# Find the values of $k$ such that the equation $x^3+x^2-x+2=k$ has three distinct real solutions

Find the values of $k$ such that the equation $$x^3+x^2-x+2=k$$ has three distinct real solutions.

Please explain how to find the solution! Thanks

• Have you drawn a graph of $y=x^3+x^2-x+2$? – Angina Seng Apr 24 '17 at 15:14
• Welcome to MSE. You have to show your own work in your questions in order to get answers. – Edu Apr 24 '17 at 15:15
• The issue is I don't know where to start. How do I find three roots with a cubic function when it doesn't factorize – K Jane Apr 24 '17 at 15:21
• It's worth emphasizing that this is a problem that is fairly simple if one knows a little calculus (as per Zain's answer below). If you want a solution that doesn't use calculus, then the matter is not nearly so obvious. – Semiclassical Apr 24 '17 at 15:23
• If you'd like to avoid using calculus, take note that a 3rd degree polynomial can have at most 3 roots. Play around with input values for the function $f(x) = x^3+x^2-x+2-k$ and exploit Intermediate value theorem – infinitylord Apr 24 '17 at 15:27

## 2 Answers

Note that the $x$-coordinates of the stationary points satisfy $3x^2 + 2x - 1 = 0$ which factorises as $(3x-1)(x+1) =0$.

So the stationary points are $(-1, 3)$ (a local maxima) and $(1/3, 49/27)$ (a local minimum). The nature of these stationary points are easily found by examining the sign of the derivative around it or looking at the second derivative.

A quick sketch now show that the horizontal line $y=k$ will only intersect the curve thrice if it lies strictly between the two extrema. So $\frac{49}{27} < k < 3$.

We can sketch the graph $y=x^3+x^2-x+2$ as below (I used WolframAlpha): Now we imagine drawing horizontal lines across this graph (setting $y=k$ in the original equation)

We can see that the only points where the line crosses the equation exactly three times are any points where $$y\text{ value of minimum point}<k<y\text{ value of maximum point}$$

Therefore, to solve the problem, we must calculate the turning points of the graph $y=x^3+x^2-x+2$

We do this by computing the derivative first:

$$y'=3x^2+2x-1$$

We then set this equal to zero to find turning points of the graph

\begin{align}3x^2+2x-1&=0\\ (x+1)(3x-1)&=0\end{align}

Therefore we have turning points at $x=-1$ and $x=\frac 13$

We can then find the $y$ co-ordinates of these points by substiuting them back into the equation:

When $x=-1$, \begin{align}y&=(-1)^3+(-1)^2-(-1)+2\\ &=-1+1+1+2\\ &=3\end{align}

When $x=\frac 13$, \begin{align}y&=\left(\frac13\right)^3+\left(\frac13\right)^2-\frac13+2\\ &=\frac 1{27}+\frac 19-\frac 13+2\\ &=\frac{49}{27}\end{align}

We know that $\frac{49}{27}<3$ and therefore we can say that $$\frac{49}{27}<k<3$$