Solve revenue and distance applications of quadratic equations

A common application of quadratics comes from revenue and distance problems. Both are set up almost identical to each other so they are both included together. Once they are set up, we will solve them in exactly the same way we solved the simultaneous product equations. Revenue problems are problems where a person buys a certain number of items for a certain price per item. If we multiply the number of items by the price per item we will get the total paid. To help us organize our information we will use the following table for revenue problems

The price column will be used for the individual prices, the total column is used for the total paid, which is calculated by multiplying the number by the price. Once we have the table filled out we will have our equations which we can solve. This is shown in the following examples.

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A man buys several fish for $56. After three fish die, he decides to sell the rest at a profit of $5 per fish. His total profit was $4. How many fish did he buy to begin with?


Using our table, we don’t know the number he bought, or at what price, so we use variables

n and p. Total price was $56.

When he sold, he sold 3 less (n – 3), for $5 more (p + 5). Total profit was $4, combined

with $56 spent is $60

Find equations by multiplying number by price

n p = 56

These are a simultaneous product

(n – 3)(p + 5) = 60

Solving for number, divide by n or (n -3)

p=\frac{56}{n} and p+5=\frac{60}{n-3}

Substitute \frac{56}{n} for p in second equation


Multiply each term by LCD: n(n – 3)


Reduce fractions

56(n – 3) + 5n(n – 3) = 60n

Combine like terms


Move all terms to one side



Solve with quadratic formula

n=\frac{19\pm \sqrt{(-19)^2-4(5)(-168)}}{2(5)}


n=\frac{19\pm \sqrt{3721}}{10}=\frac{19\pm 61}{10}

We don’t want negative solutions, only do +

n=\frac{80}{10}=8 this is our n

Our solution is 8 fish.

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