If the equations $a x^{2} + b x + c = 0$ and $x^{3} + 3 x^{2} + 3 x + 2 = 0$ have two common roots, then show that $a = b = c$ .

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Solution

$x^{3} + 3 x^{2} + 3 x + 2 = (a x^{2} + b x + c) (\frac{x}{a} + \frac{2}{c})$
The second factor is so chosen keeping in view the coefficients of $x^{3}$ and constant term.
Comparing coefficients of $x^{2}$ and $x$ ,
$\frac{b}{a} + \frac{2 a}{c} = 3, \frac{c}{a} + \frac{2 b}{c} = 3$
$b c + 2 a^{2} = 3 a c, c^{2} + 2 a b = 3 a c$
$2 a^{2} = c (3 a - b), c (c - 3 a) + 2 b = 0$
Eliminating c, we get
$7 a^{3} - 12 a^{2} b + 6 a b^{2} - b^{3} = 0$
or $(a - b) (7 a^{2} - 5 a b + b^{2}) = 0$
$∴$ $a = b$ (other factor gives imaginary as $Δ < 0)$
Putting in $2 a^{2} = c (3 a - b)$ , we get $a = c$ ,
$∴$ $a = b = c$

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