Case Study Class 10 Polynomials

Case Study Class 10 Polynomials | Free Online Test

Case Study Class 10 Polynomials

Students preparing for exams often search for case study math questions for class 10. These exercises help strengthen concepts in polynomials and their applications. Our online practice sets include interactive polynomial case study questions class 10 that focus on real-life problem solving. Practicing these questions regularly improves accuracy and speed. Additionally, students develop problem-solving skills while applying formulas like factorization and the remainder theorem in practical situations.

Importance of Case Study Class 10 Polynomials

Math case study questions class 10 on polynomials encourage analytical thinking and logical reasoning. For instance, questions based on finding zeros of polynomials or applying algebraic identities provide deeper understanding. Furthermore, solving these problems enhances critical thinking. Short exercises help reinforce key formulas and theorems. Therefore, students gain confidence and clarity in class 10 polynomials case study questions through consistent practice.

Benefits of Online Test Practice

Our polynomials case study class 10 online tests provide instant feedback and performance tracking. Students can quickly identify mistakes and improve their approach. The class 10 maths case study questions on polynomials cover multiple difficulty levels, from basic factorization to application-based sums. Consequently, learners strengthen concepts efficiently and are better prepared for board exams. Regular practice ensures mastery of fundamental polynomial concepts and boosts exam performance.

Case Study 3

A group of engineers is working on the design of a water fountain in a newly constructed park. The stream of water from the fountain follows a parabolic path, and the engineers model this using quadratic polynomials. Suppose the water jet touches the ground at two different points, creating a curve in the form of a parabola. The equation of this parabola can be represented by \( p(x) = ax^2 + bx + c \), where the zeros of the polynomial represent the points at which the stream touches the ground. The maximum height of the fountain is determined by analyzing the vertex of the parabola, and the symmetry of the stream depends on the relationship between the zeros of the polynomial. By studying these mathematical models, the engineers can adjust the design to ensure the water stream looks aesthetic and lands precisely within the fountain boundary.

Important relations for quadratic polynomials:

If \( \alpha \) and \( \beta \) are the zeros of the polynomial \( p(x) = ax^2 + bx + c \), then: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a} \]

1. The polynomial representing the fountain’s water stream is \( p(x) = x^2 – 6x + 8 \). The sum of the zeros is:

  • A) 2
  • B) 4
  • C) 6
  • D) 8
Answer: C) 6
Solution: For \( p(x) = x^2 – 6x + 8 \), \( \alpha+\beta = -\frac{-6}{1} = 6 \).

2. If the zeros of \( p(x) = 2x^2 – 5x + 2 \) are \( \alpha \) and \( \beta \), then the product \( \alpha \beta \) is:

  • A) \( \tfrac{5}{2} \)
  • B) 2
  • C) \( \tfrac{2}{2} \)
  • D) 1
Answer: D) 1
Solution: \( \alpha \beta = \frac{c}{a} = \tfrac{2}{2} = 1 \).

3. The polynomial having zeros 2 and 5 is:

  • A) \( x^2 + 7x + 10 \)
  • B) \( x^2 – 7x + 10 \)
  • C) \( x^2 – 10x + 7 \)
  • D) \( x^2 + 10x + 7 \)
Answer: B) \( x^2 – 7x + 10 \)
Solution: \( (x-2)(x-5) = x^2 – 7x + 10 \).

4. For \( p(x) = x^2 – 3x – 10 \), if \( \alpha \) and \( \beta \) are the zeros, then \( \alpha^2 + \beta^2 = \):

  • A) 9
  • B) 19
  • C) 13
  • D) 29
Answer: D) 29
Solution: \( \alpha+\beta = 3 \), \( \alpha \beta = -10 \). Then \( \alpha^2 + \beta^2 = (\alpha+\beta)^2 – 2\alpha\beta = 9 – 2(-10) = 29 \).

5. The quadratic polynomial whose zeros are \( \tfrac{1}{4} \) and \( \tfrac{1}{2} \) is:

  • A) \( 4x^2 – 3x + \tfrac{1}{2} \)
  • B) \( 8x^2 – 6x + 1 \)
  • C) \( 4x^2 + 3x + 1 \)
  • D) \( 2x^2 – 3x + 1 \)
Answer: B) \( 8x^2 – 6x + 1 \)
Solution: Required polynomial \( = \left(x – \tfrac{1}{4}\right)\left(x – \tfrac{1}{2}\right) = x^2 – \tfrac{3}{4}x + \tfrac{1}{8} \). Multiplying through by 8 gives \( 8x^2 – 6x + 1 \).

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