Solve the equation: $x^3+7x^2+16x+5=(1-2x)\sqrt[3]{-3x^2-7x+5}$

I used wolframalpha.com and get the solution: $x\in\{-3;2\sqrt2-3\}$

When $x=-3$, $\sqrt[3]{-3x^2-7x+5}=-1$

When $x=2\sqrt2-3$, $\sqrt[3]{-3x^2-7x+5}=2\sqrt2-1$

So I guess we can prove that: $x+2=\sqrt[3]{-3x^2-7x+5}$

I tried to use function method (use function's monotonous) but didn't get any result.

| cite | improve this question | | | | |
  • 2
    $\begingroup$ where did you get this???????? $\endgroup$ – Will Jagy Jun 10 '15 at 1:58
  • $\begingroup$ $x=0 , x+2 \neq {(-3x²-7x+5)}^{(1/3)}$ .they are not the same . $\endgroup$ – zeraoulia rafik Jun 10 '15 at 2:06
  • $\begingroup$ $x = -3-2 \sqrt{2}$ seems to fit too. $\endgroup$ – Alexey Burdin Jun 10 '15 at 2:08
  • $\begingroup$ and no theoritical way to solve equation that have deg>=3 , Galois theory ,only numerical method (iteratives methods ) .or only way to make a transformation to your equation as a simple dynamical system with initial conditions $\endgroup$ – zeraoulia rafik Jun 10 '15 at 2:10
  • $\begingroup$ pleas verifie your solutions i don't think that you have 3 real solution, i think you have 1 real solution and 2 complex : $x_{1}=2\sqrt(2) -3$ and $x_{2}=-3.99+3.26i $ ,$x_{3}=-3.99 -3.26i$ $\endgroup$ – zeraoulia rafik Jun 10 '15 at 2:19

$$x^3+7x^2+16x+5=(1-2x)\sqrt[3]{-3x^2-7x+5}$$ The roots of the above equation belong to the set of roots of the next equation : $$(x^3+7x^2+16x+5)^3=(1-2x)^3(-3x^2-7x+5)$$ Expanding and factoring lead to : $$(x+3)(x^2+6x+1)P(x)=0$$ where $P(x)=x^6+12x^5+68x^4+187x^3+295x^2+159x+40$


The three terms are positive. Hense $P(x)>0$ any $x$.

So the real roots of the initial equation could only be among the roots of $(x+3)(x^2+6x+1)=0$ $$\begin{cases} x_1=-3\\ x_2=-3+2\sqrt{2}\\ x_3=-3-2\sqrt{2} \end{cases}$$ Bringing back those possible roots into the initial equation, we observe that the three are convenient. Hense the initial equation has the three above real roots.

Note : This is in considering the real cubic root of $-1$, that is : $\sqrt[3]{-1}=-1$ : $$\sqrt[3]{-3x^2-7x+5}=-\sqrt[3]{3x^2+7x-5}$$

The figure below is the graphical representation of the equation and roots.

enter image description here

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.