# How can I prove this equation has no solution?

Solve the equation $$-x^3 + x + 2 =\sqrt{3x^2 + 4x + 5.}$$ I tried. The equation equavalent to $$\sqrt{3x^2 + 4x + 5} - 2 + x^3 - x=0.$$ $$\dfrac{3x^2+4x+1}{\sqrt{3x^2 + 4x + 5} + 2}+x^3 - x=0.$$ $$\dfrac{(x+1)(3x+1)}{\sqrt{3x^2 + 4x + 5} + 2}+ (x+1) x (x-1)=0.$$ $$(x+1)\left [\dfrac{3x+1}{\sqrt{3x^2 + 4x + 5} + 2}+ x (x-1)=0\right ]=0.$$ How can I prove the equation $$\dfrac{3x+1}{\sqrt{3x^2 + 4x + 5} + 2}+ x (x-1)=0$$ has no solution?

• @Deepak Please see the last equation in my tried. Mar 9, 2016 at 23:53
• Oh i see what you're doing. Never mind. Mar 9, 2016 at 23:57
• While not formal, I always tend to get my graphing tool and graph both sides as functions. In this case it would be clear... Mar 9, 2016 at 23:57
• @imranfat Using graph is not a solution. Mar 10, 2016 at 0:01
• @minthao_2011 True, like I said, it is a rather informal approach, however, IF the curves intersect, what does that mean? It's a start... Mar 10, 2016 at 0:02

I would start by squaring both sides of the equation:

$$(-x^2 + x +2)^2 = 3x^2 +4x +5$$

$$x^6 -2x^4-4x^3-2x^2-1 =0$$

As suggested by Deepak, $x = -1$ is a solution. You can factorise fully by dividing the above polynomial by $x+1$ to obtain other factors and solutions (if any).

• OK. It is not simple $(x+1) \left(x^5-x^4-x^3-3 x^2+x-1\right)=0$ Mar 9, 2016 at 23:58
• How can I prove the equation $x^5-x^4-x^3-3 x^2+x-1=0$ has no solution? Mar 10, 2016 at 0:01
• Yes, it appears that polynomial has no real roots. Mar 10, 2016 at 0:04
• I think It is not simple. Can you prove it? Mar 10, 2016 at 0:05
• You may want to see this post for more-math.stackexchange.com/questions/529227/…. Mar 10, 2016 at 0:06

hint: edit 2: when $x>1$, there is no solution, when $-\dfrac{1}{3} \le x \le 0$, there is no solution also.

when $0\le x\le 1, x(1-x) < f(x)=\dfrac{x}{3}+\dfrac{1}{2+\sqrt{5}}$

and $\dfrac{3x+1}{\sqrt{3x^2 + 4x + 5} + 2}\ge \dfrac{x}{3}+\dfrac{1}{2+\sqrt{5}}$

so only possible is $x <-\dfrac{1}{3}$

now there is a $g(x)=\dfrac{4}{3}(3x+1)$

you need to prove $\dfrac{3x+1}{\sqrt{3x^2 + 4x + 5} + 2}> g(x) >x(1-x)$ which is easy.

• Thank you very much. You have a good idea. I change by $g(x) = \dfrac{1}{2}(3x+1)$. And a new prolem is if $x\in (0,1)$, we have not yet prove the equation has no solution. Please try again. Mar 10, 2016 at 3:38
• @minthao_2011 ya, sorry for the mistake. I add new part. Mar 10, 2016 at 5:40

This is not an answer since based on approximations.

Consider the function $$f(x)=\frac{-x^3 + x + 2 -\sqrt{3x^2 + 4x + 5}}{x+1}$$ Around $x=0$ it looks as a parabola; so the function can be approximated by a Taylor series to third order. So, around $x=0$, we have $$f(x)\approx \left(2-\sqrt{5}\right)+\left(\frac{3}{\sqrt{5}}-1\right) x+\left(1-\frac{41}{10 \sqrt{5}}\right) x^2+O\left(x^3\right)$$ So, if the approximating function is $$g(x)= \left(2-\sqrt{5}\right)+\left(\frac{3}{\sqrt{5}}-1\right) x+\left(1-\frac{41}{10 \sqrt{5}}\right) x^2$$The quadratic does not show any real root and then $g(x)$ shows a maximum for $x_*=-\frac{5 \left(11 \sqrt{5}-73\right)}{1181}\approx 0.204925$ leading to $$g(x_*)=-\frac{9 \left(116 \sqrt{5}-233\right)}{1181}\approx -0.201063$$ which corresponds to the maximum. Using the full definition of $f(x)$, we should find $$f(x_*)\approx -0.200890$$

Numerically, the maximum value of $f(x)$ is found as $\approx -0.200889$ corresponding to $x=0.206212$.

Since the maximum value of $f(x)<0$, no other roots beside $x=-1$.