If equation $(a^{2} - 4) x^{2} + (a^{2} - 8 a + 16) x + (a^{2} - 3 a + 2) = 0$ has more than two roots then $a$ is

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Solution

$(a^{2} - 4) x^{2} + (a^{2} - 8 a + 16) x + (a^{2} - 3 a + 2) = 0$
$(a - 2) (a + 2) x^{2} + (a^{2} - 4 a - 4 a + 16) x + (a^{2} - 2 a - a + 2) = 0$
$\Rightarrow (a - 2) (a + 2) x^{2} + (a - 4) (a - 4) x + (a - 2) (a - 1) = 0$ as it is an identity.
$\Rightarrow (a - 2) (a + 2) = 0 \Rightarrow a = 2, - 2$
$\Rightarrow (a - 4) (a + 4) = 0 \Rightarrow a = 4, - 4$
$\Rightarrow (a - 2) (a - 1) = 0 \Rightarrow a = 2, 1$
So, ${2, - 2} \cap {4, - 4} \cap {2, 1} = ϕ$
Hence there is no solution.

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If equation (a2−4)x2+(a2−8a+16)x+(a2−3a+2)=0 has more than two roots then a is

If equation $(a^{2} - 4) x^{2} + (a^{2} - 8 a + 16) x + (a^{2} - 3 a + 2) = 0$ has more than two roots then $a$ is