Solving Problems Involving Quadratic Functions

Quantities: The values or amounts mentioned in a problem.
Variables: Letters used to represent unknown values.
Conditions: Given facts or requirements that must be followed to solve the problem.
Method: Set up equations using geometric formulas such as $\text{Area} = \text{Length} \times \text{Width}$ , rearrange to standard form $a x^2 + b x + c = 0$ , factor to solve for the variable, and reject negative roots because physical dimensions cannot be negative.

Given Conditions: Area is $60\,\text{cm}^2$ ; length is $4\,\text{cm}$ longer than width.
Variables:
- Width: $W = x$
- Length: $L = x + 4$
Equation Setup:
- $A = L \times W$
- $60 = (x + 4)(x)$
- $60 = x^2 + 4x$
- $x^2 + 4x - 60 = 0$
Factoring and Solution:
- $(x + 10)(x - 6) = 0$
- Root $x = -10$ is rejected because width cannot be negative.
- Root $x = 6$
Final Dimensions:
- $\text{Width} = 6\text{ meters}$
- $\text{Length} = 6 + 4 = 10\text{ meters}$
- Checking: $10 \times 6 = 60$

Given Conditions: Area is $48\,\text{m}^2$ ; length is $2\text{ meters}$ longer than width.
Variables:
- Width: $x$
- Length: $x + 2$
Equation Setup:
- $A = L \times W$
- $48 = (x + 2)(x)$
- $48 = x^2 + 2x$
- $x^2 + 2x - 48 = 0$
Factoring and Solution:
- $(x + 8)(x - 6) = 0$
- Root $x = -8$ is rejected because width cannot be negative.
- Root $x = 6$
Final Dimensions:
- $\text{Width} = 6\text{ meters}$
- $\text{Length} = 6 + 2 = 8\text{ meters}$
- Checking: $8 \times 6 = 48\,\text{m}^2$

Given Conditions: Area is $20\,\text{m}^2$ ; length is $3$ less than twice the width.
Variables:
- Width: $x$
- Length: $2x - 3$
Equation Setup:
- $A = L \times W$
- $20 = (2x - 3)(x)$
- $20 = 2x^2 - 3x$
- $2x^2 - 3x - 20 = 0$
Factoring and Solution:
- $(2x + 5)(x - 4) = 0$
- Root $x = -\frac{5}{2}$ is rejected because width cannot be negative.
- Root $x = 4$
Final Dimensions:
- $\text{Width} = 4\text{ meters}$
- $\text{Length} = 2(4) - 3 = 8 - 3 = 5\text{ meters}$
- Checking: $5 \times 4 = 20\,\text{m}^2$