If equation $(a - 3) x^{2} + (a^{2} - 5 a + 6) x + (a^{2} - 9) = 0$ has more than two roots then $a$ is

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Solution

$(a - 3) x^{2} + (a^{2} - 5 a + 6) x + (a^{2} - 9) = 0$
$\Rightarrow (a - 3) x^{2} + (a^{2} - 3 a - 2 a + 6) x + (a - 3) (a + 3) = 0$
$\Rightarrow (a - 3) x^{2} + (a - 3) (a - 2) x + (a - 3) (a + 3) = 0$ as it is an identity.
$\Rightarrow a - 3 = 0 \Rightarrow a = 3$
$\Rightarrow (a - 3) (a - 2) = 0 \Rightarrow a = 3, 2$
$\Rightarrow (a - 3) (a + 3) = 0 \Rightarrow a = - 3, 3$
So, ${3} \cap {3, 2} \cap {- 3, 3} = 3$
So $a = 3$

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If equation (a–3)x2+(a2−5a+6)x+(a2–9)=0 has more than two roots then a is

(a–3)x2+(a2−5a+6)x+(a2–9)=0⇒(a−3)x2+(a2−3a−2a+6)x+(a−3)(a+3)=0⇒(a−3)x2+(a−3)(a−2)x+(a−3)(a+3)=0 as it is an identity.⇒a−3=0⇒a=3⇒(a−3)(a−2)=0⇒a=3,2⇒(a−3)(a+3)=0⇒a=−3,3So,{3}∩{3,2}∩{−3,3}=3So a=3

If equation $(a - 3) x^{2} + (a^{2} - 5 a + 6) x + (a^{2} - 9) = 0$ has more than two roots then $a$ is