Animated Solution for Mathematics - Quadratic Equations: The smallest positive integral value of $a$, for which all the roots of $x^4-a{x}^2+9=0$ are real and distinct, is equal to

We are examining the biquadratic equation: Our goal is to determine the smallest positive integer $a$ such that this equation possesses four distinct real roots.

Since the equation contains only even powers of $x$, we employ the substitution $t=x^2$. This transforms the fourth-degree polynomial into a quadratic equation:

For the original equation to have four distinct real roots, the quadratic $f(t)$ must yield two distinct, positive values for $t$. If $t$ were negative, $x$ would be imaginary; if $t$ were zero, we would not have four distinct roots.

To ensure $f(t)=0$ has two distinct positive roots, we must satisfy three specific conditions:

1. The Discriminant ($D>0$): For the roots to be real and distinct, the discriminant must be strictly positive. Setting $a^2-36>0$ results in $(a-6)(a+6)>0$, which implies $a\in (-\infty ,-6)\cup (6,\infty )$.

2. The Product of Roots: By Vieta's formulas, the product of the roots is: Since $9>0$, the product is inherently positive, satisfying the requirement for roots to have the same sign.

3. The Sum of Roots: For both roots to be positive, their sum must also be positive. Thus, we must satisfy the condition $a>0$.

We now combine our constraints: $a>6$ (or $a<-6$) from the discriminant, and $a>0$ from the sum of the roots. The intersection of these conditions yields the requirement:

We seek the smallest positive integer $a$ that satisfies $a>6$. The set of integers greater than 6 is $\{7,8,9,\dots \}$.

Therefore, the smallest positive integer is 7.