The value of $a$ , for which
$(a^{2} + 4 a + 3) x^{2} + (a^{2} - a - 2) x + a (a + 1) = 0$ is satisfied by more than two distinct values of $x$ , is

1

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- 1

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none of these

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Solution

The correct option is C $- 1$
If any quadratic equation $a x^{2} + b x + c$ has
more then two solutions, it becomes
an identity and has a = b = c = 0
$\Rightarrow a^{2} + 4 a + 3 = 0 \Rightarrow a^{2} + 3 a + a + 3 = 0 \Rightarrow a (a + 3) + (a + 3) = 0$
$\Rightarrow (a + 3) (a + 1) = 0 \Rightarrow a = - 1, - 3$
And $a^{2} - a - 2 = 0 \Rightarrow a^{2} - 2 a + a - 2 \Rightarrow a (a - 2) + (a - 2) = 0$
$\Rightarrow (a + 1) (a - 2) = 0 \Rightarrow a = - 1, 2$
and $a (a + 1) = 0 \Rightarrow a = 0, - 1$
only common solution is $a = - 1 \Rightarrow (c)$

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