a special structure. Consider a two-digit number where the product of digits is 6, and adding 9 reverses the digits. Let the tens digit be x and units digit be y. The number itself equals 10x + y. From the product condition, xy = 6. From the reversal condition, 10x + y + 9 = 10y + x, which simplifies to y = x + 1. Substituting, x(x+1) = 6, giving x² + x − 6 = 0. This factors as (x + 3)(x − 2) = 0. Since digits must be positive, x = 2 and y = 3, making the number 23.
Problems based on ratios require careful setup. Suppose three positive numbers are in the ratio 1/2 : 1/3 : 1/4, and the sum of their squares is 244. First, simplify the ratio by multiplying by 12, the LCM of denominators, obtaining 6 : 4 : 3. Let the numbers be 6x, 4x, and 3x. The sum of squares condition gives 36x² + 16x² + 9x² = 244, so 61x² = 244. Thus x² = 4, and since the numbers are positive, x = 2. The numbers are 12, 8, and 6.
Time and work problems follow a consistent pattern. Imagine two workers where one takes 6 hours less than the other for the same work, and together they finish in 13 hours 20 minutes. Let the slower worker take x hours, so the faster takes x − 6 hours. Converting 13 hours 20 minutes to 40/3 hours, their combined rate gives 1/x + 1/(x−6) = 3/40. Combining and cross-multiplying leads to 3x² − 98x + 240 = 0. Factorising as (x − 30)(3x − 8) = 0, we get x = 30 or x = 8/3. The second value would make the faster worker's time negative, so we reject it. The slower worker takes 30 hours.
Cost price and selling price problems often create elegant quadratic relationships. Suppose selling an article for rupees 24 results in a loss percentage equal to the cost price itself. Let the cost price be x rupees. The loss equals x% of x, which is x²/100 rupees. Since selling price equals cost price minus loss, we have x − x²/100 = 24. Rearranging: x² − 100x + 2400 = 0. Solving this quadratic will yield the cost price.
Pipe and cistern problems mirror time and work problems. Consider two taps filling a tank together in 9 3/8 hours, with the larger tap taking 10 hours less than the smaller one alone. Let the smaller tap take x hours, so the larger takes x − 10 hours. Converting 9 3/8 to 75/8 hours, the combined rate gives 1/x + 1/(x−10) = 8/75. This simplifies to 4x² − 115x + 375 = 0. Factorising yields (x − 25)(4x − 15) = 0. The value x = 15/4 would make the larger tap's time negative, so we take x = 25. The taps take 25 hours and 15 hours respectively.
Age problems require careful attention to time references. Suppose five years ago, a woman's age was the square of her son's age, and ten years hence, her age will be twice her son's age. Let the son's age five years ago be x years. Then the woman's age five years ago was x² years. Their present ages are x + 5 and x² + 5 respectively. Ten years hence, the condition gives x² + 15 = 2(x + 15). This simplifies to x² − 2x − 15 = 0, which factors as (x − 5)(x + 3) = 0. Rejecting the negative value, the son was 5 years old five years ago, and the woman is now 30 years old.
Distance, time, and speed problems often involve the fundamental relationship: time equals distance divided by speed. Consider a car travelling 72 kilometres. If its speed increased by 10 kilometres per hour, the journey would take 36 minutes less. Let the original speed be x km/h. The original time is 72/x hours, and the new time is 72/(x+10) hours. Converting 36 minutes to 3/5 hours, we get 72/x − 72/(x+10) = 3/5. Solving yields x = 30 km/h as the original speed, rejecting the negative solution.
For boats in streams, remember: downstream speed equals speed in still water plus stream speed, while upstream speed equals speed in still water minus stream speed. A motor boat with speed 9 km/h in still water goes 12 km downstream and returns in 3 hours total. Let the stream speed be x km/h. The equation becomes 12/(9+x) + 12/(9−x) = 3. Combining and simplifying gives x² = 9, so the stream speed is 3 km/h.
Geometrical problems bring algebra and geometry together. Consider a right triangle with hypotenuse 13 centimetres, where the longer leg exceeds the shorter by 7 centimetres. Let the shorter leg be x cm, so the longer is x + 7 cm. By Pythagoras theorem: x² + (x+7)² = 13². This expands to 2x² + 14x + 49 = 169, simplifying to x² + 7x − 60 = 0. Factorising as (x + 12)(x − 5) = 0, we take x = 5, giving sides of 5 cm and 12 cm.
For rectangles where numerical area equals numerical perimeter, let breadth be x and length be x + 3. The condition x(x+3) = 2(x + x + 3) simplifies to x² − x − 6 = 0. This gives x = 3, so the dimensions are 3 metres by 6 metres.
Finally, consider miscellaneous problems involving cost and quantity relationships. A piece of cloth costs 2000 rupees. If it were 10 metres longer with each metre costing 10 rupees less, the total cost would remain unchanged. Let the original length be x metres. The original rate is 2000/x rupees per metre. The new condition gives (x+10)(2000/x − 10) = 2000. Expanding and simplifying yields x² + 10x − 2000 = 0. This factors as (x + 50)(x − 40) = 0, giving length 40 metres and original rate 50 rupees per metre.
Another classic type: a shopkeeper buys books for 960 rupees. If each book cost 8 rupees less, he could buy 4 more books for the same amount. Let the original cost per book be x rupees. The number of books is 960/x. The alternative scenario gives 960/(x−8) − 960/x = 4. Simplifying leads to x² − 8x − 1920 = 0. Factorising as (x − 48)(x + 40) = 0, we obtain the original cost as 48 rupees per book.
Let us recap the essential takeaways from this chapter.
First, always begin by defining your variable clearly, representing the unknown quantity in the problem.
Second, translate every given condition into mathematical equations, paying careful attention to words like "sum," "difference," "product," "reciprocal," and "exceeds."
Third, simplify your equation to standard quadratic form ax² + bx + c = 0 before attempting to solve.
Fourth, after obtaining solutions, always verify them against the original problem constraints — reject negative values for lengths, ages, or counts, and ensure fractions remain positive when required.
Fifth, familiarise yourself with common problem structures: numbers and their reciprocals, two-digit numbers as 10x + y, time and work as reciprocal rates, and relative speeds for boats and streams.
Sixth, remember that many problems offer two equivalent formulations — choose the one that leads to simpler algebra.
Mathematics is not merely about finding answers; it is about seeing patterns, choosing elegant paths, and verifying that your solutions make sense in the real world. With quadratic equations as your tool, you can unlock problems across numbers, ages, speeds, work, and geometry. Practice translating words into equations, and equations back into meaningful interpretations. Until next time, keep exploring, keep questioning, and enjoy the beauty of mathematical thinking.