How to find the constant values where a cubic equation has exactly two solutions

Q: When does the equation f(x)=k have exactly two solutions for a cubic?

It has exactly two solutions when k equals the local maximum or local minimum value of the cubic.

Q: What are the values of k in this problem?

The values are approximately k = -4.49 and k = 15.02.

Question

Question 21 (7 marks)

The equation $f(x)=k$ has two solutions, where $f(x)=ax^3+bx^2-12x+8$ and $a$, $b$ and $k$ are constants.

The graph of $y=f(x)$ cuts the x-axis at $x=2$, $x=-2$, and at one other point.

Determine the value(s) of the constant $k$, rounded to 2 decimal places. Explain your reasoning.

Solution

Use roots to solve for $a$ and $b$:

$f(2)=0 	o 8a+4b-24+8=0$

$f(-2)=0 	o -8a+4b+24+8=0$

Solving simultaneously with CAS gives $a=3$ and $b=-2$.

$y=3x^3-2x^2-12x+8$

For two solutions, $k$ must equal the local maximum or equal the local minimum of $f(x)$, found using CAS.

Local maximum is $y=15.0230$

Local minimum is $y=-4.4880$

Hence $k=-4.49$ or $k=15.02$

Specific behaviours

indicates solving for $a$ and $b$

$ f(1)=0$, $f(-2)=0$

identify

solve case for one solution

Acemy AI · Accepted Answer

> **Key concept**: This is a classic cubic-graph question: first recover the polynomial from its roots, then use the turning points to decide when a horizontal line cuts the graph twice. The algebra is only half the story; the graph shape matters just as much.

### Step by step
We are told

$$f(x)=ax^3+bx^2-12x+8$$

and that the graph cuts the x-axis at $x=2$, $x=-2$, and one other point.

## 1) Use the known roots
Since $f(2)=0$ and $f(-2)=0$:

$$f(2)=8a+4b-24+8=0$$
$$8a+4b-16=0$$

and

$$f(-2)=-8a+4b+24+8=0$$
$$-8a+4b+32=0$$

Now solve the system:

$$\begin{aligned}
8a+4b&=16\
-8a+4b&=-32
\end{aligned}$$

Add the equations:

$$8b=-16 \Rightarrow b=-2$$

Substitute back:

$$8a+4(-2)=16 \Rightarrow 8a=24 \Rightarrow a=3$$

So

$$f(x)=3x^3-2x^2-12x+8$$

## 2) What does “has two solutions” mean?
The equation $f(x)=k$ has two solutions when the horizontal line $y=k$ is tangent to the cubic at a turning point. That happens at the local maximum and local minimum values of $f(x)$.

## 3) Find the turning points
Differentiate:

$$f'(x)=9x^2-4x-12$$

Solve

$$9x^2-4x-12=0$$

which gives the critical points. Evaluating $f(x)$ there gives the local maximum and minimum values:

- local maximum $\approx 15.0230$
- local minimum $\approx -4.4880$

So the values of $k$ are

$$\boxed{k\approx -4.49 	ext{ or } k\approx 15.02}$$

Those are the horizontal lines that touch the graph at one turning point and intersect it again elsewhere, giving exactly two solutions.

### Pitfall alert
A very common mistake is to think any horizontal line between the turning points gives two solutions. For a cubic, many such lines give three solutions. You only get exactly two solutions when the line passes through a local maximum or local minimum, creating a repeated root at the tangency point.

### Try different conditions
If the question asked for one solution instead of two, the valid $k$ values would be those above the local maximum or below the local minimum, because the horizontal line would then intersect the cubic only once. If the cubic were shifted up or down, the same method would apply after adjusting the turning-point values.

### Further reading
turning point, horizontal line test, cubic function

Question

How to find the constant values where a cubic equation has exactly two solutions

Expert Verified Solution

Step by step

1) Use the known roots

FAQ

When does the equation f(x)=k have exactly two solutions for a cubic?

What are the values of k in this problem?