The equation $(a^{2} - a - 2) x^{2} + (a^{2} - 4) x + a^{2} - 3 a + 2 = 0$ will have more than two solutions, if $a$ is equal to

2

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Solution

The correct option is A $2$
Given quadratic equation
$(a^{2} - a - 2) x^{2} + (a^{2} - 4) x + a^{2} - 3 a + 2 = 0$
We know that $A x^{2} + B x + C = 0$ will have more than two solution, if
$A = B = C = 0$
So, $a^{2} - a - 2 = 0$
$\Rightarrow a = 2, - 1$
$a^{2} - 4 = 0$
$\Rightarrow a = - 2, 2$
$a^{2} - 3 a + 2 = 0$
$\Rightarrow (a - 1) (a - 2) = 0$
$\Rightarrow a = 1, 2$
Common value of $a = 2$ .

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