Find the relation between $q$ and $r$ in order that the equation $x^{3} + q x + r = 0$ may be put into the form $x^{4} = (x^{2} + a x + b)^{2}$ .
Hence solve the equation $8 x^{3} - 36 x + 27 = 0$ .

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Solution

Given equation, $x^{3} + q x + r = 0$

$x^{4} = (x^{2} + a x + b)^{2} ⟹ x^{3} + (\frac{a^{2} + 2 b}{2 a}) x^{2} + b x + \frac{b^{2}}{2 a} = 0$

Comparing the coefficients of the 2 equations, we have the following relations

$a = \frac{q^{2}}{2 r}; b = q; q^{3} + 8 r^{2} = 0$

Hence, the equation $8 x^{3} - 36 x^{2} + 27 = 0$ , can be written as:
$x^{4} = {(x^{2} + 3 x - \frac{9}{2})}^{2}$
$⟹ (x^{2} - x^{2} - 3 x + \frac{9}{2}) (x^{2} + x^{2} + 3 x - \frac{9}{2}) = 0$
$⟹ (3 x - \frac{9}{2}) = 0 or (4 x^{2} + 6 x - 9) = 0$
$⟹ x = \frac{3}{2}, \frac{- 3 \pm 3 \sqrt{5}}{4}$

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