Solve the equation $24 x^{3} - 14 x^{2} - 63 x + 45 = 0$ , one root being double another.

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Solution

Denote the roots by $a, 2 a, b$ ; then we have
$3 a + b = \frac{7}{12}, 2 a^{2} + 3 a b = - \frac{21}{8}, 2 a^{2} b = - \frac{15}{8}$ .
From the first two equations, we obtain
$8 a^{2} - 2 a - 3 = 0$
$∴ a = \frac{3}{4}$ or $- \frac{1}{2}$ and $b = - \frac{5}{3}$ or $\frac{25}{12}$ .
It will be found on trial that the values $a = - \frac{1}{2}, b = \frac{25}{12}$ do not satisfy the third equation $2 a^{2} b = - \frac{15}{8}$ ;
hence we are restricted to the values
$a = \frac{3}{4}, b = - \frac{5}{3}$ .
Thus the roots are $\frac{3}{4}, \frac{3}{2}, - \frac{5}{3}$ .

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