Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot is one more than twice its breadth. Find the length and breadth.
(ii) The product of two consecutive positive integers is 306. Find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed were 8 km/h less, the journey would take 3 hours more. Find the speed of the train.
(i) \(2x^2 + x - 528 = 0\), where \(x\) is the breadth (in metres).
(ii) \(x^2 + x - 306 = 0\), where \(x\) is the smaller integer.
(iii) \(x^2 + 32x - 273 = 0\), where \(x\) (in years) is Rohan’s present age.
(iv) \(u^2 - 8u - 1280 = 0\), where \(u\) (in km/h) is the speed of the train.
Goal: Convert each real-life condition into a mathematical statement and simplify it to the standard form of a quadratic equation.
1. Let the breadth of the plot be \(x\) metres.
2. The length is given as “one more than twice the breadth,” so length = \(2x + 1\).
3. Area of a rectangle = length × breadth. Therefore, \(x(2x + 1) = 528\).
4. Expand the product: \(2x^2 + x = 528\).
5. Bring all terms to one side: \(2x^2 + x - 528 = 0\).
1. Let the smaller integer be \(x\).
2. The next consecutive integer is \(x + 1\).
3. Their product is 306: \(x(x + 1) = 306\).
4. Expand: \(x^2 + x = 306\).
5. Rearrange to standard form: \(x^2 + x - 306 = 0\).
1. Let Rohan’s present age be \(x\) years.
2. Mother’s present age = \(x + 26\).
3. After 3 years, Rohan’s age = \(x + 3\). Mother’s age = \(x + 29\).
4. Their age product after 3 years is 360, so: \((x + 3)(x + 29) = 360\).
5. Expand the LHS: \(x^2 + 29x + 3x + 87 = x^2 + 32x + 87\).
6. Equation becomes: \(x^2 + 32x + 87 = 360\).
7. Move 360 to the LHS: \(x^2 + 32x - 273 = 0\).
1. Let the speed of the train be \(u\) km/h.
2. Time = distance ÷ speed. So original time = \(480/u\).
3. Reduced speed = \(u - 8\), so new time = \(480/(u - 8)\).
4. According to the problem, the slower speed increases travel time by 3 hours: \(\frac{480}{u - 8} = \frac{480}{u} + 3\).
5. Multiply both sides by \(u(u - 8)\) to clear denominators: \(480u = 480(u - 8) + 3u(u - 8)\).
6. Expand and simplify: \(480u = 480u - 3840 + 3u^2 - 24u\).
7. Combine like terms to get: \(3u^2 - 24u - 3840 = 0\).
8. Divide through by 3: \(u^2 - 8u - 1280 = 0\).
Final: Each situation now gives a quadratic equation in the required standard form.