Linear and Quadratic Equation Word Problems

Word problems require you to translate real-world situations into mathematical equations. In algebra, these scenarios usually result in either linear or quadratic equations. The main challenge isn't just solving the math—it's figuring out how to set up the equation correctly and ensuring your final answer makes sense in the real world.

5 Steps to Solve Word Problems

1. Read and Understand: Identify what the problem is asking you to find.
2. Define the Variable: Choose a letter (like $x$ or $t$) to represent your unknown value.
3. Set Up the Equation: Translate the words from the problem into mathematical expressions.
4. Solve the Equation: Use algebraic methods (like factoring or the quadratic formula) to find the value of your variable.
5. Check the Context: Does your answer make sense? In the real world, lengths, areas, and time cannot be negative. You will often need to reject negative solutions.

Example 1: Area and Dimensions (Quadratic)

The Problem: The area of a rectangle is $72\text{ m}^2$. The length is $2\text{ m}$ more than the width. Find the dimensions.

The Solution: First, define the variables. Let the width be $w$. The length is $2\text{ m}$ more than the width, so the length is $w + 2$.

The formula for the area of a rectangle is $\text{Area} = \text{length} \times \text{width}$. Set up the equation: $w(w + 2) = 72$

Expand and set the equation to zero to solve the quadratic equation: $w^2 + 2w - 72 = 0$

Factor the quadratic equation: $(w + 9)(w - 8) = 0$

This gives us two possible solutions for $w$: $w = -9 \quad \text{or} \quad w = 8$

Since a width cannot be negative, we reject $-9$. Therefore, the width is $8\text{ m}$. The length is $8 + 2 = 10\text{ m}$.

Answer: The dimensions are $8\text{ m}$ by $10\text{ m}$.

Example 2: Projectile Motion (Quadratic)

The Problem: A ball is thrown upward, and its height $h$ in feet after $t$ seconds is given by the equation $h(t) = -16t^2 + 48t + 5$. When does it hit the ground?

The Solution: The ball hits the ground when its height is zero. So, we set $h(t) = 0$: $-16t^2 + 48t + 5 = 0$

Because this doesn't factor easily, we use the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plug in $a = -16$, $b = 48$, and $c = 5$: $t = \frac{-48 \pm \sqrt{48^2 - 4(-16)(5)}}{2(-16)}$ $t = \frac{-48 \pm \sqrt{2304 + 320}}{-32}$ $t = \frac{-48 \pm \sqrt{2624}}{-32}$

Calculate the two possible values for $t$: $t \approx \frac{-48 + 51.22}{-32} \approx -0.1 \text{ s}$ $t \approx \frac{-48 - 51.22}{-32} \approx 3.1 \text{ s}$

Time cannot be negative, so we reject $-0.1$.

Answer: The ball hits the ground after approximately $3.1$ seconds.