Related Formula
For a quadratic At² + Bt + C = 0 to have distinct positive roots:$$\text{For a quadratic } At^2 + Bt + C = 0 \text{ to have distinct positive roots:}$$
D > 0, -(B)/(2A) > 0, (C)/(A) > 0$$D > 0, \quad -\frac{B}{2A} > 0, \quad \frac{C}{A} > 0$$
Core Logic
Substitute x² = t$x^2 = t$. The equation becomes a quadratic in t$t$:
t² - at + 9 = 0 (2)$$t^2 - at + 9 = 0 \quad \dots (2)$$
For the original quartic equation x⁴ - ax² + 9 = 0$x^4 - ax^2 + 9 = 0$ to have 4 real and distinct roots, the quadratic equation in t$t$ must have 2 distinct positive real roots (since x = ± √(t)$x = \pm \sqrt{t}$ requires t > 0$t > 0$).
Step 1: Discriminant Condition
Condition 1: Roots must be real and distinct (D > 0$D > 0$)
D = a² - 4(1)(9) > 0$$D = a^2 - 4(1)(9) > 0$$
a² - 36 > 0$a^2 - 36 > 0$
a in (-∞, -6) (6, ∞)$$a \in (-\infty, -6) \cup (6, \infty)$$
Step 2: Location of Roots Condition
Condition 2: Sum of roots must be positive (since both roots are positive)
-(-a)/(1) > 0 a > 0$$-\frac{-a}{1} > 0 \implies a > 0$$
Condition 3: Product of roots must be positive
f(0) > 0 9 > 0 (This is always true, a in R)$$f(0) > 0 \implies 9 > 0 \quad \text{(This is always true, } a \in \mathbb{R}\text{)}$$
Step 3: Intersection and Conclusion
Taking the intersection of all conditions:
a in (-∞, -6) (6, ∞)$a \in (-\infty, -6) \cup (6, \infty)$ AND a > 0$a > 0$.
Intersection yields: a in (6, ∞)$a \in (6, \infty)$.
The smallest positive integral value in this interval is 7$7$.
Pattern Recognition
Bi-quadratic equations x⁴ + Bx² + C = 0$x^4 + Bx^2 + C = 0$ map cleanly to t² + Bt + C = 0$t^2 + Bt + C = 0$. The nature of x$x$ roots depends entirely on the signs of t$t$ roots. 4 real distinct x$x$ roots ≡$\equiv$ 2 positive distinct t$t$ roots.
Chapter Mix
Class 11 Maths: Quadratic Equations
