# Number of real solutions to $x^7 + 2x^5 + 3x^3 + 4x = 2018$

Find the number of real solutions of $$x^7 + 2x^5 + 3x^3 + 4x = 2018$$?

What is the general approach to solving this kind of questions? I am interested in the thought process.

Few of my thoughts after seeing this question: since $$x$$ has all odd powers so, it can not have any negative solution. 2018 is semiprime; not much progress here. We can sketch the curve but graphing a seven order polynomial is difficult.

• Do you know differentiation? Apr 15, 2019 at 17:39
• yes, I know @badjohn Apr 15, 2019 at 17:42
• – lhf
Apr 15, 2019 at 17:42
• So, differentiate it and have a look at the result. Apr 15, 2019 at 17:45

If $$x\le 0$$ the left hand side is negative therefore no solution. We suppose $$x>0$$ and we consider $$f(x)=x^7+2x^5+3x^3+4x$$, then $$f$$ is the sum of increasing functions therefore increasing. Since $$f(0,\infty)=(0,\infty)$$ this equation has only one solution.

This can be done by differentiation which give a more simple proof since the derivative is clearly positive.

Render

$$x^7+0x^6+2x^5+0x^4+3x^3+0x^2+4x-2018=0$$

Descartes' Rule of Signs forces exactly one positive root and no negative roots. That's it!