If equation $(a - 1) x^{2} + (a^{2} - 2 a + 1) x + (a^{2} - 1) = 0$ has more than two roots then $a$ is

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Solution

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

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If equation (a−1)x2+(a2−2a+1)x+(a2–1)=0 has more than two roots then a is

(a−1)x2+(a2−2a+1)x+(a2–1)=0(a−1)x2+(a2−a−a+1)x+(a−1)(a+1)=0⇒(a−1)x2+(a−1)(a−1)x+(a−1)(a+1)=0 as it is an identity.⇒a−1=0⇒a=1⇒(a−1)(a−1)=0⇒a=1,1⇒(a−1)(a+1)=0⇒a=−1,1So,{1}∩{1}∩{−1,1}=1So a=1

If equation $(a - 1) x^{2} + (a^{2} - 2 a + 1) x + (a^{2} - 1) = 0$ has more than two roots then $a$ is