Worksheet
Explore the world of quadratic equations, learning to solve them using various methods like factoring, completing the square, and the quadratic formula.
Quadratic Equations - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
This worksheet covers essential long-answer questions to help you build confidence in Quadratic Equations from Mathematics for Class X (Mathematics).
Basic comprehension exercises
Strengthen your understanding with fundamental questions about the chapter.
Questions
Explain the concept of quadratic equations and how they differ from linear equations. Provide examples to illustrate the difference.
Recall the definitions and general forms of both types of equations.
Solution
Quadratic equations are polynomial equations of the second degree, meaning the highest power of the variable is two. The general form is ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. Linear equations, on the other hand, are first-degree equations with the general form ax + b = 0. The key difference lies in the degree of the variable; quadratic equations can have two solutions, while linear equations have one. For example, x² - 5x + 6 = 0 is a quadratic equation with solutions x = 2 and x = 3, whereas 2x + 3 = 0 is a linear equation with solution x = -1.5. Quadratic equations often arise in areas like physics, engineering, and economics to model scenarios involving acceleration, area, and optimization.
Describe the method of factorisation to solve quadratic equations. Use the equation x² - 5x + 6 = 0 as an example.
Identify two numbers that multiply to the constant term and add to the coefficient of x.
Solution
Factorisation involves expressing the quadratic equation as a product of two binomials set to zero. For x² - 5x + 6 = 0, we look for two numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of x). These numbers are -2 and -3. Thus, the equation can be written as (x - 2)(x - 3) = 0. Setting each factor equal to zero gives x - 2 = 0 or x - 3 = 0, leading to solutions x = 2 and x = 3. This method is efficient for equations that can be easily factored but may not be suitable for all quadratics, especially those with irrational or complex roots.
What is the quadratic formula, and how is it derived? Demonstrate its use by solving 2x² + 4x - 6 = 0.
Recall the process of completing the square and how it leads to the quadratic formula.
Solution
The quadratic formula, x = [-b ± √(b² - 4ac)] / (2a), is derived by completing the square on the general quadratic equation ax² + bx + c = 0. For 2x² + 4x - 6 = 0, a = 2, b = 4, and c = -6. Plugging these into the formula gives x = [-4 ± √(16 + 48)] / 4 = [-4 ± √64] / 4 = [-4 ± 8] / 4. This yields two solutions: x = (-4 + 8)/4 = 1 and x = (-4 - 8)/4 = -3. The quadratic formula is universally applicable, providing solutions even when factorisation is not feasible.
Explain the significance of the discriminant in quadratic equations. How does it determine the nature of the roots?
Consider the expression under the square root in the quadratic formula.
Solution
The discriminant, D = b² - 4ac, is a key component of the quadratic formula that indicates the nature of the roots of a quadratic equation. If D > 0, there are two distinct real roots. If D = 0, there is exactly one real root (a repeated root). If D < 0, there are no real roots, but two complex conjugate roots. For example, in x² - 4x + 4 = 0, D = 16 - 16 = 0, indicating a repeated root at x = 2. The discriminant thus provides a quick way to assess the nature of solutions without solving the equation fully.
How can quadratic equations be applied to real-life situations? Provide an example involving area calculation.
Think about how area problems can lead to quadratic equations.
Solution
Quadratic equations model various real-life scenarios, such as calculating areas, projectile motion, and profit maximization. For instance, suppose a rectangular garden's length is 5 meters more than its width, and the area is 150 m². Let the width be x meters; then the length is (x + 5) meters. The area equation is x(x + 5) = 150, leading to x² + 5x - 150 = 0. Solving this quadratic equation (by factorisation, completing the square, or the quadratic formula) gives x = 10 (since width cannot be negative). Thus, the garden is 10m wide and 15m long. This example shows how quadratics can solve practical problems involving dimensions and areas.
Solve the quadratic equation 3x² - 2x - 1 = 0 using the method of completing the square.
Remember to adjust the equation so the coefficient of x² is 1 before completing the square.
Solution
To complete the square for 3x² - 2x - 1 = 0, first divide by the coefficient of x²: x² - (2/3)x - 1/3 = 0. Move the constant term: x² - (2/3)x = 1/3. Take half of the coefficient of x, square it, and add to both sides: x² - (2/3)x + (1/3)² = 1/3 + (1/3)² → x² - (2/3)x + 1/9 = 4/9. Write the left side as a square: (x - 1/3)² = 4/9. Take square roots: x - 1/3 = ±2/3. Solve for x: x = 1/3 ± 2/3, giving x = 1 or x = -1/3. This method transforms the equation into a perfect square, facilitating solution.
Discuss the conditions under which a quadratic equation has no real roots. Provide an example.
Examine the discriminant's value to determine the nature of the roots.
Solution
A quadratic equation ax² + bx + c = 0 has no real roots when the discriminant D = b² - 4ac is negative. This means the parabola represented by the equation does not intersect the x-axis. For example, x² + x + 1 = 0 has D = 1 - 4(1)(1) = -3 < 0, indicating no real roots. The solutions are complex: x = [-1 ± √(-3)]/2 = [-1 ± i√3]/2. Such cases often occur in problems where physical quantities cannot be negative or complex, implying no feasible solution under given constraints.
Find the roots of the equation x² + 4x + 5 = 0 and explain the nature of these roots.
Calculate the discriminant to assess the nature of the roots before solving.
Solution
For x² + 4x + 5 = 0, the discriminant D = 16 - 20 = -4 < 0, indicating no real roots. The solutions are complex: x = [-4 ± √(-4)]/2 = [-4 ± 2i]/2 = -2 ± i. Thus, the roots are x = -2 + i and x = -2 - i, which are complex conjugates. This illustrates that when D < 0, the quadratic equation has two complex roots, reflecting scenarios where real-number solutions do not exist, such as certain physical systems or geometric constraints.
A train travels 300 km at a uniform speed. If the speed were increased by 5 km/h, the journey would take 2 hours less. Formulate a quadratic equation to find the original speed of the train.
Relate time, distance, and speed to form the equation.
Solution
Let the original speed be x km/h. Time taken at this speed is 300/x hours. At increased speed (x + 5) km/h, time is 300/(x + 5) hours. According to the problem, 300/x - 300/(x + 5) = 2. Multiply through by x(x + 5) to eliminate denominators: 300(x + 5) - 300x = 2x(x + 5) → 1500 = 2x² + 10x. Rearrange: 2x² + 10x - 1500 = 0 → x² + 5x - 750 = 0. This quadratic equation can be solved to find x = 25 (since speed cannot be negative), so the original speed was 25 km/h.
Explain how the sum and product of the roots of a quadratic equation relate to its coefficients. Use the equation 2x² - 8x + 6 = 0 as an example.
Recall Vieta's formulas connecting the roots to the coefficients.
Solution
For any quadratic equation ax² + bx + c = 0, the sum of the roots (α + β) = -b/a, and the product (αβ) = c/a. For 2x² - 8x + 6 = 0, sum = -(-8)/2 = 4, and product = 6/2 = 3. Indeed, solving the equation gives roots x = 1 and x = 3, with sum 4 and product 3. These relationships, known as Vieta's formulas, are useful for checking solutions or constructing equations with given roots. They highlight the symmetric properties of quadratics and their roots.
Quadratic Equations - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Quadratic Equations to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
Questions
A charity trust decides to build a prayer hall with a carpet area of 300 square meters. The length is one meter more than twice its breadth. Formulate the quadratic equation representing this situation and find its roots.
Start by expressing the length in terms of breadth and then form the area equation.
Solution
Let the breadth be x meters. Then, length = (2x + 1) meters. Area = length × breadth = (2x + 1)x = 2x² + x. Given area = 300 m², so 2x² + x - 300 = 0. Using the quadratic formula, x = [-1 ± √(1 + 2400)]/4 = [-1 ± √2401]/4 = [-1 ± 49]/4. Thus, x = 12 or x = -12.5. Since breadth can't be negative, x = 12 meters.
Compare and contrast the methods of solving quadratic equations by factorization and by using the quadratic formula. Include examples.
Consider the applicability and steps involved in each method.
Solution
Factorization involves expressing the quadratic equation as a product of two linear factors and solving for x. It's straightforward when the equation can be easily factorized. Example: x² - 5x + 6 = 0 factors to (x-2)(x-3)=0, giving x=2,3. The quadratic formula, x = [-b ± √(b² - 4ac)]/(2a), works for any quadratic equation, even when factorization is complex. Example: For 2x² - 4x - 6 = 0, x = [4 ± √(16 + 48)]/4 = [4 ± √64]/4 = [4 ± 8]/4, giving x=3, -1.
Find the discriminant of the quadratic equation 3x² - 2x + 1/3 = 0 and determine the nature of its roots.
Recall that the discriminant reveals the nature of the roots without solving the equation.
Solution
The discriminant D = b² - 4ac = (-2)² - 4×3×(1/3) = 4 - 4 = 0. Since D=0, the equation has two equal real roots. The roots are x = [2 ± √0]/6 = 2/6 = 1/3.
A train travels 480 km at a uniform speed. If the speed were 8 km/h less, it would take 3 hours more. Formulate the quadratic equation and find the original speed.
Relate the time difference to the speed change to form the equation.
Solution
Let original speed be x km/h. Time taken = 480/x hours. Reduced speed = (x-8) km/h, new time = 480/(x-8). Given, 480/(x-8) - 480/x = 3. Solving, 480x - 480(x-8) = 3x(x-8) → 3840 = 3x² - 24x → x² - 8x - 1280 = 0. Using quadratic formula, x = [8 ± √(64 + 5120)]/2 = [8 ± √5184]/2 = [8 ± 72]/2. Thus, x=40 or x=-32. Speed can't be negative, so x=40 km/h.
Prove that the quadratic equation x² + (a+b)x + ab = 0 always has real roots for all real values of a and b.
Calculate the discriminant and show it's always non-negative.
Solution
The discriminant D = (a+b)² - 4×1×ab = a² + 2ab + b² - 4ab = a² - 2ab + b² = (a - b)². Since squares are always non-negative, D ≥ 0. Thus, the equation always has real roots.
A rectangular park's perimeter is 80 m and area is 400 m². Is this possible? If so, find its dimensions.
Use the perimeter and area to form two equations and solve simultaneously.
Solution
Let length = l, breadth = b. Perimeter = 2(l + b) = 80 → l + b = 40. Area = l × b = 400. The quadratic equation formed is x² - (l+b)x + lb = 0 → x² - 40x + 400 = 0. Discriminant D = 1600 - 1600 = 0. Thus, l = b = 20 m. So, a square park of side 20 m satisfies the conditions.
Explain why the quadratic equation x² + 1 = 0 has no real roots.
Consider the properties of square numbers and the discriminant.
Solution
The equation can be written as x² = -1. For real numbers, x² is always non-negative, so x² = -1 has no real solutions. The discriminant D = 0 - 4×1×1 = -4 < 0 confirms no real roots.
Find the value of k for which the equation 2x² + kx + 3 = 0 has equal roots.
Set the discriminant to zero and solve for k.
Solution
For equal roots, discriminant D = 0. Here, D = k² - 4×2×3 = k² - 24 = 0 → k² = 24 → k = ±2√6.
The product of two consecutive positive integers is 306. Formulate the quadratic equation and find the integers.
Express the product in terms of x and form the quadratic equation.
Solution
Let the integers be x and x+1. Then, x(x+1) = 306 → x² + x - 306 = 0. Solving, x = [-1 ± √(1 + 1224)]/2 = [-1 ± √1225]/2 = [-1 ± 35]/2. Thus, x=17 or x=-18. Since integers are positive, x=17. The integers are 17 and 18.
A right triangle's hypotenuse is 13 cm, and one side is 7 cm less than the other. Formulate the quadratic equation and find the sides.
Use the Pythagorean theorem to relate the sides and form the equation.
Solution
Let one side be x cm, then the other is (x-7) cm. By Pythagoras, x² + (x-7)² = 13² → x² + x² -14x +49 =169 → 2x² -14x -120=0 → x² -7x -60=0. Solving, x = [7 ± √(49 + 240)]/2 = [7 ± √289]/2 = [7 ± 17]/2. Thus, x=12 or x=-5. Side length can't be negative, so x=12 cm. The sides are 12 cm and 5 cm.
Quadratic Equations - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Quadratic Equations in Class X.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
Questions
A charity trust decides to build a prayer hall with a carpet area of 300 square meters, where the length is one meter more than twice its breadth. Formulate the quadratic equation representing this scenario and find the dimensions of the hall.
Start by expressing the length in terms of breadth and then use the area formula to form the equation.
Solution
Let the breadth be x meters. Then, the length is (2x + 1) meters. The area is x(2x + 1) = 300, leading to the quadratic equation 2x² + x - 300 = 0. Solving this equation by factorization or quadratic formula gives x = 12 (breadth) and length = 25 meters.
Explain how the Babylonians solved quadratic equations of the form x² - px + q = 0 and compare it with the modern quadratic formula.
Consider the relationship between the sum and product of roots and the coefficients of the quadratic equation.
Solution
Babylonians used geometric methods to find two numbers with a given sum (p) and product (q), equivalent to solving x² - px + q = 0. The modern quadratic formula, x = [p ± √(p² - 4q)]/2, provides a direct algebraic solution. Both methods aim to find roots but differ in approach.
Given the quadratic equation 3x² - 2x + 1/3 = 0, find its discriminant and determine the nature of its roots. If real, find them.
Recall that the discriminant reveals the nature of the roots: positive for two distinct real roots, zero for equal roots, and negative for no real roots.
Solution
The discriminant D = (-2)² - 4*3*(1/3) = 4 - 4 = 0. Since D = 0, the equation has two equal real roots. The roots are x = [2 ± √0]/6 = 1/3.
A train travels 480 km at a uniform speed. If the speed were 8 km/h less, it would take 3 hours more. Formulate the quadratic equation and find the original speed.
Express the time difference in terms of speed and set up the equation based on the relationship between speed, distance, and time.
Solution
Let the original speed be x km/h. Time taken is 480/x hours. Reduced speed scenario: 480/(x - 8) = 480/x + 3. This leads to the equation x² - 8x - 1280 = 0. Solving gives x = 40 km/h (original speed).
Discuss the significance of the discriminant in determining the nature of the roots of a quadratic equation, with examples.
The discriminant is part of the quadratic formula under the square root, affecting the root's reality and nature.
Solution
The discriminant (D = b² - 4ac) determines the nature of roots: (1) D > 0: two distinct real roots (e.g., x² - 5x + 6 = 0, D = 1, roots 2 and 3). (2) D = 0: one real root (e.g., x² - 4x + 4 = 0, D = 0, root 2). (3) D < 0: no real roots (e.g., x² + x + 1 = 0, D = -3).
Find the value of k for which the quadratic equation 2x² + kx + 3 = 0 has equal roots.
Set the discriminant to zero and solve for k to ensure the equation has exactly one real root.
Solution
For equal roots, the discriminant must be zero: D = k² - 4*2*3 = 0 → k² = 24 → k = ±2√6.
Is it possible to design a rectangular park with perimeter 80 m and area 400 m²? Justify your answer mathematically.
Formulate equations for perimeter and area, then derive a quadratic equation to check for real solutions.
Solution
Let length be l and breadth be b. 2(l + b) = 80 → l + b = 40. Area lb = 400. The quadratic equation is x² - 40x + 400 = 0. Discriminant D = 1600 - 1600 = 0, so l = b = 20 m. Yes, a square park of 20 m sides meets the criteria.
The product of two consecutive positive integers is 306. Form the quadratic equation and find the integers.
Consecutive integers differ by 1. Set up the product equation and solve the resulting quadratic.
Solution
Let the integers be x and x + 1. The equation is x(x + 1) = 306 → x² + x - 306 = 0. Solving gives x = 17 (since x must be positive), so the integers are 17 and 18.
Analyze the quadratic equation x² - 55x + 750 = 0 derived from a toy production scenario, where x is the number of toys produced. Interpret the roots in this context.
Consider that quadratic equations can have two real roots, both potentially meaningful in the given context.
Solution
The equation models the cost of production. Solving x² - 55x + 750 = 0 gives x = 25 or 30. Both roots are valid, representing production levels where total cost is 750. The scenario shows two possible production quantities leading to the same total cost.
A right triangle's hypotenuse is 13 cm, and the altitude is 7 cm less than the base. Formulate and solve the quadratic equation to find the base and altitude.
Apply the Pythagorean theorem to relate the sides of the triangle and form the quadratic equation.
Solution
Let base be x cm. Altitude is (x - 7) cm. By Pythagoras, x² + (x - 7)² = 13² → 2x² - 14x - 120 = 0 → x² - 7x - 60 = 0. Solving gives x = 12 cm (base) and altitude = 5 cm.
Real Numbers encompass all rational and irrational numbers, forming a complete and continuous number line essential for various mathematical concepts.
Explore the world of Polynomials, understanding their types, degrees, and operations to solve algebraic expressions and equations effectively.
Explore the methods to solve a pair of linear equations in two variables, including graphical, substitution, elimination, and cross-multiplication techniques.
A chapter that explores sequences where each term after the first is obtained by adding a constant difference, focusing on their properties, nth term, and sum formulas.
