Based on pure intuition, the root is 0. It’s very obvious. But what is the proper way to solve it? Also how do we know it’s the only possible root (the answer is 1 root only, but how can I confirm?)
Asked
Active
Viewed 322 times
3
-
4Same answer as: number of positive solutions of $X^4+X^3-4 X^2+X+1 = 0$. Obviously $X=1$ is a positive solution. What about others? Do you know Descartes' Rule of Signs? – GEdgar Jan 10 '20 at 13:18
-
@GEdgar I have heard of it, but my knowledge doesn’t extend that far at the current stage – Aditya Jan 10 '20 at 13:24
-
1In general problems like this, you will need Descartes. But here you do not. After you factor out all rational roots, it will be obvious how many positive roots remain. – GEdgar Jan 10 '20 at 13:26
2 Answers
8
as @GEdgar said here we have:
$$x^{4}+x^{3}-4x^{2}+x+1=\left(x-1\right)^{2}\left(x^{2}+3x+1\right)$$
I just divided $x^{4}+x^{3}-4x^{2}+x+1$ by $x-1$
so either $(x-1)=0$ or $\left(x^{2}+3x+1\right)=0$
both roots of the second one are negative hence the only root is $x=0$
also consider that $e^x$ is always positive for all real $x$.
1
Set $e^x=y>0$
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
As $y\ne0$
Divide both sides by $y^{4/2}$
$$0=y^2+y-4+\dfrac1y+\dfrac1{y^2}$$
$$\implies\left(y+\dfrac1y\right)^2+\left(y+\dfrac1y\right)-6=0$$
Now replace $y+\dfrac1y=u$ to form a quadratic equation in $u$
and use AM-GM inequality to find $u\ge2$ for real$x$
Can you take it from here?
lab bhattacharjee
- 274,582