If $3 b = a + c$ , the roots of $a x^{2} + 3 b x + c = 0$ are

1, \frac{c}{a}

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- 1, \frac{c}{a}

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- 1, - \frac{c}{a}

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1, - \frac{c}{a}

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Solution

The correct option is C $- 1, - \frac{c}{a}$
$a x^{2} + 3 b x + c = 0$
$x = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A}$
$= \frac{- 3 b \pm \sqrt{9 b^{2} - 4 a c}}{2 a}$
Since
$3 b = a + c$
Hence
$9 b^{2} = (a + c)^{2}$
Therefore
$\frac{- 3 b \pm \sqrt{9 b^{2} - 4 a c}}{2 a}$
$= \frac{- (a + c) \pm \sqrt{(a + c)^{2} - 4 a c}}{2 a}$
$= \frac{- (a + c) \pm \sqrt{(a - c)^{2}}}{2 a}$
$= \frac{- (a + c) \pm (a - c)}{2 a}$
$x = \frac{- a - c + a - c}{2 a}$ and $x = \frac{- a - c - a + c}{2 a}$
$x = \frac{- c}{a}$ and $x = - 1$

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