Solve the following equation:

$x^{3} - 13 x^{2} + 15 x + 189 = 0$ if one root exceeds other by $2$ . Find its smallest root?

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Solution

Let the roots be $α, α + 2, β$
$S_{1} = Σ α = 2 α + β + 2 = 13 \Rightarrow β = 11 - 2 α . . . (1)$

$S_{2} = Σ α β = α (α + 2) + (α + 2) β + β α \Rightarrow α^{2} + 2 α + 2 (α + 1) β = 15 . . . (2)$

$S_{3} = α β γ \Rightarrow α β (α + 2) = - 189$

Eliminating $β$ between (1) and (2), we get
$α^{2} + 2 α + 2 (α + 1) (11 - 2 α) = 15 \Rightarrow 3 α^{2} - 20 α - 7 = 0$
$\Rightarrow (α - 7) (3 α + 1) = 0 \Rightarrow α = 7, - \frac{1}{3} \Rightarrow β = - 3, \frac{35}{3}$
Out of these values $α = 7, β = - 3$ satisfy the third relation $α β (α + 2) = - 189 \Rightarrow (- 21) (9) = - 189$
Hence the roots are $7, 7 + 2, - 3 \Rightarrow 7, 9, - 3.$

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