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

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Solution

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

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

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

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