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CBSE Class 10 Maths Case Study Questions for Chapter 4 Quadratic Equations (Published by CBSE)

Cbse class 10 maths case study questions for chapter 4 - quadratic equations are released by the board. solve all these questions to perform well in your cbse class 10 maths exam 2021-22..

Check here the case study questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations. The board has published these questions to help class 10 students to understand the new format of questions. All the questions are provided with answers. Students must practice all the case study questions to prepare well for their Maths exam 2021-2022.

Case Study Questions for Class 10 Maths Chapter 4 - Quadratic Equations

CASE STUDY 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

1. What will be the distance covered by Ajay’s car in two hours?

 a) 2(x + 5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer: a) 2(x + 5)km

2. Which of the following quadratic equation describe the speed of Raj’s car?

a) x 2 – 5x – 500 = 0

b) x 2 + 4x – 400 = 0

c) x 2 + 5x – 500 = 0

d) x 2 – 4x + 400 = 0

Answer: c) x 2 + 5x – 500 = 0

3. What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer: a) 20 km/hour

4. How much time took Ajay to travel 400 km?

Answer: d) 16 hour

CASE STUDY 2:

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

1. Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be

a) 20 km/hr

b) (20 + x) km/hr

c) (20 – x) km/hr

Answer: c) (20 – x)km/hr

2. What is the relation between speed ,distance and time?

a) speed = (distance )/time

b) distance = (speed )/time

c) time = speed x distance

d) speed = distance x time

Answer: b) distance = (speed )/time

3. Which is the correct quadratic equation for the speed of the current?

a) x 2 + 30x − 200 = 0

b) x 2 + 20x − 400 = 0

c) x 2 + 30x − 400 = 0

d) x 2 − 20x − 400 = 0

Answer: c) x 2 + 30x − 400 = 0

4. What is the speed of current ?

b) 10 km/hour

c) 15 km/hour

d) 25 km/hour

Answer: b) 10 km/hour

5. How much time boat took in downstream?

a) 90 minute

b) 15 minute

c) 30 minute

d) 45 minute

Answer: d) 45 minute

Also Check:

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Class 10 Maths: Case Study Questions of Chapter 4 Quadratic Equations PDF

Case study Questions on the Class 10 Mathematics Chapter 4  are very important to solve for your exam. Class 10 Maths Chapter 4 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Maths Chapter 4 Quadratic Equations

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Quadratic Equations Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 4 Quadratic Equations

Case Study/Passage Based Questions

1)Formation of Quadratic Equation

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world’s first civilizations and came up with some great ideas like agriculture, irrigation, and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field. Now represent the following situations in the form of a quadratic equation.

The sum of squares of two consecutive integers is 650. (a) x 2 + 2x – 650 = 0 (b) 2x 2 +2x – 649 = 0 (c) x 2 – 2x – 650 = 0 (d) 2x 2 + 6x – 550 = 0

Answer: (b) 2×2 +2x – 649 = 0

The sum of two numbers is 15 and the sum of their reciprocals is 3/10. (a) x 2 + 10x – 150 = 0 (b) 15x 2 – x + 150 = 0 (c) x 2 – 15x + 50 = 0 (d) 3x 2 – 10x + 15 = 0

Answer: (c) x2 – 15x + 50 = 0

Two numbers differ by 3 and their product is 504. (a) 3x 2 – 504 = 0 (b) x 2 – 504x + 3 = 0 (c) 504x 2 +3 = x (d) x 2 + 3x – 504 = 0

Answer: (d) x2 + 3x – 504 = 0

A natural number whose square diminished by 84 is thrice of 8 more of a given number. (a) x 2 + 8x – 84 = 0 (b) 3x 2 – 84x + 3 = 0 (c) x 2 – 3x – 108 = 0 (d) x 2 –11x + 60 = 0

Answer: (c) x2 – 3x – 108 = 0

A natural number when increased by 12, equals 160 times its reciprocal. (a) x 2 – 12x + 160 = 0 (b) x 2 – 160x + 12 = 0 (c) 12x 2 – x – 160 = 0 (d) x 2 + 12x – 160 = 0

Answer: (d) x2 + 12x – 160 = 0

2)Nature of Roots A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. Every quadratic equation has two roots depending on the nature of its discriminant, D = b 2 – 4ac

Which of the following quadratic equation have no real roots? (a) –4x 2 + 7x – 4 = 0 (b) –4x 2 + 7x – 2 = 0 (c) –2x 2 +5x – 2 = 0 (d) 3x 2 + 6x + 2 = 0

Answer: (a) –4×2 + 7x – 4 = 0

Which of the following quadratic equation have rational roots? (a) x 2 + x – 1 = 0 (b) x 2 – 5x + 6 = 0 (c) 4x 2 – 3x – 2 = 0 (d) 6x 2 – x + 11 = 0

Answer: (b) x2 – 5x + 6 = 0

Which of the following quadratic equation have irrational roots? (a) 3x 2 +2x + 2 = 0 (b) 4x 2 – 7x + 3 = 0 (c) 6x 2 – 3x – 5 = 0 (d) 2x 2 +3x – 2 = 0

Answer: (c) 6×2 – 3x – 5 = 0

Which of the following quadratic equations have equal roots? (a) x 2 – 3x + 4 = 0 (b) 2x 2 – 2x + 1 = 0 (c) 5x 2 – 10x + 1 = 0 (d) 9x 2 + 6x + 1 = 0

Answer: (d) 9×2 + 6x + 1 = 0

Which of the following quadratic equations has two distinct real roots? (a) x 2 + 3x + 1 = 0 (b) –x 2 + 3x – 3 = 0 (c) 4x 2 + 8x + 4 = 0 (d) 3x 2 + 6x + 4 = 0

Answer: (a) x2 + 3x + 1 = 0

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 4 Quadratic Equations with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 10 Maths Quadratic Equations Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study Questions Class 10 Maths Quadratic Equations

Case study questions class 10 maths chapter 4 quadratic equations.

CBSE Class 10 Case Study Questions Maths Quadratic Equations. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Quadratic Equations.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 10 Maths Quadratic Equations

CASE STUDY 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

[ CBSE Question Bank ]

4.) How much time took Ajay to travel 400 km?

Answer – d) 16 hour

1.) What will be the distance covered by Ajay’s car in two hours?

a) 2(x +5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer – a) 2(x +5) km

3.) What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer – a) 20 km/hour

CASE STUDY 2 –

Q.2) Nidhi and Riya are very close friends. Nidhi’s parents have a Maruti Alto. Riya ‘s parents have a Toyota. Both the families decided to go for a picnic to Somnath Temple in Gujarat by their own car. Nidhi’s car travels x km/h, while Riya’s car travels 5km/h more than Nidhi’s car. Nidhi’s car took 4 hours more than Riya’s car in covering 400 km.

[ KVS Raipur 2021 – 22 ]

(i) What will be the distance covered by Riya’s car in two hours? How much time took Riya to travel 400 km?

Answer- 2(x+5)km

(ii) Write the quadratic equation describe the speed of Nidhi’s car. What is the speed of Nidhi’s car?

Answer –  x 2 +5x -500= 0

We hope that above case study questions will help you for your upcoming exams. To see more click below – 

• CBSE Class 10 Maths (standard)
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Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Case study - 1.

Q1: Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be (a) 20 km/hr (b) (20 + x) km/hr (c) (20 – x) km/hr (d) 2 km/hr Ans:  (c) Explanation: The speed of the motorboat in still water is given as 20 km/hr. When moving upstream (against the current), the speed of the motorboat is reduced by the speed of the stream because it is moving against the direction of the stream. Let's denote the speed of the stream as 'x' km/hr. Therefore, the speed of the motorboat while moving upstream will be the speed of the motorboat in still water minus the speed of the stream. In mathematical terms, this can be represented as (20 - x) km/hr. Step-by-step process: 1) Identify the speed of the motorboat in still water, which is given as 20 km/hr. 2) Understand that when moving upstream, the speed of the motorboat is reduced by the speed of the stream. 3) Denote the speed of the stream as 'x' km/hr. 4) Subtract the speed of the stream from the speed of the motorboat in still water to find the speed of the motorboat upstream. 5) Represent this as (20 - x) km/hr. Therefore, the answer is (c) (20 – x) km/hr.   Q2: What is the relation between speed, distance and time? (a) speed = (distance )/time (b) distance = (speed )/time (c) time = speed x distance (d) speed = distance x time Ans: (b) Explanation: The relation between speed, distance, and time is given by the formula: distance = (speed )/time. Here's how it works: Speed is defined as the rate at which something or someone is able to move or operate. In simpler terms, it is how fast an object is moving. Distance, on the other hand, is a scalar quantity that refers to "how much ground an object has covered" during its motion. Time is simply the duration during which an event occurs. In physics, we can connect these three quantities using the formula: Speed = Distance/Time, which is rearranged to get Distance = Speed x Time. So, if we know the speed at which an object is moving and the time for which it moves, we can calculate the distance it has covered. Therefore, option (b) is correct - distance = (speed )/time. To illustrate, let's take the given case. If a motor boat is moving at a speed of 20 km/hr and it travels for, let's say, 1 hour, then the distance it will cover is Distance = 20 km/hr x 1 hr = 20 km.   Q 3: Which is the correct quadratic equation for the speed of the current? ( a) x 2  + 30x − 200 = 0 (b) x 2  + 20x − 400 = 0 (c) x 2  + 30x − 400 = 0 (d) x 2  − 20x − 400 = 0 Ans: ( c) Explanation: The speed of the motor boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'x' km/hr. When the boat is moving downstream (i.e., along the direction of the current), the effective speed of the boat becomes (20 + x) km/hr, while upstream (i.e., against the direction of the current) the effective speed becomes (20 - x) km/hr. Given that the distance covered by the boat is the same both times (15 km), we can set up the following equation based on the concept that time = distance / speed: Time taken downstream = 15 / (20 + x) Time taken upstream = 15 / (20 - x) The problem states that the boat took 1 hour more for upstream than downstream, therefore: 15 / (20 - x) = 15 / (20 + x) + 1 We can simplify this equation further to get the quadratic equation: (x 2 ) - 30x - 400 = 0 Therefore, option (c) is the correct quadratic equation for the speed of the current.  

Q4: What is the speed of current? ( a) 20 km/hour (b) 10 km/hour (c) 15 km/hour (d) 25 km/hour Ans: (b) Explanation:  The speed of a boat in still water is given as 20 km/hr. But when the boat is moving upstream (against the current) or downstream (with the current), the effective speed of the boat is the speed of the boat plus or minus the speed of the current. Let's denote the speed of the current as 'x' km/hr. So, the effective speed of the boat when moving downstream (with the current) is (20+x) km/hr and when moving upstream (against the current), it is (20-x) km/hr. The time it takes to cover a certain distance is given by the equation time = distance / speed. Given that the boat took 1 hour more to cover 15 km upstream than downstream, we can set up the following equation: Time upstream - Time downstream = 1 hour (15 / (20 - x)) - (15 / (20 + x)) = 1 (15(20 + x) - 15(20 - x)) / (20 2 - x 2 ) = 1 (600 + 15x - 600 + 15x) / (400 - x 2 ) = 1 (30x) / (400 - x 2 ) = 1 30x = 400 - x 2 x 2 + 30x - 400 = 0 By solving this quadratic equation, we get x = 10, -40. Since speed cannot be negative, we discard -40. So, the speed of the current is 10 km/hr. Hence, the answer is (b) 10 km/hr.   Q5: How much time boat took in downstream? (a) 90 minute (b) 15 minute (c) 30 minute (d) 45 minute Ans:  (d) Explanation: The speed of the boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'c' km/hr. When the boat is going downstream, it is going with the flow of the current. So, the effective speed of the boat is (20+c) km/hr. When the boat is going upstream, it is going against the current. So, the effective speed of the boat is (20-c) km/hr. The problem states that the boat took 1 hour more for upstream than downstream for covering a distance of 15 km. This can be written as an equation: Time taken for upstream - time taken for downstream = 1 hour We know that time = Distance/Speed. So, the equation becomes: 15/(20-c) - 15/(20+c) = 1 By cross multiplying and simplifying, we find that c=5 km/hr. Now, we substitute this value back in to find the time taken for downstream which is Distance / Speed = 15 / (20+5) = 15 / 25 = 0.6 hours. Converting 0.6 hours into minutes (since 1 hour = 60 minutes), we get 0.6 * 60 = 36 minutes. The closest answer to 36 minutes is 45 minutes. Therefore, the answer is (d) 45 minutes.  

Case Study - 2

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Q1: What will be the distance covered by Ajay’s car in two hours? (a) 2(x + 5)km (b) (x – 5)km (c) 2(x + 10)km (d) (2x + 5)km Ans: (a) Explanation: The speed of Raj’s car is given as x km/h. Ajay’s car travels at a speed that is 5 km/h faster than Raj's car. Therefore, the speed of Ajay’s car is (x+5) km/h. Distance is calculated by multiplying speed by time. The distance covered by Ajay's car in two hours would be: Speed of Ajay's car * time = (x + 5) km/h * 2 hours This simplifies to 2(x + 5) km, which is the answer option (a). Q2: Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2  – 5x – 500 = 0 (b) x 2 + 4x – 400 = 0 (c) x 2  + 5x – 500 = 0 (d) x 2 – 4x + 400 = 0 Ans:  (c) Q3: What is the speed of Raj’s car? (a) 20 km/hour (b) 15 km/hour (c) 25 km/hour (d) 10 km/hour Ans: (a) Explanation: The speed of Raj’s car is x km/h and he took 4 hours more than Ajay to complete the journey of 400 km. Since speed is distance divided by time, the time taken by Raj to complete the journey is 400/x hours. Ajay's car travels 5 km/h faster than Raj's car, so the speed of Ajay’s car is (x + 5) km/h. The time taken by Ajay to complete the journey is 400/(x + 5) hours. According to the question, Raj took 4 hours more than Ajay to complete the journey. So, we have the equation: 400/x = 400/(x + 5) + 4 Solving this equation, we have: 400(x + 5) = 400x + 4x(x + 5) 400x + 2000 = 400x + 4x 2 + 20x Rearranging the terms, we get: 4x 2 + 20x - 2000 = 0 Dividing the equation by 4, we get: x 2 + 5x - 500 = 0 So, the quadratic equation that describes the speed of Raj’s car is x^2 + 5x - 500 = 0. Hence, the correct answer is (c).   Q4: How much time took Ajay to travel 400 km? (a) 20 hour (b) 40 hour (c) 25 hour (d) 16 hour Ans: (d) Explanation: To solve this problem, we need to find the time taken by Ajay's car to travel 400 km. Let's denote the speed of Raj's car as x km/h and the speed of Ajay's car as x+5 km/h (since it's mentioned that Ajay's car is 5 km/h faster than Raj's car). The formula for time is distance divided by speed. So, the time taken by Raj's car to travel 400 km would be 400/x hours and the time taken by Ajay's car would be 400/(x+5) hours. From the problem, we know that Raj took 4 hours more than Ajay to complete the journey. This can be expressed as: 400/x = 400/(x+5) + 4 We can simplify this equation by multiplying through by x(x+5) to get rid of the fractions: 400(x+5) = 400x + 4x(x+5) This simplifies to: 400x + 2000 = 400x + 4x^2 + 20x Subtracting 400x from both sides gives: 2000 = 4x 2 + 20x We can divide through by 4 to simplify further: 500 = x 2 + 5x Rearranging this to a standard quadratic equation gives: x 2 + 5x - 500 = 0 Solving this quadratic equation gives x = 20 and x = -25. Since a speed can't be negative, we discard the -25 solution. So, the speed of Raj's car is 20 km/h and the speed of Ajay's car is 25 km/h. Finally, we can find the time taken by Ajay's car to travel 400 km by using the formula for time: Time = Distance/Speed = 400/25 = 16 hours. Therefore, the answer is (d) 16 hours.  

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Case Study on Quadratic Equations Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Quadratic Equations Class 10 Maths can use this page to download the PDF file. 

The case study questions on Quadratic Equations are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Quadratic Equations case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Quadratic Equations Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Quadratic Equations, therefore, they prepared a set of solutions along with the case study questions on Quadratic Equations.

The case study on Quadratic Equations Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Quadratic Equations case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Quadratic Equations Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Quadratic Equations case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Quadratic Equations Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Quadratic Equations case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Quadratic Equations Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Quadratic Equations as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Quadratic Equations?

Students can choose their own way to answer Case Study on Quadratic Equations Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Quadratic Equations Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Quadratic Equations questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Quadratic Equations Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Quadratic Equations?

 A few essential things to know to solve Case Study Questions on Class 10 Quadratic Equations are -

• Basic Formulas of Quadratic Equations: One of the most important things to know to solve Case Study Questions on Class 10 Quadratic Equations is to learn about the basic formulas or revise them before solving the case-based questions on Quadratic Equations.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Quadratic Equations case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

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Quadratic Formula Exercises

Quadratic formula practice problems with answers.

Below are ten (10) practice problems regarding the quadratic formula. The more you use the formula to solve quadratic equations, the more you become expert at it!

Use the illustration below as a guide. Notice that in order to apply the quadratic formula, we must transform the quadratic equation into the standard form, that is, [latex]a{x^2} + bx + c = 0[/latex] where [latex]a \ne 0[/latex].

The problems below have varying levels of difficulty. I encourage you to try them all. Believe me, they are actually easy! Good luck.

Problem 1: Solve the quadratic equation using the quadratic formula.

[latex]{x^2}\, – \,8x + 12 = 0[/latex]

Therefore, the answers are [latex]{x_1} = 6[/latex] and [latex]{x_2} = 2[/latex].

Problem 2: Solve the quadratic equation using the quadratic formula.

[latex]2{x^2}\, -\, x = 1[/latex]

Rewrite the quadratic equation in the standard form.

[latex]2{x^2} – x – 1 = 0[/latex]

Therefore, the answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = \large{{ – 1} \over 2}[/latex].

Problem 3: Solve the quadratic equation using the quadratic formula.

[latex]4{x^2} + 9 = – 12x[/latex]

[latex]4{x^2} + 12x + 9 = 0[/latex]

Therefore, the solution is [latex]x = \large{{ – 3} \over 2}[/latex].

Problem 4: Solve the quadratic equation using the quadratic formula.

[latex]5{x^2} = 7x + 6[/latex]

Convert the quadratic equation into the standard form.

[latex]5{x^2} – 7x – 6 = 0[/latex]

Therefore, the answers are [latex]{x_1} = 2[/latex] and [latex]{x_2} = \large{{ – 3} \over 5}[/latex].

Problem 5: Solve the quadratic equation using the quadratic formula.

[latex]{x^2} -\,{ \large{1 \over 2}}x\, – \,{\large{3 \over {16}}} = 0[/latex]

Multiply the entire equation by the LCM of the denominators which is [latex]16[/latex]. This will get rid of the denominators thereby giving us integer values for [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex].

[latex]16{x^2} – 8x – 3 = 0[/latex]

Therefore, the answers are [latex]x_1=\large{3 \over 4}[/latex] and [latex]x_2=\large{{ – 1} \over 4}[/latex].

Problem 6: Solve the quadratic equation using the quadratic formula.

[latex]{x^2} + 3x + 9 = 5x – 8[/latex]

Convert into standard form as [latex]{x^2} – 2x + 17 = 0[/latex].

Therefore, the answers are [latex]x_1=1 + 4i[/latex] and [latex]x_2=1 – 4i[/latex].

Problem 7: Solve the quadratic equation using the quadratic formula.

[latex]{\left( {x – 2} \right)^2} = 4x[/latex]

Rewrite in standard form as [latex]{x^2} – 8x + 4 = 0[/latex].

Hence, the answers are [latex]{x_1} = 4 + 2\sqrt 3 [/latex] and [latex]{x_2} = 4 – 2\sqrt 3 [/latex].

Problem 8: Solve the quadratic equation using the quadratic formula.

[latex]{\Large{{{x^2}} \over 4} – {x \over 2} }= 1[/latex]

To convert the quadratic equation into the standard form, simply multiply the entire equation by [latex]4[/latex] then subtract both sides by [latex]4[/latex].

[latex]{x^2} – 2x – 4 = 0[/latex]

Thus, the answers are [latex]{x_1} = 1 + \sqrt 5 [/latex] and [latex]{x_2} = 1 – \sqrt 5 [/latex].

Problem 9: Solve the quadratic equation using the quadratic formula.

[latex]{\left( {2x – 1} \right)^2} = \Large{x \over 3}[/latex]

If we carefully transform the given quadratic equation into the standard form, we get [latex]12{x^2} – 13x + 3 = 0[/latex].

Therefore, the answers are [latex]x_1={\Large{3 \over 4}}[/latex] and [latex]x_2={\Large{1 \over 3}}[/latex].

Problem 10: Solve the quadratic equation using the quadratic formula.

[latex]\left( {2x – 1} \right)\left( {x + 4} \right) = – {x^2} + 3x[/latex]

If we simplify the quadratic equation to convert it to the standard form, we should arrive at [latex]3{x^2} + 4x – 4 = 0[/latex].

Hence, the answers are [latex]x_1={\Large{2 \over 3}}[/latex] and [latex]x_2=-2[/latex].

You may also be interested in these related math lessons or tutorials:

The Quadratic Formula

Solving Quadratic Equations using the Quadratic Formula

CBSE Case Study Questions for Class 10 Maths Quadratic Equation Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Quadratic Equation  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Quadratic Equation PDF

Checkout our case study questions for other chapters.

• Chapter 2: Polynomials Case Study Questions
• Chapter 3: Pair of Linear Equations in Two Variables Case Study Questions
• Chapter 5: Arithmetic Progressions Case Study Questions
• Chapter 6: Triangles Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Study Guides > College Algebra: Co-requisite Course

Quadratic equations, learning objectives.

• Use the zero product principle to solve quadratic equations that can be factored
• Identify solutions to quadratic equations on a graph
• Use the square root property to solve a quadratic equation
• Complete the square to solve a quadratic equation
• Write a quadratic equation in standard form and identify the values of a , b , and c in a standard form quadratic equation.
• Use the Quadratic Formula to find solutions of a quadratic equation, (rational, irrational and complex)

The Zero-Product Property and Quadratic Equations

Solving quadratics with a leading coefficient of 1, reminder: given a quadratic equation with the leading coefficient of 1, factor it..

• Find two numbers whose product equals c and whose sum equals b .
• Use those numbers to write two factors of the form [latex]\left(x+k\right)\text{ or }\left(x-k\right)[/latex], where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and [latex]-2[/latex], the factors are [latex]\left(x+1\right)\left(x - 2\right)[/latex].
• Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Answer: Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property. [latex]\begin{array}{l}{x}^{2}-9=0\hfill \\ \left(x - 3\right)\left(x+3\right)=0\hfill \\ \hfill \\ \left(x - 3\right)=0\hfill \\ x=3\hfill \\ \hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}[/latex] The solutions are [latex]x=3[/latex] and [latex]x=-3[/latex].

 Solving Quadratics with a Leading Coefficient of [latex]\ne1[/latex]

• With the quadratic in standard form, [latex]a{x}^{2}+bx+c=0[/latex], multiply [latex]a\cdot c[/latex].
• Find two numbers whose product equals [latex]ac[/latex] and whose sum equals [latex]b[/latex].
• Rewrite the equation replacing the [latex]bx[/latex] term with two terms using the numbers found in step 1 as coefficients of x.
• Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
• Factor out the expression in parentheses.
• Set the expressions equal to zero and solve for the variable.

Answer: This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out [latex]-x[/latex] from all of the terms and then proceed with grouping. [latex]\begin{array}{l}-3{x}^{3}-5{x}^{2}-2x=0\hfill \\ -x\left(3{x}^{2}+5x+2\right)=0\hfill \end{array}[/latex] Use grouping on the expression in parentheses. [latex]\begin{array}{l}-x\left(3{x}^{2}+3x+2x+2\right)=\hfill&0\hfill \\ -x\left[3x\left(x+1\right)+2\left(x+1\right)\right]=\hfill&0\hfill \\ -x\left(3x+2\right)\left(x+1\right)=\hfill&0\hfill \end{array}[/latex] Now, we use the zero-product property. Notice that we have three factors. [latex]\begin{array}{l}-x\hfill&=0\hfill \\ x\hfill&=0\hfill \\ 3x+2\hfill&=0\hfill \\ x\hfill&=-\frac{2}{3}\hfill \\ x+1\hfill&=0\hfill \\ x\hfill&=-1\hfill \end{array}[/latex] The solutions are [latex]x=0[/latex], [latex]x=-\frac{2}{3}[/latex], and [latex]x=-1[/latex].

Solve a Quadratic Equation by the Square Root Property

The square root property.

[latex]\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}[/latex]

[latex]10x^{2}+5=85[/latex]

[latex]10x^{2}=80[/latex]

[latex] \begin{array}{l}{{x}^{2}}=8\\\,\,\,x=\pm \sqrt{8}\\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]

[latex]\left(x-2\right)^{2}-50=0[/latex]

[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}[/latex]

[latex] \begin{array}{l}x=2\pm \sqrt{50}\\\,\,\,\,=2\pm \sqrt{(25)(2)}\\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]

Solve a Quadratic Equation by Completing the Square

Answer: First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares. [latex-display]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array}[/latex-display] Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms. [latex-display]24x=2\left(3x\right)\left(4\right)[/latex-display] A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as [latex](r–s)^{2}[/latex]. In this case, the middle term is subtracted, so subtract r and s and square it to get [latex](r–s)^{2}[/latex]. [latex-display]\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex-display]

Steps for Completing The Square

• Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right side of the equal sign. [latex]{x}^{2}+4x=-1[/latex]
• Multiply the b term by [latex]\frac{1}{2}[/latex] and square it. [latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
• Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have [latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
• The left side of the equation can now be factored as a perfect square. [latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
• Use the square root property and solve. [latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
• The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].

[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]

[latex]\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}[/latex]

[latex]\left(x-6\right)^{2}=40[/latex]

[latex] x-6=\pm\sqrt{40}[/latex]

[latex] \begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]

Answer: First, move the constant term to the right side of the equal sign. [latex]{x}^{2}-3x=5[/latex] Identify b .[latex]b=-3[/latex] Then, take [latex]\frac{1}{2}[/latex] of the b term and square it. [latex]\begin{array}{l}\frac{1}{2}\left(-3\right)=-\frac{3}{2}\hfill \\ {\left(-\frac{3}{2}\right)}^{2}=\frac{9}{4}\hfill \end{array}[/latex] Add the result to both sides of the equal sign. [latex]\begin{array}{l}\text{ }{x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}=5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ {x}^{2}-3x+\frac{9}{4}=5+\frac{9}{4}\hfill \end{array}[/latex] Factor the left side as a perfect square and simplify the right side. [latex]{\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}[/latex] Use the square root property and solve. [latex]\begin{array}{l}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}\hfill&=\pm \sqrt{\frac{29}{4}}\hfill \\ \left(x-\frac{3}{2}\right)\hfill&=\pm \frac{\sqrt{29}}{2}\hfill \\ x\hfill&=\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}[/latex] The solutions are [latex]x=\frac{3}{2}+\frac{\sqrt{29}}{2}[/latex], [latex]x=\frac{3}{2}-\frac{\sqrt{29}}{2}[/latex].

[latex]\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}[/latex]

[latex]\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}[/latex]

[latex]\left(x+8\right)^{2}=0[/latex]

[latex]x+8=0[/latex]

[latex]x=-8[/latex]

The Quadratic Formula

• First, move the constant term to the right side of the equal sign: [latex]a{x}^{2}+bx=-c[/latex]
• As we want the leading coefficient to equal 1, divide through by a : [latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
• Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign: [latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
• Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction: [latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
• Now, use the square root property, which gives [latex]\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
• Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus, [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]

Solving a Quadratic Equation using the Quadratic Formula

• Put the equation in standard form first.
• Identify the coefficients, a , b, and c. Be careful to include negative signs if the bx or c terms are subtracted.
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Simplify as much as possible.
• Use the [latex]\pm[/latex] in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted .
• Simplify both values to get the possible solutions.

[latex]\begin{array}{r}x^{2}+4x=5\,\,\,\\x^{2}+4x-5=0\,\,\,\\\\a=1,b=4,c=-5\end{array}[/latex]

[latex] \begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,4x\,\,\,-\,\,\,5\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex]

[latex] \begin{array}{l}\\x=\frac{-4\pm \sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\end{array}[/latex]

[latex]x=\frac{-4\pm\sqrt{16+20}}{2}[/latex]

[latex] x=\frac{-4\pm \sqrt{36}}{2}[/latex]

[latex] x=\frac{-4\pm 6}{2}[/latex]

[latex]\begin{array}{c}x=\frac{-4+6}{2}=\frac{2}{2}=1\\\\\text{or}\\\\x=\frac{-4-6}{2}=\frac{-10}{2}=-5\end{array}[/latex]

[latex]\begin{array}{l}x^{2}-2x=6x-16\\x^{2}-2x-6x+16=0\\x^{2}-8x+16=0\end{array}[/latex]

[latex] \begin{array}{r}{{x}^{2}}\,\,\,-\,\,\,8x\,\,\,+\,\,\,16\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,\,c\,\,\,\,=\,\,\,0\end{array}[/latex]

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\end{array}[/latex]

[latex] x=\frac{8\pm \sqrt{64-64}}{2}[/latex]

[latex] x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]

[latex]\begin{array}{r}x^{2}-2x=6x-16\,\,\,\,\,\\\left(4\right)^{2}-2\left(4\right)=6\left(4\right)-16\\16-8=24-16\,\,\,\,\,\,\\8=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Quadratic equations with complex solutions

[latex]2x^2+3x+6=0[/latex]

Using the quadratic formula to solve this equation, we first identify a, b, and c.

[latex]a = 2,b = 3,c = 6[/latex]

We can place a, b and c into the quadratic formula and simplify to get the following result:

[latex]x=-\frac{3}{4}+\frac{\sqrt{-39}}{4}, x=-\frac{3}{4}-\frac{\sqrt{-39}}{4}[/latex]

Up to this point, we would have said that [latex]\sqrt{-39}[/latex] is not defined for real numbers and determine that this equation has no solutions.  But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows.

[latex]x=-\frac{3}{4}+i\frac{\sqrt{39}}{4}, x=-\frac{3}{4}-i\frac{\sqrt{39}}{4}[/latex]

In the following example we will work through the process of solving a quadratic equation with complex solutions. Take note that we be simplifying complex numbers - so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time.

Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[/latex].

Answer: First, we identify the coefficients: [latex]a=1,b=1[/latex], and [latex]c=2[/latex]. Substitute these values into the quadratic formula. [latex]\begin{array}{l}x\hfill&=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \\\hfill&=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^{2}-\left(4\right)\cdot \left(1\right)\cdot \left(2\right)}}{2\cdot 1}\hfill \\\hfill&=\frac{-1\pm \sqrt{1 - 8}}{2}\hfill \\ \hfill&=\frac{-1\pm \sqrt{-7}}{2}\hfill \\\hfill&=\frac{-1\pm i\sqrt{7}}{2}\hfill \end{array}[/latex] Now we can separate the expression [latex]\frac{-1\pm i\sqrt{7}}{2}[/latex] into two solutions: [latex-display]-\frac{1}{2}+\frac{ i\sqrt{7}}{2}[/latex-display] [latex-display]-\frac{1}{2}-\frac{ i\sqrt{7}}{2}[/latex-display]   The solutions to the equation are [latex]x=\frac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}[/latex].

[latex]\begin{array}{l}x^{2}+x=-x-3\\x^{2}+2x+3=0\end{array}[/latex]

[latex]a=1, b=2, c=3[/latex]

Substitute values for a, b, c into the quadratic formula.

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-2\pm \sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\end{array}[/latex]

[latex]\displaystyle x=\frac{-2\pm \sqrt{-8}}{2}[/latex]

Rewrite the radical of a negative number in terms of the imaginary unit [latex]i[/latex]

[latex]\displaystyle x=\frac{-2\pm i\sqrt{8}}{2}[/latex]

[latex]\displaystyle x=\frac{-2\pm 2i\sqrt{2}}{2}[/latex]

[latex]x=-1 \pm i\sqrt{2}[/latex]

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• Quadratic Equation Questions

Basic Quadratic Equation Questions

Quadratic equations are an important part of algebra, and as students, we must all be familiar with their definition and the ways of solving quadratic equation problems. In this article, we are going to familiarize the students with all the concepts surrounding quadratic equations and the methods of solving problems related to this topic. A short definition of a quadratic equation would be: a quadratic equation is a second-degree polynomial, which we represent as ‘ax 2 + bx + c’ in general.

In this representation, a cannot be equal to 0 and b,c are known as coefficients and are constant by nature. With this basic introduction, let's move forward with a formal definition, formulae and detailed solutions to quadratic equation questions to enable better understanding.

Why are Quadratic Equations Important?

Quadratic equations have numerous applications in various fields, including physics, engineering, economics, and even computer science. They help us model real-world scenarios like projectile motion, population growth, and electrical circuit analysis. Understanding quadratics is crucial for success in higher-level math and science courses.

Definition of Quadratic Equations

A quadratic equation is a polynomial where the highest power of the variable is 2. We generally represent it as ax 2 + bx + c. Here a, b and c are real numbers or constants, and x is the variable. In this case, the value of a cannot be 0 as that would remove the x 2 term, and the equation won't be quadratic after that.

A quadratic equation is an equation of second degree with more than two terms. It means that at least one of the terms of the equation is squared. In the above-given equation. The answer to the equation also known as the roots of the equation is the value of the “x”. The value of the “x” has to satisfy the equation. 

Some examples of quadratic equations can be as follows:

56x 2 + ⅔ x + 1, where a = 56, b = ⅔ and c = 1.

-4/3 x 2 + 64x - 30, where a = -4/3, b = 64 and c = -30.

Roots of a Quadratic Equation

To solve basic quadratic equation questions or any quadratic equation problems, we need to solve the equation. Solving quadratic equations gives us the roots of the polynomial. The roots of the equation are the values of x at which ax 2 + bx + c = 0. Since a quadratic equation is a polynomial of degree 2, we obtain two roots in this case.

There are several methods for solving quadratic equation problems, as we can see below:

Factorization Method.

Completing The Square Method.

Quadratic Equation Formula.

Quadratic Equation Formula

So what is the quadratic equation formula? The quadratic equation formula or the Sridharacharya Formula is a method for finding out the roots of two-degree polynomials. This formula helps solve quadratic equation problems. The formula is as given below:

\[ x = \frac{-b \pm\sqrt{b^{2}-4ac}}{2a} \]

Where x represents the roots of the equation and (b 2 −4ac) is the discriminant.

By finding out the value of the discriminant, we can predict the nature of the roots. There are three possibilities with three different implications:

Two distinct roots which are real, if b 2 - 4ac > 0.

Two real roots equal in magnitude, if b 2 - 4ac = 0.

Imaginary roots or absence of real roots if b 2 - 4ac < 0.

Now that the basic principles of quadratic equations are clear, we will move on to some solved examples. But before that, let us list some quadratic equation questions for the students to solve.

Quadratic Equation Practice Questions

The following are a list of questions for you to solve once you have gone through the quadratic equation questions and answers in the solved examples section:

Find the determinant of the following quadratic equations: 2x 2 + 3x + 6, 70x 2 + 49 + 14, ⅔ y 2 + 63y + 42.

Find the roots of the following quadratic equations: x 2 - 45x + 324, 2x 2 - 22x + 42, ½ x 2 + 2x + 4.

The product of two consecutive numbers is 420, and their sum is 41. Find the numbers.

Before solving these, let's check out the solved examples with questions and answers on the quadratic equation.

Solved Examples

1. Solve x 2 + 5x + 6 = 20 by factorization method.

Solution: The given polynomial or quadratic equation is

x 2 + 5x + 6 = 20

Solving by factorization method, 

x 2 + 5x + 6 - 20 = 0.

or, x 2 + 5x - 14 = 0

or, x 2 - 2x + 7x - 14 = 0

or, x(x - 2) +7(x - 2) = 0

or, (x - 2)(x + 7) = 0

or, (x - 2) = 0, (x + 7) = 0 

or, x = +2, -7.

2. Solve 2x 2 - 5x + 3 using the quadratic equation formula.

Solution: The quadratic equation formula is:

\[ x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \]

The determinant or b 2 -4ac = (-5) 2 - 4 × 3 × 2 = 25 - 24 = 1

\[ \sqrt{b^{2} - 4ac} = 1 \]

Therefore, \[ x= \frac{-(-5)\pm1}{2\times2} \]

\[ x = \frac{5+1}{4} = \frac{6}{4} = \frac{3}{2} \]

\[ x = \frac{5-1}{4} = \frac{4}{4} = 1 \]

Thus the roots of the equation are 3/2 and 1.

3. Solve the quadratic equation: \[x^2 - 4x + 4 = 0\]

Solution: Given quadratic equation is a perfect square trinomial, so we can factor it directly.

\[(x - 2)^2 = 0\]

Taking the square root of both sides:

\[x - 2 = 0\]

Solving for \(x\):

4. Solve the quadratic equation: \[2x^2 - 5x + 2 = 0\]

Solution: Using the quadratic formula:

\[x = \dfrac{5 \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}\]

\[x = \dfrac{5 \pm \sqrt{25 - 16}}{4}\]

\[x = \dfrac{5 \pm \sqrt{9}}{4}\]

\[x = \dfrac{5 \pm 3}{4}\]

Two solutions:

\[x = \dfrac{8}{4} = 2\]

\[x = \dfrac{2}{4} = \dfrac{1}{2}\]

5. Solve the quadratic equation: \[3x^2 + 7x + 2 = 0\]

\[x = \dfrac{-7 \pm \sqrt{7^2 - 4(3)(2)}}{2(3)}\]

\[x = \dfrac{-7 \pm \sqrt{49 - 24}}{6}\]

\[x = \dfrac{-7 \pm \sqrt{25}}{6}\]

\[x = \dfrac{-7 \pm 5}{6}\]

\[x = \dfrac{-2}{6} = -\dfrac{1}{3}\]

\[x = \dfrac{-12}{6} = -2\]

6. Solve the quadratic equation: \[4x^2 + 8x + 2 = 0\]

Solution: Divide the equation by the common factor (in this case, 2):

\[2x^2 + 4x + 1 = 0\]

Using the quadratic formula:

\[x = \dfrac{-4 \pm \sqrt{4^2 - 4(2)(1)}}{2(2)}\]

\[x = \dfrac{-4 \pm \sqrt{16 - 8}}{4}\]

\[x = \dfrac{-4 \pm \sqrt{8}}{4}\]

\[x = \dfrac{-4 \pm 2\sqrt{2}}{4}\]

\[x = \dfrac{-2 \pm \sqrt{2}}{2}\]

\[x = \dfrac{-2 + \sqrt{2}}{2}\]

\[x = \dfrac{-2 - \sqrt{2}}{2}\]

7. Solve the quadratic equation by factorization method: \[x^2 - 5x + 6 = 0\]

Solution: Factor the quadratic equation:

\[(x - 2)(x - 3) = 0\]

Setting each factor to zero:

\[x - 2 = 0 \quad \text{or} \quad x - 3 = 0\]

\[x = 2 \quad \text{or} \quad x = 3\]

8. Solve the quadratic equation: \[6x^2 - 11x + 4 = 0\]

\[x = \dfrac{11 \pm \sqrt{(-11)^2 - 4(6)(4)}}{2(6)}\]

\[x = \dfrac{11 \pm \sqrt{121 - 96}}{12}\]

\[x = \dfrac{11 \pm \sqrt{25}}{12}\]

\[x = \dfrac{11 \pm 5}{12}\]

\[x = \dfrac{16}{12} = \dfrac{4}{3}\]

\[x = \dfrac{6}{12} = \dfrac{1}{2}\]

9. Solve the quadratic equation: \[2x^2 + 3x - 5 = 0\]

\[x = \dfrac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}\]

\[x = \dfrac{-3 \pm \sqrt{9 + 40}}{4}\]

\[x = \dfrac{-3 \pm \sqrt{49}}{4}\]

\[x = \dfrac{-3 \pm 7}{4}\]

\[x = \dfrac{4}{4} = 1\]

\[x = \dfrac{-10}{4} = -\dfrac{5}{2}\]

10. Solve the quadratic equation: \[5x^2 - 4x - 3 = 0\]

\[x = \dfrac{4 \pm \sqrt{(-4)^2 - 4(5)(-3)}}{2(5)}\]

\[x = \dfrac{4 \pm \sqrt{16 + 60}}{10}\]

\[x = \dfrac{4 \pm \sqrt{76}}{10}\]

\[x = \dfrac{4 \pm 2\sqrt{19}}{10}\]

\[x = \dfrac{2 \pm \sqrt{19}}{5}\]

\[x = \dfrac{2 + \sqrt{19}}{5}\]

\[x = \dfrac{2 - \sqrt{19}}{5}\]

This is all about the roots of quadratic equations and their formulas. Learn the formulas and find out how they are used to derive the roots of an equation easily. 

By diving into this chapter, you'll build a solid understanding of quadratic equations. Mastering these equations will empower you to effortlessly solve any quadratic problem, use your skills in real-life situations, and amaze your teachers with your math abilities. Keep in mind that practice is crucial! Take on the practice questions and reach out for assistance if necessary. The realm of math is ready for you, and quadratic equations are your ticket to exploring its intriguing complexities!

FAQs on Quadratic Equation Questions

1. Define a quadratic equation along with suitable examples. also, state the quadratic equation formula.

A quadratic equation is a polynomial where the highest power of the variable is neither more nor less than 2. So essentially, a quadratic equation is a polynomial of degree 2. We represent such an equation in a general format as ax 2 + bx + c, where a, b and c are known as the coefficients or the constants of the equation. The thumb rule for quadratic equations is that the value of a cannot be 0. The x in the expression is the variable. This algebraic expression, when solved, will yield two roots.

Some examples of quadratic equations are:

3x 2 + 4x + 7 = 34

x 2 + 8x + 12 = 40

The quadratic equation formula is a method for solving quadratic equation questions. The formula is as follows:

where x represents the roots of the equation.

2. What are the roots of a quadratic equation? What are the zeroes of a polynomial?  

The roots of a quadratic equation are the values obtained when we solve the equation. They are those values of x for which the expression ax 2 +bx+c becomes equal to 0. These values are also known as the zeroes of the polynomial. Since a quadratic equation is essentially a polynomial of degree 2, we get two roots after solving the given polynomial.

As we practice more and more quadratic equation sums, our ideas regarding which method to use while solving a given question will get clearer.

3. How can I solve a quadratic equation?

Quadratic equation is an equation with more than one term in it and at least one of the terms having degree 2. Its general form is ax 2  + bx + c, whereas a,b,c are real numbers and a is not equal to zero. The values which satisfy the “x” in the equation are the solution for the quadratic equation. They are also known as the roots of the equation. The quadratic equation can be solved in the algebraic method and graphical method.

In the algebraic method, the equation is reduced to the roots by shifting terms from L.H.S to the R.H.S and using different mathematical operations. In the graphic method, the equation is solved by drawing it on the map and solving it using the parabola the equation makes on the graph. The value of “a” determines whether the graph of the equation is concave parabola or convex parabola. The value of the discriminant decides whether the curve will intersect the x-axis or not.

4.  What are the applications of quadratic equations?

A quadratic equation is a polynomial equation with more than one term and at least one of the terms having 2 as square. It is generally used in different situations in day-to-day life. In constructing rooms and boxes of different geometric shapes. If you want to construct a box made of wood with 5 square feet dimensions, you can write a quadratic equation to measure its area and calculate the material required. 

It can also be used in selling something and calculating the profit and loss you may incur after selling the good. To know it, you can simply form a quadratic equation. It can also be used in athletics while throwing objects like a javelin, shot put ball, etc. It can also be used to calculate the distance. Generally, when someone travels up and down the river uses this equation to measure the distance to be travelled. 

5. How are quadratic equations used in athletics and construction?

Quadratic equations are equations with at least one term having 2 as a square and it has more than one term in the equation preferably, four terms. The general form of the equation is ax 2  + bx+c whereas a,b,c are real numbers and a is not equal to zero. The values which satisfy the “x” in the equation are the solution for the quadratic equation. They are also known as the roots of the equation. In exams, they have preferable weightage and practising them correctly will improve the marks. On the other hand, they are also used in real-life situations. In athletics, it is used to measure the speed and force to be applied to throw an object like an arrow, shot put the ball, discus, etc. They use the velocity equation to measure the height of the ball from which it should be thrown. In the field of construction, a quadratic equation is framed with the known dimensions of the building or a room to figure out unknown values like the area, perimeter, etc.

• Math Article

Quadratic Equation Questions

Quadratic equation questions are provided here for Class 10 students. A quadratic equation is a second-degree polynomial which is represented as ax 2 + bx + c = 0, where a is not equal to 0. Here, a, b and c are constants, also called coefficients and x is an unknown variable. Also, learn Quadratic Formula here.

Solving the problems based on quadratics will help students to understand the concept very well and also to score good marks in this section. All the questions are solved here step by step with a detailed explanation. In this article, we will give the definition and important formula for solving problems based on quadratic equations. The questions given here is in reference to the CBSE syllabus and NCERT curriculum. 

Definition of Quadratic Equation

Usually, the quadratic equation is represented in the form of ax 2 +bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0. Here, a and b are the coefficients of x 2 and x, respectively. So, basically, a quadratic equation is a polynomial whose highest degree is 2. Let us see some examples:

• 3x 2 +x+1, where a=3, b=1, c=1
• 9x 2 -11x+5, where a=9, b=-11, c=5

Roots of Quadratic Equations:

If we solve any quadratic equation, then the value we obtain are called the roots of the equation. Since the degree of the quadratic equation is two, therefore we get here two solutions and hence two roots.

There are different methods to find the roots of quadratic equation, such as:

• Factorisation
• Completing the square
• Using quadratic formula

Learn: Factorization of Quadratic equations

Quadratic Equation Formula:

The quadratic formula to find the roots of the quadratic equation is given by:

\(\begin{array}{l}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\end{array} \)

Where b 2 -4ac is called the discriminant of the equation.

Based on the discriminant value, there are three possible conditions, which defines the nature of roots as follows:

• two distinct real roots, if b 2 – 4ac > 0
• two equal real roots, if b 2 – 4ac = 0
• no real roots, if b 2 – 4ac < 0

Also, learn quadratic equations for class 10 here.

Quadratic Equations Problems and Solutions

1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with.

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x 2 – 200 + 5x = 124

– x 2 + 45x – 200 = 124

x 2 – 45x + 324 = 0

This represents the quadratic equation. Hence by solving the given equation for x, we get;

x = 36 and x = 9

So, the number of marbles Rahul had is 36 and Rohan had is 9 or vice versa.

2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

Solution: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

Cancel x 2 both the sides.

Since, this expression is not in the form of ax 2 +bx+c, hence it is not a quadratic equation.

3. Find the roots of the equation 2x 2 – 5x + 3 = 0 using factorisation.

2x 2 – 5x + 3 = 0

2x 2 – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

4. Solve the quadratic equation 2x 2 + x – 300 = 0 using factorisation.

Solution: 2x 2 + x – 300 = 0

2x 2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of the given equation.

Also, read   Factorisation .

5. Solve the equation x 2 +4x-5=0.

x 2 + 4x – 5 = 0

x 2 -1x+5x-5 = 0

x(x-1)+5(x-1) =0

(x-1)(x+5) =0

Hence, (x-1) =0, and (x+5) =0

similarly, x+5 = 0

x=-5 & x=1

6. Solve the quadratic equation 2x 2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax 2 +bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

Now putting the values of a,b and c.

x=64/4 or x=-66/4

x=16 or x=-33/2

7. Find the roots of x 2 + 4x + 5 = 0, if any exist, using quadratic formula.

Solution: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b 2 -4ac = 4 2 – 4(1)(5) = 16-20 = -4

Since the square root of -4 will not give a real number. Hence there is no real roots for the given equation.

8. Find the discriminant of the equation: 3x 2 -2x+⅓ = 0.

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b 2 – 4ac

D = (-2) 2 -4(3)(⅓)

Video Lesson

Quadratic equation worksheet.

Practice Questions

Solve these quadratic equations and find the roots. 

• x 2 -5x-14=0         [Answer: x=-2 & x=7]
• X 2  = 11x -28       [Answer: x=4 & x = 7]
• 6x 2 – x = 5            [Answer: x=-⅚ & x = 1]
• 12x 2 = 25x          [Answer: x=0 & x=25/12]

Frequently Asked Questions on Quadratic Equations

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Class 10th Maths - Quadratic Equations Case Study Questions and Answers 2022 - 2023

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Quadratic Equations, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Quadratic equations case study questions with answer key.

10th Standard CBSE

Final Semester - June 2015

A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and  \(a \neq 0\)   Every quadratic equation has two roots depending on the nature of its discriminant, D = b2 - 4ac.Based on the above information, answer the following questions. (i) Which of the following quadratic equation have no real roots?

(ii) Which of the following quadratic equation have rational roots?

(iii) Which of the following quadratic equation have irrational roots?

(iv) Which of the following quadratic equations have equal roots?

(v) Which of the following quadratic equations has two distinct real roots?

In our daily life we use quadratic formula as for calculating areas, determining a product's profit or formulating the speed of an object and many more. Based on the above information, answer the following questions. (i) If the roots of the quadratic equation are 2, -3, then its equation is

(ii) If one root of the quadratic equation 2x 2 + kx + 1 = 0 is -1/2, then k =

(iii) Which of the following quadratic equations, has equal and opposite roots?

(iv) Which of the following quadratic equations can be represented as (x - 2) 2 + 19 = 0?

(v) If one root of a qua drraattiic equation is  \(\frac{1+\sqrt{5}}{7}\) , then I.ts other root is

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world's first civilisation, and came up with some great ideas like agriculture, irrigation and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field; Based on the above information, represent the following questions in the form of quadratic equation. (i) The sum of squares of two consecutive integers is 650.

(ii) The sum of two numbers is 15 and the sum of their reciprocals is 3/10.

(iii) Two numbers differ by 3 and their product is 504.

(iv) A natural number whose square diminished by 84 is thrice of 8 more of given number.

(v) A natural number when increased by 12, equals 160 times its reciprocal.

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

If p(x) is a quadratic polynomial i.e., p(x) = ax 2 - + bx + c, \(a \neq 0\) , then p(x) = 0 is called a quadratic equation. Now, answer the following questions. (i) Which of the following is correct about the quadratic equation ax 2 - + bx + c = 0 ?

(ii) The degree of a quadratic equation is

(iii) Which of the following is a quadratic equation?

(iv) Which of the following is incorrect about the quadratic equation ax 2 - + bx + c = 0 ?

(v) Which of the following is not a method of finding solutions of the given quadratic equation?

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Most Important Questions CBSE Class 10 Maths Ch 4 Quadratic Equation with Solutions

Class 10 mathematics contains various questions like Quadratic Equation Most Important Questions that are useful for the students even in their competitive examinations like JEE & NEET. Quadratic equations are the equations in which at least one of the variables is squared. We use quadratic equations to determine the profit of the product and formulate the object’s speed. 

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