The question says find all values for $m$, such that the below equation has 4 real roots in Arithmetic Progression.
My approach:
\begin{gather}
x^4-(3m+2)x^2+m^2 = 0 \\
\\
\text{Let roots be }\\
\beta+d \\
\beta-d \\
\beta+2d \\
\beta-2d \\
\\
x^4+0x^2-(3m+2)x^2+0x+m^2 = 0\\
\implies 0 = (\beta-2d) + (\beta-d) + (\beta+2d) + (\beta+d) = 4\beta \\
\implies \beta = 0\\
\\
-(3m+2) = (-d)(d)+(-d)(2d)+(-d)(-2d)+(d)(2d)+(d)(-2d)+(2d)(-2d)\\
\implies -(3m+2) = -d^2 -2d^2 + 2d^2 +2d^2 -2d^2 -4d^2 = -5d^2\\
\implies \frac{3m+2}{5} = d^2 \\
\\
m^2 = (-2d)(2d)(-d)(d) = 4d^4\\
\implies m^2 = 4\cdot\frac{(3m+2)^2}{25}\\
\implies 25m^2 = 4\cdot(9m^2 +12m + 4) \\
\implies 0 = 36m^2 -25m^2 +48m + 16 \\
\implies 0 = 11m^2+48m+16 \\
\implies m = -4 \text{ or } m = -\frac{4}{11} \\
\end{gather}
However, when you plug the values back into the equation, you get complex roots for $m = -4$ and only 3 real roots for $m = -\frac{4}{11}$.
I am stuck here, I cannot see how to move forward. I also do not know if what I did was correct or not.
Edit: I am stupid and it is $\beta -3d$ and $\beta + 3d$.
Best Answer
In case others need help in a similar question, I am putting up the solution.
\begin{gather} x^4-(3m+2)x^2+m^2 = 0 \\ \\ \text{Let roots be }\\ \beta+d \\ \beta-d \\ \beta+3d \\ \beta-3d \\ \\ x^4+0x^2-(3m+2)x^2+0x+m^2 = 0\\ \implies 0 = (\beta-3d) + (\beta-d) + (\beta+3d) + (\beta+d) = 4\beta \\ \implies \beta = 0\\ \\ -(3m+2) = (-d)(d)+(-d)(3d)+(-d)(-3d)+(d)(3d)+(d)(-3d)+(3d)(-3d)\\ \implies -(3m+2) = -d^2 -3d^2 + 3d^2 +3d^2 -3d^2 -9d^2 = -10d^2\\ \implies \frac{3m+2}{10} = d^2 \\ \\ m^2 = (-3d)(3d)(-d)(d) = 9d^4\\ \implies m^2 = 9\cdot\frac{(3m+2)^2}{100}\\ \implies 100m^2 = 9\cdot(9m^2 +12m + 4) \\ \implies 0 = -19m^2 + 108m + 36 \\ \implies m = 6 \text{ or } m = -\frac{6}{19} \\ \end{gather}