Problem Solving with Quadratics

There are many situations in science, engineering, economics and other fields where quadratic equations are a vital part of the mathematics used. Some real word problems can be solved using a quadratic equation. We are generally only interested in any real solutions which result. We will look at some of these applications.

We employ the following general problem-solving method when solving worded problems involving quadratic equations.

  1. Define the terms if required.
  2. Convert the sentences or sentence parts into algebraic terms.
  3. Look for keywords such as equal, subtracted from, multiplied by … and so on.
  4. Write an equation in the form $ax^2+bx+c=0$ form.
  5. Solve the equation by factorising, completing the square or quadratic formula.
  6. Decide which solutions are valid
  7. Answer the question in a sentence.

Any answer we obtain must be checked to see if it is reasonable. For example,

  • if we are finding a length it must be positive and we reject any negative solutions.
  • if we are finding ‘how many people are present’ then clearly the answer must be an integer.

Example 1

A rectangle has length $4$ cm longer than its width. If the area is $45$ cm2. Find its width.

If the width is $x$ cm, then the length is $(x+4)$ cm.
\( \begin{align} \displaystyle
x(x+4) &= 45 \\
x^2 + 4x &= 45 \\
x^2 + 4x – 45 &= 0 \\
(x+9)(x-5) &= 0 \\
x+9 &= 0 \text{ or } x-5 = 0 \\
\therefore x &= -9 \text{ or } 5 \\
\end{align} \)
We reject the negative solution as lengths are positive.
Thus the width is $5$ cm.

Example 2

The product of two consecutive odd numbers is $323$. Find the numbers.

Let two consecutive odd numbers be $x$ and $(x+2)$.
\( \begin{align} \displaystyle
x(x+2) &= 323 \\
x^2 + 2x &= 323 \\
x^2 + 2x – 323 &= 0 \\
(x+19)(x-17) &= 0 \\
x+19 &= 0 \text{ or } x-17 = 0 \\
x &= -19 \text{ or } 17 \\
\therefore x &= 17 \text{ and } x+2 = 19\\
\end{align} \)
We reject the negative solution as odd numbers are positive.
Thus two odd numbers are $17$ and $19$.

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