Word problems that lead to quadratics

Four steps for every word problem

1. Name the unknown. Say in words what $x$ stands for, with its unit.
2. Write the equation. Turn each sentence of the problem into algebra.
3. Solve. Put the equation in standard form and factorise.
4. Check the meaning. Keep only the roots that fit the problem, and check them against the words.

Step 4 is new. A quadratic equation can have two roots, but a problem may allow only one. A width of $-5\ \text{m}$ or an age of $-3$ years makes no sense.

Areas

A worked example. A rectangular garden has a perimeter of $46\ \text{m}$ and an area of $132\ \text{m}^2$. Find its sides.

1. Name. Let the width be $w$ metres. Half the perimeter is $23\ \text{m}$, so the length is $23 - w$ metres.
2. Equation. $w(23 - w) = 132$, that is $23w - w^2 = 132$.
3. Standard form. $w^2 - 23w + 132 = 0$. The pair with product $132$ and sum $-23$ is $-11$ and $-12$, so $(w - 11)(w - 12) = 0$.
4. Meaning. $w = 11$ or $w = 12$. Both make sense, and both give the same garden: $11\ \text{m}$ by $12\ \text{m}$.

Check. $2 \times (11 + 12) = 46\ \text{m}$ and $11 \times 12 = 132\ \text{m}^2$.

Numbers

A worked example. The product of two consecutive positive odd numbers is $143$. Find them.

1. Name. Let the smaller be $x$. The next odd number is $x + 2$.
2. Equation. $x(x + 2) = 143$, so $x^2 + 2x - 143 = 0$.
3. Solve. The pair with product $-143$ and sum $2$ is $13$ and $-11$: $(x + 13)(x - 11) = 0$.
4. Meaning. $x = 11$ or $x = -13$. The numbers must be positive, so they are $11$ and $13$.

Speed and time

Time is distance divided by speed. A problem that compares two journeys often leads to a quadratic equation.

A worked example. A boat goes $24\ \text{km}$ downstream at a speed $4\ \text{km/h}$ more than its usual speed, and saves half an hour compared with the usual time. Find the usual speed $v$.

1. Equation. $\dfrac{24}{v} - \dfrac{24}{v + 4} = \dfrac{1}{2}$.
2. Clear the fractions. Multiply by $2v(v + 4)$: $48(v + 4) - 48v = v(v + 4)$, so $192 = v^2 + 4v$.
3. Solve. $v^2 + 4v - 192 = 0$, so $(v + 16)(v - 12) = 0$.
4. Meaning. A speed cannot be negative, so $v = 12\ \text{km/h}$.

Ages

Write every age at the same moment in terms of one unknown. Five years from now, a person aged $a$ years will be $a + 5$ years old. Three years ago they were $a - 3$ years old.

Practise with new numbers

Try it in Python

A computer can simply try every whole width. This program finds the sides of the garden of the worked example.

Common mistakes

• Not saying what $x$ is. Write "Let the width be $w$ metres" before you start.
• Keeping an impossible root. Lengths, speeds, ages and counts cannot be negative.
• Mixing units. Change minutes to hours before you use a speed in $\text{km/h}$.
• Forgetting the second number. Once you find $x$, answer the whole question: both sides, or both numbers.

Remember these

• Name the unknown, write the equation, solve, then check the meaning.
• Reject the roots that do not fit the situation.
• Time $=$ distance $\div$ speed. Clear fractions by multiplying by every denominator.
• Check the final answer against the words of the problem.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The product of two consecutive positive whole numbers is $110$. Find them.

2. A rectangle has a perimeter of $30\ \text{m}$ and an area of $56\ \text{m}^2$. Find its sides.

3. The square of a number is $20$ more than the number. Find the number.

4. Why does a problem about the width of a field never keep a negative root?