Quadratic Equation and Inequalities Part 2

Concept:Put $x^2=t$ to reduce the quartic to a quadratic in $t$.For all four roots of the original equation to be real and distinct, this quadratic must have two distinct positive roots.Explanation:Let $x^2=t$, then $x^4-ax^2+9=0$ becomes $t^2-at+9=0$.If its roots are $t_1,t_2$, the roots of the original equation are $\pm\sqrt{t_1}$ and $\pm\sqrt{t_2}$.These four numbers are real and distinct exactly when $t_1,t_2$ are positive and $t_1\ne t_2$.Distinct real roots of the $t$-quadratic require $\Delta=(-a)^2-4(1)(9)=a^2-36>0$, i.e. $a<-6$ or $a>6$.Positivity requires $t_1+t_2=a>0$ and $t_1t_2=9>0$; the product condition holds automatically.Combining both conditions gives $a>6$.Indeed for $a>6$, $t=\frac{a\pm\sqrt{a^2-36}}{2}$ are both positive since $\frac{a-\sqrt{a^2-36}}{2}>0$ as $a>\sqrt{a^2-36}$.Hence the smallest positive integral value of $a$ is $7$.Answer:Option A ($7$).