Solve the equation $|2x^2+x-1|=|x^2+4x+1|$ Find the sum of all the solutions of the equation  $|2x^2+x-1|=|x^2+4x+1|$
Though I tried to solve it in desmos.com and getting the requisite answer but while solving it manually it is getting very lengthy. 
I tried to construct the two parabola and mirror image the region below y axis but still getting it is getting complicated. 

Is there any easy method to solve it and get the sum of all the solutions ?

 A: We are asked for the sum of the roots; we don't necessarily have to find the roots.
Squaring, we get $(2x^2+x-1)^2=(x^2+4x-1)^2$.
So $4x^4+4x^3...=x^4+8x^3...$.
So $\color{blue}3x^4-\color{blue}4x^3....=0$.
By Vieta's formulas, the answer is $\dfrac43$.
A: Hint : Solve the two equations $$2x^2+x-1=x^2+4x+1$$ and $$2x^2+x-1=-x^2-4x-1$$
A: The expressions between the absolute value bars have the same or opposite signs. Hence there are two independent cases (by addition and subtraction):
$$3x^2+5x=0$$ and $$x^2-3x-2=0.$$
Then by Vieta,
$$-\frac53+3.$$

For complete rigor, one should show that no root is repeated. This is true, because the polynomials have no double root, and their $\text{gcd}$ is $1$.
A: $$
|2x^2+x-1|=|x^2+4x+1|\\
(2x^2+x-1)^2=(x^2+4x+1)^2\\
(2x^2+x-1)^2-(x^2+4x+1)^2=0\\
[(2x^2+x-1)+(x^2+4x+1)]\cdot[(2x^2+x-1)-(x^2+4x+1)]=0\\
(3x^2+5x)\cdot(x^2-3x-2)=0\\
x\cdot(3x+5)\cdot(x^2-3x-2)=0\\
Solving\space for\space all\space cases,\space we\space get:\\
x=0\\
x=-\frac{5}{3}\\
x=\frac{3\pm\sqrt{17}}{2}
$$
A: Squaring and factorizing we get
$$x(3x+5)(x^2-3x-2)=0$$
The solutions are given by $$x=-\frac{5}{3}\lor x=0\lor x=\frac{1}{2} \left(3-\sqrt{17}\right)\lor
   x=\frac{1}{2} \left(3+\sqrt{17}\right)$$
