How do I find/estimate the unknown given that the equation has exactly $2$ solutions?

Question

My problem:

The equation $x(x^2-7) = 2x+c$ has exactly $2$ solutions. Estimate the value(s) of $c$, writing you answer(s) in the form $c_1< c <c_2$, where $c_1$ and $c_2$ are integers.

My question: How do "$2$ solutions" contribute to finding $c$? Like I couldn't find any relation between them.

Thank you.

Answer 1

For a method not involving calculus, let

$$f(x) = x^3 - 9x - c = 0 \tag{1}\label{eq1}$$

Also, as the comment by Claude Leibovici says, for there to be $2$ solutions means $1$ of them must be repeated as there can't be just $1$ non-real solution as each such one comes in conjugate pairs. Thus, let $r_1$ be the repeated root and $r_2$ be the other root. Using Vieta's formulas for a cubic polynomial (you can also see this by fully expanding $\left(x - r_1\right)^2 \left(x - r_2\right)$, gives

$$2r_1 + r_2 = 0 \; \Rightarrow \; r_2 = -2r_1 \tag{2}\label{eq2}$$ $$r_1^2 + 2r_1 r_2 = -9 \tag{3}\label{eq3}$$ $$r_1^2 r_2 = c \tag{4}\label{eq4}$$

Substituting \eqref{eq2} into \eqref{eq3} gives

$$-3r_1^2 = -9 \; \Rightarrow \; r_1^2 = 3 \; \Rightarrow \; r_1 = \mp \sqrt{3} \tag{5}\label{eq5}$$

Thus, from \eqref{eq2}

$$r_2 = \pm 2\sqrt{3} \tag{6}\label{eq6}$$

Thus, using \eqref{eq5} and \eqref{eq6} int \eqref{eq4} gives

$$c = \pm 6\sqrt{3} \tag{7}\label{eq7}$$

I trust you can now finish the rest.

Update: From how the equation is stated and the request to just estimate the values of $c$ between $2$ integers, I now believe the intent was to graph the $2$ functions of $f(x) = x\left(x^2 - 7\right)$ and $g(x) = 2x + c$ to roughly see where they cross at only $2$ values of $x$.

Answer 2

Consider $f(x)=x^3-9x$.

Now, $$f'(x)=3x^2-9=3(x-\sqrt3)(x+\sqrt3).$$

For $f'(x)=0$ we obtain $x=\sqrt3$ or $x=-\sqrt3$.

For $x=\sqrt3$ we obtain $c=-6\sqrt3$ and $$x(x^2-7)-2x-c=0$$ it's $$(x-\sqrt3)^2(x+2\sqrt3)=0,$$ which has two different roots.

For $x=-\sqrt3$ we obtain $c=6\sqrt3$ and the following equation. $$(x+\sqrt3)^2(x-2\sqrt3)=0.$$

Answer 3

Here is a more visual approach to this problem.

You are being asked to find the value(s) of $c$ for which

$x^3 - 9x=c$

has just two distinct solutions.

If you draw the graph of $y=x^3-9x$ you will see that it has a "hill" at $x=-\sqrt{3}$ and a "valley" at $x=\sqrt{3}$. The solutions to $x^3-9x=c$ are the values of $x$ where the horizontal line $y=c$ crosses this graph.

If $|c|$ is large i.e. the line $y=c$ is far from the $x$ axis then it will only cross $x^3-9x$ at one point so $x^3-9x=c$ only has one solution.

On the other hand if $|c|$ is small i.e. the line $y=c$ is close to the $x$ axis then it will cross $x^3-9x$ at three points so $x^3-9x=c$ has three solutions.

To get just two solutions the line $y=c$ must just touch the top of the hill or the bottom of tha valley in the graph of $x^3-9x$.

At the top of the hill $c=(-\sqrt{3})^3+9\sqrt{3}=6\sqrt3$.

And at bottom of the valley $c=(\sqrt{3})^3-9\sqrt{3}=-6\sqrt3$.