Question

If the equation \[ x^4-ax^2+9=0 \] has four real and distinct roots, then the least possible integral value of $a$ is

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

For biquadratic equations, always convert to quadratic in $x^2$ and check positivity of roots.

Answer 1

The Correct Option is C

To solve the equation \(x^4 - ax^2 + 9 = 0\) and find the least possible integral value of \(a\) for which the equation has four real and distinct roots, let us use some algebraic manipulation and properties of polynomials.

We need to ensure that all roots are real and distinct. Therefore, the discriminant conditions and nature of roots should be explored.

Consider substituting \(y = x^2\). This transforms the equation to a quadratic in terms of \(y\):

\(y^2 - ay + 9 = 0\) 

For \(x^4 - ax^2 + 9 = 0\) to have four distinct real roots, the quadratic equation \(y^2 - ay + 9 = 0\) must have two distinct positive roots.

The roots \(y_1\) and \(y_2\) of the equation can be found using the quadratic formula:

\(y = \frac{a \pm \sqrt{a^2 - 36}}{2}\)

In order for the roots \(y_1\) and \(y_2\) to be real and distinct, the discriminant must be positive:

\(a^2 - 36 > 0 \Rightarrow a^2 > 36 \Rightarrow a > 6 \text{ or } a < -6\)

Additionally, since \(y\) represents \(x^2\), the roots must be positive, meaning that both roots \(y_1\) and \(y_2\) are positive. Therefore, we only consider \(a > 6\) (ignoring \(a < -6\) as it doesn't satisfy positivity of both roots).

The smallest integer \(a\) satisfying \(a > 6\) is \(a = 7\).

Thus, the least possible integral value of \(a\) such that the equation \(x^4 - ax^2 + 9 = 0\) has four real and distinct roots is:

\(a = 7\)

Therefore, the correct answer is 7.