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

Hint:

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

Answer 1

Correct Answer: 7

We want the least integer value of \(a\) such that

\[ x^4 - ax^2 + 9 = 0 \]

has four real and distinct roots.

Put \(y=x^2\) (so \(y \ge 0\)). Then the equation becomes

\[ y^2 - ay + 9 = 0. \]

For \(x\) to have four real and distinct roots, this quadratic in \(y\) must have two distinct positive roots \(y_1, y_2\) (so that \(x=\pm\sqrt{y_1}, \pm\sqrt{y_2}\)).

• Distinct roots in \(y\): Discriminant \(a^2 - 36 > 0 \Rightarrow |a| > 6\).
• Positive roots in \(y\): Product \(y_1y_2 = 9 > 0\) and sum \(y_1+y_2=a\). For both to be positive, we need \(a > 0\).

Combining: \(a > 6\). Hence the least integer value is

\(a = 7\).