Math Case Study Class 10 Quadratic Equation

Math Case Study Class 10 Quadratic Equation

Understanding Math Case Study Class 10 Quadratic Equation

Chapter: Math Case Study Class 10 Quadratic Equation helps students apply concepts through real-world scenarios. These problems support deeper reasoning and structured analysis. Moreover, they also reinforce skills needed to answer Case Study math questions for class 9, math case study questions class 9, and other math case study questions effectively.

Applying Quadratic Equations in Case-Based Questions

Quadratic equation case studies require identifying variables and forming equations from practical situations. Students learn to evaluate roots and interpret results. Additionally, this approach improves accuracy and exam readiness. Many concepts also connect to Case Study math questions for class 9, helping younger learners build strong fundamentals.

Benefits of Practicing Quadratic Case Studies

Practice builds confidence. These case studies improve logical thinking and problem-solving. They also help students understand real applications. Furthermore, they support long-term conceptual growth.

Case Studies: Quadratic Equations for Grade 10

Mathematics Assessment | Focus: Application of Quadratic Equations (Forming, Solving, Properties)

Case Study 1: The Garden Design Problem 🪴

A landscape architect is designing a rectangular flower garden for a park. The garden’s **length** is **5 meters more than its width**. The **area** of the garden is **84 square meters**. The architect needs to determine the dimensions of the garden and analyze some mathematical properties related to its design. The situation can be modeled using quadratic equations.
Q1. What quadratic equation represents this situation if the width is represented by $x$ meters?
Answer: (A) $x^2 + 5x – 84 = 0$
Solution: Width = $x$, Length = $x + 5$. Area = Length $\times$ Width. $x(x + 5) = 84$. Expanding and setting to zero: $x^2 + 5x – 84 = 0$.
Q2. What are the actual dimensions of the garden?
Answer: (B) Width = 7 m, Length = 12 m
Solution: Solving $x^2 + 5x – 84 = 0$ by factoring: $(x + 12)(x – 7) = 0$. The roots are $x = -12$ or $x = 7$. Since width must be positive, $x = 7$ m (Width). Length $= x + 5 = 7 + 5 = 12$ m.
Q3. What is the discriminant ($\Delta$) of the quadratic equation $x^2 + 5x – 84 = 0$?
Answer: (A) 361
Solution: The discriminant is $\Delta = b^2 – 4ac$. For $x^2 + 5x – 84 = 0$, $a=1, b=5, c=-84$. $\Delta = 5^2 – 4(1)(-84) = 25 – (-336) = 25 + 336 = 361$.
Q4. If the area were increased to 96 square meters with the same length-width relationship ($L=x+5$), what would be the new quadratic equation?
Answer: (A) $x^2 + 5x – 96 = 0$
Solution: New Area = $x(x + 5) = 96$. Expanding and setting to zero: $x^2 + 5x – 96 = 0$.
Q5. What is the sum of the roots of the original equation ($x^2 + 5x – 84 = 0$)?
Answer: (B) -5
Solution: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is given by the formula $-\frac{b}{a}$. For $x^2 + 5x – 84 = 0$, $a=1$ and $b=5$. Sum $= -\frac{5}{1} = -5$. (Product of roots is $c/a = -84/1 = -84$).

Results Summary 💡

Total Questions: 5

Correct Answers: 0

Percentage Score: 0%

Review the case study questions above to see the correct solutions and detailed mathematical explanations.

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