Answer :
Let's analyze each of the given problems to determine whether they illustrate quadratic equations. We will also provide the mathematical sentences representing each situation.
### 1. The area of a triangle
Problem: The area of a triangle is [tex]\(50 \, \text{cm}^2\)[/tex] whose base is 2 cm longer than its height.
Mathematical Sentence: Let the height be [tex]\(h\)[/tex] cm. Then, the base is [tex]\((h + 2)\)[/tex] cm. The area of the triangle can be expressed as:
[tex]\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Given the area is [tex]\(50 \, \text{cm}^2\)[/tex]:
[tex]\[ 50 = \frac{1}{2} \times (h + 2) \times h \][/tex]
Simplifying, we get:
[tex]\[ 50 = \frac{1}{2}h(h + 2) \][/tex]
Multiplying both sides by 2 to clear the fraction, we have:
[tex]\[ 100 = h(h + 2) \][/tex]
Rewriting this equation, we get:
[tex]\[ h^2 + 2h - 100 = 0 \][/tex]
Conclusion: This is indeed a quadratic equation.
Mathematical Equation:
[tex]\[ h^2 + 2h - 100 = 0 \][/tex]
### 2. Cleaning a garden together
Problem: It takes 3 hours for Susie to clean a rectangular garden. Peter can clean the same garden in 1.5 hours. How long will it take Peter and Susie to clean the garden together?
Mathematical Sentence: Let [tex]\(t\)[/tex] be the time (in hours) it takes for them to clean the garden together. Susie's rate of cleaning is [tex]\(\frac{1}{3}\)[/tex] gardens per hour, and Peter's rate is [tex]\(\frac{1}{1.5}\)[/tex] or [tex]\( \frac{2}{3}\)[/tex] gardens per hour. Together, their combined rate is:
[tex]\[ \frac{1}{3} + \frac{2}{3} = \frac{1}{t} \][/tex]
Combining these rates:
[tex]\[ \frac{1}{3} + \frac{2}{3} = \frac{1}{t} \][/tex]
Simplifying, we find:
[tex]\[ 1 = \frac{1}{t} \][/tex]
Conclusion: This is not a quadratic equation; it is a linear equation.
Mathematical Equation:
[tex]\[ 1 = \frac{1}{t} \][/tex]
### 3. Rectangle with an increased side length
Problem: One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle will increase by [tex]\(18 \, \text{cm}^2\)[/tex].
Mathematical Sentence: Let one side be [tex]\(x\)[/tex] cm, and the other side be [tex]\((x - 3)\)[/tex] cm. The original area is:
[tex]\[ \text{Original Area} = x(x - 3) \][/tex]
The new dimensions after increasing each side by 1 cm are [tex]\(x + 1\)[/tex] and [tex]\(x - 2\)[/tex]. Thus, the new area is:
[tex]\[ \text{New Area} = (x + 1)(x - 2) \][/tex]
The increase in area is given by:
[tex]\[ (x + 1)(x - 2) - x(x - 3) = 18 \][/tex]
Expanding and simplifying this, we get:
[tex]\[ x^2 - x - 2 - (x^2 - 3x) = 18 \][/tex]
[tex]\[ x^2 - x - 2 - x^2 + 3x = 18 \][/tex]
[tex]\[ 2x - 20 = 0 \][/tex]
Conclusion: This simplifies to a linear equation in the context of this problem.
Mathematical Equation:
[tex]\[ 2x - 20 = 0 \][/tex]
### 4. Product equals sum
Problem: One number is 3 more than another number. Their product equals their sum.
Mathematical Sentence: Let one number be [tex]\(n\)[/tex]. The other number is [tex]\(n + 3\)[/tex]. Express the problem as:
[tex]\[ n(n + 3) = n + (n + 3) \][/tex]
Simplifying, we get:
[tex]\[ n^2 + 3n = 2n + 3 \][/tex]
Rearranging to form a standard quadratic equation:
[tex]\[ n^2 + 3n - 2n - 3 = 0 \][/tex]
[tex]\[ n^2 + n - 3 = 0 \][/tex]
Conclusion: This is a quadratic equation.
Mathematical Equation:
[tex]\[ n^2 + n - 3 = 0 \][/tex]
---
In summary:
1. Quadratic Equation: [tex]\(h^2 + 2h - 100 = 0\)[/tex]
2. Linear Equation: [tex]\(1 = \frac{1}{t}\)[/tex]
3. Linear Equation: [tex]\(2x - 20 = 0\)[/tex]
4. Quadratic Equation: [tex]\(n^2 + n - 3 = 0\)[/tex]
### 1. The area of a triangle
Problem: The area of a triangle is [tex]\(50 \, \text{cm}^2\)[/tex] whose base is 2 cm longer than its height.
Mathematical Sentence: Let the height be [tex]\(h\)[/tex] cm. Then, the base is [tex]\((h + 2)\)[/tex] cm. The area of the triangle can be expressed as:
[tex]\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Given the area is [tex]\(50 \, \text{cm}^2\)[/tex]:
[tex]\[ 50 = \frac{1}{2} \times (h + 2) \times h \][/tex]
Simplifying, we get:
[tex]\[ 50 = \frac{1}{2}h(h + 2) \][/tex]
Multiplying both sides by 2 to clear the fraction, we have:
[tex]\[ 100 = h(h + 2) \][/tex]
Rewriting this equation, we get:
[tex]\[ h^2 + 2h - 100 = 0 \][/tex]
Conclusion: This is indeed a quadratic equation.
Mathematical Equation:
[tex]\[ h^2 + 2h - 100 = 0 \][/tex]
### 2. Cleaning a garden together
Problem: It takes 3 hours for Susie to clean a rectangular garden. Peter can clean the same garden in 1.5 hours. How long will it take Peter and Susie to clean the garden together?
Mathematical Sentence: Let [tex]\(t\)[/tex] be the time (in hours) it takes for them to clean the garden together. Susie's rate of cleaning is [tex]\(\frac{1}{3}\)[/tex] gardens per hour, and Peter's rate is [tex]\(\frac{1}{1.5}\)[/tex] or [tex]\( \frac{2}{3}\)[/tex] gardens per hour. Together, their combined rate is:
[tex]\[ \frac{1}{3} + \frac{2}{3} = \frac{1}{t} \][/tex]
Combining these rates:
[tex]\[ \frac{1}{3} + \frac{2}{3} = \frac{1}{t} \][/tex]
Simplifying, we find:
[tex]\[ 1 = \frac{1}{t} \][/tex]
Conclusion: This is not a quadratic equation; it is a linear equation.
Mathematical Equation:
[tex]\[ 1 = \frac{1}{t} \][/tex]
### 3. Rectangle with an increased side length
Problem: One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle will increase by [tex]\(18 \, \text{cm}^2\)[/tex].
Mathematical Sentence: Let one side be [tex]\(x\)[/tex] cm, and the other side be [tex]\((x - 3)\)[/tex] cm. The original area is:
[tex]\[ \text{Original Area} = x(x - 3) \][/tex]
The new dimensions after increasing each side by 1 cm are [tex]\(x + 1\)[/tex] and [tex]\(x - 2\)[/tex]. Thus, the new area is:
[tex]\[ \text{New Area} = (x + 1)(x - 2) \][/tex]
The increase in area is given by:
[tex]\[ (x + 1)(x - 2) - x(x - 3) = 18 \][/tex]
Expanding and simplifying this, we get:
[tex]\[ x^2 - x - 2 - (x^2 - 3x) = 18 \][/tex]
[tex]\[ x^2 - x - 2 - x^2 + 3x = 18 \][/tex]
[tex]\[ 2x - 20 = 0 \][/tex]
Conclusion: This simplifies to a linear equation in the context of this problem.
Mathematical Equation:
[tex]\[ 2x - 20 = 0 \][/tex]
### 4. Product equals sum
Problem: One number is 3 more than another number. Their product equals their sum.
Mathematical Sentence: Let one number be [tex]\(n\)[/tex]. The other number is [tex]\(n + 3\)[/tex]. Express the problem as:
[tex]\[ n(n + 3) = n + (n + 3) \][/tex]
Simplifying, we get:
[tex]\[ n^2 + 3n = 2n + 3 \][/tex]
Rearranging to form a standard quadratic equation:
[tex]\[ n^2 + 3n - 2n - 3 = 0 \][/tex]
[tex]\[ n^2 + n - 3 = 0 \][/tex]
Conclusion: This is a quadratic equation.
Mathematical Equation:
[tex]\[ n^2 + n - 3 = 0 \][/tex]
---
In summary:
1. Quadratic Equation: [tex]\(h^2 + 2h - 100 = 0\)[/tex]
2. Linear Equation: [tex]\(1 = \frac{1}{t}\)[/tex]
3. Linear Equation: [tex]\(2x - 20 = 0\)[/tex]
4. Quadratic Equation: [tex]\(n^2 + n - 3 = 0\)[/tex]
