Let (a-3)x^2+12x+(a+6)0, forall x in R and a | vedclass

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(C) For the quadratic expression $f(x) = (a-3)x^2 + 12x + (a+6)$ to be positive for all $x \in R$,we must have the coefficient of $x^2$ be positive an

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$-1, -2$

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(C) For the quadratic expression $f(x) = (a-3)x^2 + 12x + (a+6)$ to be positive for all $x \in R$,we must have the coefficient of $x^2$ be positive and the discriminant be negative.$1$. $a-3 > 0 \implies a > 3$.$2$. $D = 12^2 - 4(a-3)(a+6) $144 - 4(a^2 + 3a - 18) $36 - (a^2 + 3a - 18) $36 - a^2 - 3a + 18 $-a^2 - 3a + 54 $a^2 + 3a - 54 > 0$$(a+9)(a-6) > 0$.Since $a > 3$,the condition $a > 6$ must hold. Thus,$a \in (6, \infty)$,so $\ell = 6$.The least positive integral value of $a$ is $\alpha = 7$.Now,substitute $\alpha = 7$ and $\ell = 6$ into the equation $(\alpha-3)x^2 + 12x + (\ell+2) = 0$:$(7-3)x^2 + 12x + (6+2) = 0$$4x^2 + 12x + 8 = 0$Divide by $4$: $x^2 + 3x + 2 = 0$$(x+1)(x+2) = 0$.The roots are $x = -1, -2$.

Let $(a-3)x^2+12x+(a+6)>0, \forall x \in R$ and $a \in (\ell, \infty)$. If $\alpha$ is the least positive integral value of $a$,then the roots of $(\alpha-3)x^2+12x+(\ell+2)=0$ are

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