Find the values of k that make a cubic have exactly two real solutions

Q: What values of k make the cubic have exactly two real solutions?

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

Q: Why do turning points matter here?

A horizontal line gives exactly two solutions when it touches the cubic at a turning point and crosses it once more.

Question

Question 21
(7 marks)
A function $f(x)=k$ 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 one other point.
Determine the value(s) of the constant $k$, rounded to 2 decimal places. Explain your reasoning.

$a x^3+b x^2-12x+8$

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

$f(-2)=0$
$a(-2)^3+b(-2)^2-12(-2)+8$
$-8a+4b+24+8=0$
$-8a+4b+32=0$

$f(2)=0$
$a(2)^3+b(2)^2-12(2)+8$
$8a+4b-24+8=0$
$8a+4b-16=0$

$\left\{
\begin{aligned}
8a+4b-16&=0\
-8a+4b+32&=0
\end{aligned}
ight.$
$a=3$
$b=-2$

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

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

$x=-2\quad x=0.66$
$x=2

Acemy AI · Accepted Answer

**Key takeaway**: This kind of problem rewards careful reading. The x-intercepts pin down the cubic, and the phrase “has two solutions” is really asking about where a horizontal line touches the graph just right.

Start with

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

and the facts that the graph crosses the x-axis at $x=2$ and $x=-2$, plus one more point.

## Step 1: Use the intercepts
Because the graph has x-intercepts at $x=2$ and $x=-2$:

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

Substitute into the function:

$$8a+4b-24+8=0 \Rightarrow 8a+4b-16=0$$

$$-8a+4b+24+8=0 \Rightarrow -8a+4b+32=0$$

Solve the system:

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

Adding gives

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

Then

$$8a-8=16 \Rightarrow a=3$$

So the cubic is

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

## Step 2: Interpret “has two solutions”
The equation $f(x)=k$ has two solutions when the horizontal line $y=k$ is tangent to the cubic at a turning point. That means $k$ must equal either the local maximum or the local minimum of $f$.

## Step 3: Find the turning-point values
Differentiate:

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

Solve $f'(x)=0$ to get the critical points. Evaluating $f(x)$ there gives approximately:

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

Rounding to 2 decimal places:

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

---
**Pitfalls the pros know** 👇
Do not confuse “two solutions” with “two turning points.” A cubic can intersect many horizontal lines in three places, so you must use tangency to get exactly two solutions. Also, if your calculator gives decimal x-values for the critical points, that is fine; the final k-values still need the correct rounding.

**What if the problem changes?**
If the question asked for values of $k$ that give one solution, the answer would be any $k$ above the local maximum or below the local minimum. If it asked for three solutions, then $k$ must lie strictly between the local minimum and local maximum.

`Tags`: cubic polynomial, local maximum, local minimum

Question

Find the values of k that make a cubic have exactly two real solutions

Expert Verified Solution

Step 1: Use the intercepts

FAQ

What values of k make the cubic have exactly two real solutions?

Why do turning points matter here?