Question

Number of solutions of the equations $|2x^2+x-1|=|x^2+4x+1|$

Answer 1

$2x^2+x-1>0$ for $x\in (-\infty ,-1)\cup (1/2,\infty )$

and $x^2+4x+1>0$ for $x\in (-\infty ,-(2+\sqrt{3}\left)\right)\cup \left(\right(-2+\sqrt{3}),\infty ))$

Case 1:

If $-\infty <x<-(2+\sqrt{3})$, then

$2x^2+x-1=x^2+4x+1⟹x^2-3x-2=0$

$x=\frac{3\pm \sqrt{17}}{2}$, do not satisfy the condition for case 1.

Case 2:

$-(2+\sqrt{3})<x<-1$

$2x^2+x-1=-(x^2+4x+1)⟹3x^2+5x=0$

$x=0,-5/2$, do not satisfy the condition for case 2.

Case 3:

$-1<x<-2+\sqrt{3}$

$-(2x^2+x-1)=-(x^2+4x+1)⟹x^2-3x-2=0$

$x=\frac{3-\sqrt{17}}{2}$, satisfies the condition for case 3.

Case 4:

$-2+\sqrt{3}<x<1/2$

$-(2x^2+x-1)=x^2+4x+1⟹3x^2+5x=0$

$x=0$, satisfies the condition for case 4.

Case 5 :

$x>1/2$

$2x^2+x-1=x^2+4x+1⟹x^2-3x-2=0$

$x=\frac{3+\sqrt{17}}{2}$, satisfies the condition for case 5.

$3$ solutions are possible for $x$.