If equation $(a^{2} - 4) x^{2} + (a^{2} - 6 a + 8) x + (a^{2} - 3 a + 2) = 0$ has more than two roots then $a$ is

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Solution

$(a^{2} - 4) x^{2} + (a^{2} - 6 a + 8) x + (a^{2} - 3 a + 2) = 0$
$(a - 2) (a + 2) x^{2} + (a^{2} - 4 a - 2 a + 8) x + (a^{2} - 2 a - a + 2) = 0$
$\Rightarrow (a - 2) (a + 2) x^{2} + (a - 4) (a - 2) x + (a - 2) (a - 1) = 0$ as it is an identity.
$\Rightarrow (a - 2) (a + 2) = 0 \Rightarrow a = 2, - 2$
$\Rightarrow (a - 4) (a - 2) = 0 \Rightarrow a = 4, 2$
$\Rightarrow (a - 2) (a - 1) = 0 \Rightarrow a = 2, 1$
So, ${2, - 2} \cap {4, 2} \cap {2, 1} = 2$
So $a = 2$

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