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

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.

• where did you get this???????? – Will Jagy Jun 10 '15 at 1:58
• $x=0 , x+2 \neq {(-3x²-7x+5)}^{(1/3)}$ .they are not the same . – zeraoulia rafik Jun 10 '15 at 2:06
• $x = -3-2 \sqrt{2}$ seems to fit too. – Alexey Burdin Jun 10 '15 at 2:08
• 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 – zeraoulia rafik Jun 10 '15 at 2:10
• 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 $x_{2}=-3.99+3.26i$ ,$x_{3}=-3.99 -3.26i$ – 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 $P(x)=x^6+12x^5+68x^4+187x^3+295x^2+159x+40$
$P(x)=\left(x^3+6x^2+16x-\frac{5}{2}\right)^2+69\left(x+\frac{40}{69}\right)^2+\frac{2915}{276}$
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 : $\sqrt[3]{-1}=-1$ : $$\sqrt[3]{-3x^2-7x+5}=-\sqrt[3]{3x^2+7x-5}$$