Quadratic Equation #9

Put $x=\frac{1}{x}$ in the equation $x^2+2x+3=0$ we get $3x^2+2x+1=0$ which is same as the first equation.

So, $a+b+c=3+2+1=\boxed{6}$

Let the roots of $\color{#3D99F6}{x^2+2x+3\;be\;e,d}$ .Then the quadratic $\color{#20A900}{ax^2+bx+c}$ has roots $\color{#20A900}{\frac{1}{e},\frac{1}{d}}$ .So the quadratic is $\color{#3D99F6}{\left(x-\frac{1}{e}\right)\left(x-\frac{1}{d}\right)=x^2-\left(\frac{1}{e}+\frac{1}{d}\right)x+\frac{1}{ed}=x^2-\left(\frac{e+d}{ed}\right)x+\frac{1}{ed}}$ .From Vieta's formulas we know that $\color{#20A900}{e+d=-2,ed=3}$ .So the equation becomes $\color{#69047E}{x^2+\frac{2}{3}x+\frac{1}{3}=0\rightarrow 3x^2+2x+1=0\rightarrow a=3,b=2,c=1\rightarrow a+b+c=3+2+1=\boxed{6}}$ .Sujoy Roy's solution is the more elegant one and here is the proof.

Just reverse the coefficients.