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?
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$\begingroup$ @Deepak Please see the last equation in my tried. $\endgroup$– minthao_2011Mar 9, 2016 at 23:53
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$\begingroup$ Oh i see what you're doing. Never mind. $\endgroup$– DeepakMar 9, 2016 at 23:57
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$\begingroup$ While not formal, I always tend to get my graphing tool and graph both sides as functions. In this case it would be clear... $\endgroup$– imranfatMar 9, 2016 at 23:57
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$\begingroup$ @imranfat Using graph is not a solution. $\endgroup$– minthao_2011Mar 10, 2016 at 0:01
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$\begingroup$ @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... $\endgroup$– imranfatMar 10, 2016 at 0:02
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3 Answers
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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).
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$\begingroup$ OK. It is not simple $(x+1) \left(x^5-x^4-x^3-3 x^2+x-1\right)=0$ $\endgroup$ Mar 9, 2016 at 23:58
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$\begingroup$ How can I prove the equation $x^5-x^4-x^3-3 x^2+x-1=0$ has no solution? $\endgroup$ Mar 10, 2016 at 0:01
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$\begingroup$ Yes, it appears that polynomial has no real roots. $\endgroup$ Mar 10, 2016 at 0:04
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$\begingroup$ I think It is not simple. Can you prove it? $\endgroup$ Mar 10, 2016 at 0:05
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$\begingroup$ You may want to see this post for more-math.stackexchange.com/questions/529227/…. $\endgroup$ Mar 10, 2016 at 0:06
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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.
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$\begingroup$ 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. $\endgroup$ Mar 10, 2016 at 3:38
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$\begingroup$ @minthao_2011 ya, sorry for the mistake. I add new part. $\endgroup$– chenbaiMar 10, 2016 at 5:40
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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 $x=-1$.
answered Mar 10, 2016 at 4:50