# Quadratic Formula Word Problems Answer Key and Lesson

Mastering the quadratic formula is essential for solving many types of algebraic word problems. There are many different methods that can be used to solve geometry problems, age-related puzzles, and standard quadratic equations, but understanding how to use the quadratic formula can save time and ensure accurate results.

My goal is to provide both a lesson and a quadratic formula word problems answer key for each example I share.

Let’s take a look at some geometry problems and some real-world applications so that you can apply the knowledge you gain here to your classroom exercises or your next homework assignment.

## Understanding the Quadratic Formula

The quadratic formula is a helpful tool that can be used to solve quadratic equations in standard form, which take the form:

\[ax^2 + bx + c = 0\]

The quadratic formula is:

\[x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]

When applying this formula, it is important to make sure the equation is in standard form before solving!

One of the most common uses of the quadratic formula is to find the roots of the equation. These roots represent the x-values where the parabola (the graph of the quadratic function) crosses the x-axis.

To use the quadratic formula, we first substitute the values of *a*, *b*, and *c* into the quadratic formula. After simplifying the formula, the resulting x-value(s) will be the roots of the quadratic equation.

The term under the square root, \(b^2 – 4ac\), is known as the discriminant value. This helpful tool gives us valuable information about the nature of the roots of the quadratic equation.

Roots of quadratic equations can be:

- real numbers (positive discriminant value)
- imaginary numbers (negative discriminant value)
- repeated (a discriminant value of zero)

## Word Problems Involving the Quadratic Formula

You’ll frequently encounter quadratic equations when solving geometry problems, such as calculating the area of shapes like rectangles or triangles.

Let’s start with a simple geometry word problem involving the unknown width of a rectangle when given the length of a rectangle.

### Problem 1: Solving for the Width of a Rectangle

The length of a rectangle is 8 inches more than its width. What is the width of this rectangle if the area of this rectangle is 48 square inches?

### Problem 1: Answer Key

**Step 1: Set up the equation**

We begin be letting \(w\) represent the width of the rectangle. This means that the length will be \(w + 8\).

The area of a rectangle can be calculated by multiplying the length by the width. Therefore, our equation becomes:

\[w(w + 8) = 48\]

**Step 2: Rearrange into standard form**

Remember that our equation must be written in standard form if we want to use the quadratic formula.

Expanding the equation by distributing the *w* into the brackets results in:

\[w^2 + 8w – 48 = 0\]

Now that we have an equation in standard form, we are ready to apply the quadratic formula!

**Step 3: Apply the quadratic formula**

To use the quadratic formula, we substitute the values of \(a = 1\), \(b = 8\), and \(c = -48\) into our formula:

\[\begin{split} w &= \frac{-8 \pm \sqrt{8^2 – 4(1)(-48)}}{2(1)} \\ \\ w &= \frac{-8 \pm \sqrt{64 + 192}}{2}\\ \\ w &= \frac{-8 \pm \sqrt{256}}{2}\\ \\ w &= \frac{-8 \pm 16}{2} \end{split} \]

**Step 4: Solve for the roots**

To find the roots, we simplify the final step of our quadratic formula calculation. Note that we will have two real roots as our answers in this case, since the discriminant value is positive.

\[w = \frac{-8 + 16}{2} = 4 \quad \text{or} \quad w = \frac{-8 – 16}{2} = -12\]

Since the width must be a positive integer, we reject the negative answer. This leaves us with \(w = 4\).

**Therefore the width of this rectangle is 4 inches.**

## The Quadratic Formula in Geometry Problems

Quadratic equations appear often in geometry, especially when calculating areas of shapes like rectangles and triangles.

### Problem 2: Solving for the Height of a Triangle Using the Quadratic Formula

The base of a triangle is \( x + 4 \) units, and the height is \( x \) units. What are the possible values for \( x \), the height of the triangle, if the area of the triangle is 48 square units?

### Problem 2: Answer Key

**Step 1: Set up the equation**

If we consider the area formulas that we know, we know that the formula for the area of a triangle is:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

We are told that the area is 48 square units, the base as \( x + 4 \), and the height is \( x \). Plugging each of these pieces of information into the formula results in:

\[ 48 = \frac{1}{2} \times (x + 4) \times x \]

**Step 2: Simplify and rearrange into standard form**

Next, we multiply both sides by 2 to eliminate the fraction. We also expand the right side by distributing the \(x\) into the brackets:

\[ \begin{split} 96 &= (x + 4) \times x \\ \\ 96 &= x^2 + 4x \end{split} \]

Remember that we need to write our expression in standard form in order to be able to apply the quadratic formula.

\[ x^2 + 4x – 96 = 0 \]

**Step 3: Apply the quadratic formula**

Now that we have a quadratic equation in standard form, \(x^2 + 4x – 96 = 0\), we can apply the quadratic formula.

Substituting \(a=1\), \(b=4\), and \(c=-96\) results in:

\[ \begin{split} x &= \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\ \\ x &= \frac{-4 \pm \sqrt{4^2 – 4(1)(-96)}}{2(1)} \\ \\ x &= \frac{-4 \pm \sqrt{16 + 384}}{2} \\ \\ x &= \frac{-4 \pm \sqrt{400}}{2} \\ \\ x &= \frac{-4 \pm 20}{2} \end{split} \]

**Step 4: Solve for the roots**

The final step is to determine the two possible values of \(x\):

\[ x = \frac{-4 + 20}{2} = \frac{16}{2} = 8 \]

\[x = \frac{-4 – 20}{2} = \frac{-24}{2} = -12\]

Since the height of the triangle must be a positive value, we reject the negative solution.

**Therefore, the height of the triangle is \( x = 8 \) units.**

## Applications Beyond Geometry

The quadratic formula can also be used to solve word problems in real-world contexts, such as problems involving age, distance, or speed. Let’s take a look at a classic age problem to see how the quadratic formula can be applied in the real-world!

### Problem 3: Age Problem

The sum of the squares of the current ages of two siblings is 650. If one sibling is 5 years older than the other, what are their current ages?

### Problem 3: Answer Key

**Step 1: Set up the equation**

We can start by letting \(y\) represent the age of the younger sibling. This means that the older sibling’s age can be represented by \(y + 5\).

Our equation becomes:

\[y^2 + (y + 5)^2 = 650\]

**Step 2: Expand and simplify**

We can expand the set of brackets and simplify in order to arrive at a quadratic equation in standard form.

\[ \begin{split} y^2 + (y^2 + 10y + 25) &= 650 \\ \\ 2y^2 + 10y + 25 &= 650 \\ \\ 2y^2 + 10y – 625 &= 0 \\ \\ \end{split} \]

**Step 3: Apply the quadratic formula**

We can substitute the values of \(a = 2\), \(b = 10\), and \(c = -625\) into the quadratic formula and solve for \(y\):

\[ \begin{split} y &= \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\\\ y &= \frac{-10 \pm \sqrt{10^2 – 4(2)(-625)}}{2(2)} \\\\ y &= \frac{-10 \pm \sqrt{100 + 5000}}{4} \\\\ y &= \frac{-10 \pm \sqrt{5100}}{4} \\\\ y &= \frac{-10 \pm 71.42}{4} \\\\ \end{split} \]

**Step 4: Solve for the roots**

We can determine the roots by simplifying the result of our calculation further:

\[y = \frac{-10 + 71.42}{4} = \frac{61.42}{4} \approx 15.36\]

\[y = \frac{-10 – 71.42}{4} = \frac{-81.42}{4} \approx -20.36\]

Since age cannot be negative, we reject the negative root. Therefore, the younger sibling is approximately 15.36 years old.

**Step 5: Find the age of the older sibling**

To find the age of the older sibling, we simply add 5 to the age of the younger sibling:

\[y + 5 \approx 15.36 + 5 = 20.36\]

Therefore we can say that the younger sibling is approximately 15 years old, and the older sibling is approximately 20 years old.

## Next Steps

My hope is that the quadratic formula word problems answer key and lesson I’ve shared here helps you feel more comfortable applying this important tool.

Your next step is to practice!

Take some time to review the problems as well as the answer keys provided.

As you can see, there are a wide variety of problems that can be solved using the quadratic formula. Just remember that the quadratic formula is only one tool amongst many different methods that can be used to solve quadratic equations.

To learn more about the tools that we can use to solve quadratic equations, check out my walkthrough on how to factor quadratic trinomials!

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