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.
- Define the terms if required.
- Convert the sentences or sentence parts into algebraic terms.
- Look for keywords such as equal, subtracted from, multiplied by …, etc.
- Write an equation in the form $ax^2+bx+c=0$ form.
- Solve the equation by factorising and completing the square or quadratic formula.
- Decide which solutions are valid
- Answer the question in a sentence.
Any answer we obtain must be checked to see if it is reasonable. For example,
- if we find 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 a length \( 4 \) cm longer than its width. Suppose the area is \( 45 \) cm2. Find its width.
If the width is $x$ cm, 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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