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Quadratic Equations - Formulate and solve real-life problems leading to quadratic equations

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A quadratic equation in the variable xx is an equation of the form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,ca, b, c are real numbers and a≠0a \neq 0.

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To formulate a real-life problem into a quadratic equation, identify the unknown quantity (e.g., speed, age, length) and represent it by a variable xx.

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Translate the given conditions into algebraic expressions and set up an equation. For example, in speed-distance problems, use the relation Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}.

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Once the equation ax2+bx+c=0ax^2 + bx + c = 0 is formed, solve it using Factorization or the Quadratic Formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

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For physical quantities like length, breadth, or speed, only positive roots are usually considered. If the discriminant D=b2−4acD = b^2 - 4ac is negative, the given situation is not mathematically possible in the real number system.

📐Formulae

ax2+bx+c=0,a≠0ax^2 + bx + c = 0, a \neq 0

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

D=b2−4acD = b^2 - 4ac

Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}

Area of Rectangle=length×breadth\text{Area of Rectangle} = \text{length} \times \text{breadth}

💡Examples

Problem 1:

The area of a rectangular plot is 528 m2528 \text{ m}^2. The length of the plot (in metres) is one more than twice its breadth. Find the dimensions of the plot.

Solution:

Let the breadth of the plot be x metresx \text{ metres}. Then, the length is (2x+1) metres(2x + 1) \text{ metres}. Area of the rectangle = length×breadth\text{length} \times \text{breadth}. x(2x+1)=528x(2x + 1) = 528 2x2+x−528=02x^2 + x - 528 = 0 Using the quadratic formula where a=2,b=1,c=−528a = 2, b = 1, c = -528: D=b2−4ac=12−4(2)(−528)D = b^2 - 4ac = 1^2 - 4(2)(-528) 1+42244225\begin{array}{r} 1 \\ + 4224 \\ \hline 4225 \end{array} x=−1±42252(2)=−1±654x = \frac{-1 \pm \sqrt{4225}}{2(2)} = \frac{-1 \pm 65}{4} x=644=16x = \frac{64}{4} = 16 or x=−664=−16.5x = \frac{-66}{4} = -16.5 Since breadth cannot be negative, x=16x = 16. Breadth = 16 m16 \text{ m}, Length = 2(16)+1=33 m2(16) + 1 = 33 \text{ m}.

Explanation:

We define the breadth as xx and length as 2x+12x+1 based on the problem statement. The product of these equals the area, resulting in a quadratic equation. We ignore the negative root because dimensions must be positive.

Problem 2:

An express train takes 1 hour1 \text{ hour} less than a passenger train to travel 132 km132 \text{ km} between Mysore and Bangalore. If the average speed of the express train is 11 km/h11 \text{ km/h} more than that of the passenger train, find the average speed of the two trains.

Solution:

Let the average speed of the passenger train be x km/hx \text{ km/h}. Speed of the express train = (x+11) km/h(x + 11) \text{ km/h}. Time taken by passenger train = 132x hours\frac{132}{x} \text{ hours}. Time taken by express train = 132x+11 hours\frac{132}{x + 11} \text{ hours}. According to the problem: 132x−132x+11=1\frac{132}{x} - \frac{132}{x + 11} = 1 132(x+11−xx(x+11))=1132 \left( \frac{x + 11 - x}{x(x + 11)} \right) = 1 132(11)=x(x+11)132(11) = x(x + 11) 1452=x2+11x1452 = x^2 + 11x x2+11x−1452=0x^2 + 11x - 1452 = 0 Factorizing the equation: x2+44x−33x−1452=0x^2 + 44x - 33x - 1452 = 0 x(x+44)−33(x+44)=0x(x + 44) - 33(x + 44) = 0 (x−33)(x+44)=0(x - 33)(x + 44) = 0 x=33x = 33 or x=−44x = -44. Speed cannot be negative, so x=33 km/hx = 33 \text{ km/h}. Passenger train speed = 33 km/h33 \text{ km/h}, Express train speed = 44 km/h44 \text{ km/h}.

Explanation:

This problem uses the relationship between speed, distance, and time. By setting the difference in time equal to 11, we form a quadratic equation. Splitting the middle term helps find the average speed of the trains.