Quadradic Equation Word Problems - Examples & Practice

Quadratic Word Problems

Let's look at $2$ pretty common types of word problems that use quadratic functions. Recall that quadratic expressions follow this general form: $$y = ax^2 + bx + c$$ In a quadratic expression, $a$ and $b$ are coefficients (numbers in front of the variable $x$), and $c$ is constant (a number by itself). It's important to remember that $a \neq 0$. It's possible for $b=0$ and/or $c=0$, but they don't have to.

Example 1: Word problems involving area

Many quadratic word problems use the area of a rectangle. Here is a graphic to get you started

Here is another example.

Singh knows the total area of his garden is $50$ m$^2$, but he doesn't know the length or width. Use the diagram below to find the length and width in meters (m).

Recall that the formula for area of a rectangle is $\text{length}\cdot\text{width}$. So, let's plug the values from Singh's garden into that formula, and set it equal to $50$ (the area). $$\begin{align} \text{area} &= \text{length}\cdot\text{width} \\ 50 &= x\cdot (x-5)\end{align}$$

Now, let's multiply out this equation and get it in the standard quadratic form (written at the top of this explanation):

$$\begin{align} 50 &= x\cdot (x-5) \\ 50 &= x^2 - 5x \\ 0 &= x^2 - 5x - 50\end{align}$$

Great! How did we do this? Remember, how to multiply a polynomial by a monomial. Then, we moved everything to the right side.

The next thing we need to do is solve for $x$. Remember that, when working with quadratic equations, we can solve for $x$ by factoring.

$$(x-10)(x+5) = 0$$ So, this gives us two possible values for $x$ that make the left side of the equation equal to $0$. These two possible values are: $$x=10$$ $$x=-5$$ If we look back at the original diagram, the width of Singh's garden is written as $x-5$.

Try plugging in both possible values for $x$ into $x-5$. Notice that when we plug in $x=-5$, we find that the width of the garden is $-10$. Unless we're in Narnia, that's not possible. We can't have a negative width.

So, the value of $x$ must be the other option, $x=10$.

Great job! In summary, what we did was:

1. Set up a quadratic equation from the image.
2. Solve the quadratic equation.
3. Check the possible solutions that will accurately answer the original question.

We'll use this same pattern when we tackle future quadratic word problems!

Example 2

Jill sells hats. She has come up with a formula to figure out how much demand there is for the hats based on the price: $$D = p^2 - 50p + 1200$$ where $D$ is the demand and $p$ is the price in dollars. What should each hat cost for the demand to be $1800$ hats?

This word problem gives us a quadratic function to use. We're told that the demand needs to be for $1800$ hats. This means that, in our formula, $D=1800$. Let's plug that in. $$1800 = p^2 - 50p + 1200$$

Now we need to solve for $p$, the price per hat. Because this is a quadratic equation, we need to:

• Set one side equal to $0$
• Factor the equation to find possible values for $p$ $$\begin{align} 1800 &= p^2 - 50p + 1200 \\ 0 &= p^2 - 50p - 600 \\ 0 &= (p-60)(p+10) \\ p&= 60\,\,\text{or}\,\,p=-10 \end{align}$$ So, we're left with two prices per hat: $\$60$ or $\$-10$.

Obviously, $-10$ isn't going to work since prices are always positive. So, we can say $p=60$ and that Jill should charge $\$60$ per hat in order for the demand to be for $1800$ hats.

If we want to double check our work, we can plug in our answer ($p=60$) back into the original equation ($D = p^2 - 50p + 1200$).

In these two examples, we saw word problems dealing with area and price. You can also use these methods to solve problems with consecutive integers and other types of problems.

(Videos) Solving Geometry Word Problems Using Quadratic Equations

by PatrickJMT

This video by Patrick JMT covers a quadratic word problem based on area of a rectangle.

$\large{\text{Summary}}$

Patrick walks us through how to solve the following quadratic geometry problem:

“A picture inside a frame is $2$ inches longer than it is wide. The picture is in a frame that has width $3$ inches on each side of the picture. If the area of the pic, including the frame is $195 \text{ in.}^2$, find the dimensions of the frame.”

He goes on to explain how we get the equations for the sides of the frame, so the whole problem looks like:

We know the area of the picture frame, $195 \text{ in}^2$. We can set the two sides of the frame equal to that and find a value for $x$.

$$(x + 6)(x + 8) = 195$$

Using the FOIL method, we get an equation that looks like

$$x^2 + 14x + 48 = 195$$

We want the left side to equal $0$, so subtracting $195$ from both sides will give us

$$x^2 + 14x – 147 = 0$$

This looks pretty hard to factor. Let’s use our handy dandy quadratic formula.

Our values are going to be $a = 1, b = 14,$ and $c = -147$

$$\begin{align}x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x &= \frac{-14\pm\sqrt{14^2 – 4(1)(-147)}}{2(1)}\\ x &= \frac{-14\pm\sqrt{196 +588}}{2}\\ x &= \frac{-14 \pm \sqrt{784}}{2}\\ x &= \frac{-14 \pm 28}{2}\\ x &= -7 \pm 14 \end{align}$$

*Note: be careful with the square root!

This leaves us two values for $x$, $x = 7$ or $x = -21$. We cannot have a negative measurement, so our final answer is $x = 7$.

(Videos) Solve Projectile Problems Using Quadratics

by PatrickJMT

This video by Patrick JMT works through a word problem based on projectile motion.

$\large{\text{Summary}}$

Solving a Projectile Problem Using Quadratics

Patrick uses the following problem as an example of projectile motion:

“Larry throws a rock in the air. The height, $h$, in feet above the ground of the rock is given by: $h = -16t^2 + 123t +40.$ How long is the rock in the air?”

From this equation, at time $t = 0$, we gather that $h = 40$. This means Larry is on a cliff with height $40$. We want to know the time at which the rock reaches a height of $0$. So, plugging in $0$ for $h$, we get an equation:

$$0 = -16t^2 + 123t +40$$

This is now a quadratic equation, which we can solve for using the quadratic formula! We have values of $a =16, b = 123,$ and $c = 40$. We can plug these into the quadratic formula: $$\begin{align} t&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ t&=\frac{-123\pm\sqrt{(123)^2-4(-16)(40)}}{2(-16)}\\ t&= \frac{-123\pm\sqrt{17689}}{-32}\\ t&=\frac{123 \pm \sqrt{17689}}{32}\\ t&=\frac{123 \pm 133}{32}\\ t&= 8, -\frac{5}{16}\end{align}$$

We get two solutions: $8$ and $-\frac{5}{16}$. $t$ is time, which can't be negative, so our answer is $t=8$.