How can an open-topped box with volume 500 cm3 be made from a square piece of card 20 cm by 20 cm?
To make such a box, a net can be made by cutting off equal squares from each corner, as shown in the diagram.
Suppose each smaller square has side x cm.
The dimensions of the box will then be
x by (20 − 2x) by (20 − 2x).
For the volume to be 500 cm3 , x must satisfy the equation
x(20 − 2x)2 = 500, 0 ≤ x ≤ 10
When the expression has been multiplied out, the highest power of x in this equation is x3.
This is an example of a cubic equation.
As with a quadratic equation, it is good practice to reduce this equation to the form f(x) = 0.
x(400 − 80x + 4x2) = 500
4x3 − 80x2 + 400x − 500 = 0
x3 − 20x2 + 100x − 125 = 0, 0 ≤ x ≤ 10
Again by comparison with quadratic equations, the next thing you might want to do would be to try to factorise it. With a cubic function, however, this is not easy.
Fortunately one solution to the problem can be seen 'by inspection'. A little experimentation reveals that x = 5 will make this equation true.
Is this the only solution?
Remember that for quadratic equations when there is one answer there is always another, except when there are equal roots.
One method of solution is to draw the graph of the cubic function
f(x) = x3 − 20x2 + 100x − 125
and see where it crosses the x-axis.
This is shown opposite. You can see that it does in fact have three real roots.
You can use a graph calculator to find approximations for the other roots but we can use factorization to find values for these roots.
As x = 5 is a solution, we know that (x – 5) is a factor of the cubic equation
x3 − 20x2 + 100x − 125 = 0
and can write
x3 − 20x2 + 100x − 125 = (x – 5) × (quadratic formula)
Moreover, the quadratic must be of the form
x2 + bx + 25
Finding b is not quite so obvious but can be done as follows.
x3 − 20x2 + 100x − 125 = (x − 5) (x2 + bx + 25)
To make −20x in the cubic equation,
−20 = −5 + b ⇒ b = −15
To make +100x in the cubic equation,
100 = 25 − 5b ⇒ b = −15
Hence x3 − 20x2 + 100x − 125 = (x − 5) (x2 − 15x + 25)
So we can now solve the cubic equation,
x3 − 20x2 + 100x – 125 = 0
We can write (x − 5) (x2 − 15x + 25) = 0
Either (x − 5) = 0 or (x2 − 15x + 25) = 0
So x = 5 or, solving the quadratic,
| x | = |
| = | |
| = (working to 2 decimal places) | |
| = 13.09 and 1.91 |
Worked Examples
Find all solutions of x3 − 3x2 − 33x + 35 = 0
You can see that x = 1 is a root of this equation since f(1) = 0 .
Hence (x − 1) must be a factor, and you can write
x3 − 3x2 − 33x + 35 = (x − 1)(x2 + bx − 35)
Now to find b. The cubic equation contains −33x, so
| −33 = −b − 35 | |
| ⇒ | b = −2 |
| ⇒ | x3 − 3x2 − 33x + 35 = (x − 1)(x2 − 2x − 35) |
In this case, the quadratic factorises to (x − 7)(x + 5).
Hence
x3 − 3x2 − 33x + 35 = (x − 1)(x − 7)(x + 5)
The original equation can thus be re-written
(x − 1)(x − 7)(x + 5) = 0
giving the three solutions x = −5, x = 1 and x = 7.
Solve x3 − 14x − 15 = 0
x = −3 is a root of this equation. Hence (x + 3) is a factor.
x3 − 14x − 15 = (x + 3)(x2 + bx − 5)
Now to find b. The cubic equation has 0 x2, so
| 0 = b + 3 | |
| ⇒ | b = −3 |
| ⇒ | x3 − 14x − 15 = (x + 3)(x2 − 3x − 5) |
The original equation thus reads
(x + 3)(x2 − 3x − 5) = 0
So either x + 3 = 0 or x2 − 3x − 5 = 0
This gives x = −3, x = − 1.19 or x = 4.19, to 3 significant places.
Exercises
