If equation $x^{4} - (3 m + 2) x^{2} + m^{2} = 0 (m > 0)$ has four real solution which are in A.P. Then the value of m

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Solution

$x^{4} - (3 m + 2) x^{2} + m^{2} = 0$

Let the roots be,

$β + d, β - d, β + 3 d, β - 3 d$

$x^{4} + 0 x^{2} - (3 m + 2) x^{2} + 0 x + m^{2} = 0$

$\Rightarrow 0 = (β - 3 d) + (β - d) + (β + 3 d) + (β + d) = 4 β$

$\Rightarrow β = 0$

$- (3 m + 2) = (- d) (d) + (- d) (3 d) + (- d) (- 3 d) + (d) (3 d) + (d) (- 3 d) + (3 d) (- 3 d)$

$\Rightarrow - (3 m + 2) = - d^{2} - 3 d^{2} + 3 d^{2} + 3 d^{2} - 3 d^{2} - 9 d^{2} = - 10 d^{2}$

$\Rightarrow \frac{3 m + 2}{10} = d^{2}$

$m^{2} = (- 3 d) (3 d) (- d) (d) = 9 d^{4}$

$\Rightarrow m^{2} = 9 \frac{(3 m + 2)^{2}}{100}$

$\Rightarrow 100 m^{2} = 9 (9 m^{2} + 12 m + 4)$

$\Rightarrow 0 = - 10 m^{2} + 108 m + 36$

$\Rightarrow m = 6, m = - \frac{6}{19}$

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