Question

The smallest positive integral value of $a$, for which all the roots of $x^4-ax^2+9=0$ are real and distinct, is equal to

• A. 3
• B. 9
• C. 7
• D. 4

Hint:

For biquadratic equations, reduce degree first and apply conditions for both the substituted variable and original variable.

Answer 1

The Correct Option is C

To find the smallest positive integral value of \(a\) for which all the roots of the polynomial equation \(x^4-ax^2+9=0\) are real and distinct, follow these steps:

1. The given equation is a quadratic in terms of \(x^2\). Let \(y = x^2\). The equation becomes \(y^2 - ay + 9 = 0\).
2. For all the roots of the original quartic equation to be real, the roots of this quadratic should be non-negative (since \(y = x^2 \geq 0\)).
3. The discriminant of this quadratic equation must be non-negative for real roots. 
   The discriminant of \(y^2 - ay + 9 = 0\) is given by \(a^2 - 4 \cdot 9 = a^2 - 36\). For real and distinct roots, we need:
   • \(a^2 - 36 \gt 0 \implies a^2 \gt 36\)
   • Solving \(a^2 \gt 36\) yields \(a \gt 6\) or \(a \lt -6\).
   • Since we need the smallest positive integral value, consider \(a \gt 6\).
4. The smallest integer greater than 6 is 7. Thus, \(a = 7\).
5. Finally, verify that \(a = 7\) gives distinct and positive roots for \(y\):
   • Substitute \(a = 7\) into the equation \(y^2 - 7y + 9 = 0\).
   • Calculate the discriminant: \(7^2 - 36 = 13 \gt 0\) (ensuring distinct roots).
   • The roots of \(y^2 - 7y + 9 = 0\) are real and given by the quadratic formula:
   • \(y = \frac{7 \pm \sqrt{13}}{2}\), which are positive as required, ensuring x will have real roots.

Thus, the smallest positive integer value of \(a\) that makes all the roots real and distinct is \(7\).