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

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Solution

$(a^{2} - 9) x^{2} + (a^{2} - 5 a + 6) x + (a^{2} - 3 a + 2) = 0$
$(a - 3) (a + 3) x^{2} + (a^{2} - 3 a - 2 a + 6) x + (a^{2} - 2 a - a + 2) = 0$
$\Rightarrow (a - 3) (a + 3) x^{2} + (a - 3) (a - 2) x + (a - 2) (a - 1) = 0$ as it is an identity.
$\Rightarrow (a - 3) (a + 3) = 0 \Rightarrow a = 3, - 3$
$\Rightarrow (a - 3) (a - 2) = 0 \Rightarrow a = 3, 2$
$\Rightarrow (a - 2) (a - 1) = 0 \Rightarrow a = 2, 1$
So, ${3, - 3} \cap {3, 2} \cap {2, 1} = ϕ$
Hence there is no solution.

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If equation (a2−9)x2+(a2−5a+6)x+(a2−3a+2)=0 has more than two roots then a is

(a2−9)x2+(a2−5a+6)x+(a2−3a+2)=0(a−3)(a+3)x2+(a2−3a−2a+6)x+(a2−2a−a+2)=0⇒(a−3)(a+3)x2+(a−3)(a−2)x+(a−2)(a−1)=0 as it is an identity.⇒(a−3)(a+3)=0⇒a=3,−3⇒(a−3)(a−2)=0⇒a=3,2⇒(a−2)(a−1)=0⇒a=2,1So,{3,−3}∩{3,2}∩{2,1}=ϕHence there is no solution.

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