Consider an even polynomial p(s) given by

p(s) = s⁴ + 5s² + 4 + K

where K is an unknown real parameter. The complete range of K for which p(s) has all its roots on the imaginary axis is ________.

To determine the complete range of the real parameter K for which the given even polynomial p(s) has all its roots on the imaginary axis, we need to analyze its structure and the properties of its roots.

Polynomial Roots on the Imaginary Axis

The given polynomial is:

$$p(s)=s^4+5s^2+4+K$$

This is an even polynomial, meaning it only contains even powers of s. For an even polynomial, if a root is s = j\omega (where \omega is a real number), then s = -j\omega must also be a root. If all roots are on the imaginary axis, they will be of the form s = \pm j\omega_i.

If s = j\omega, then s^2 = (j\omega)^2 = -\omega^2. Since \omega is a real number, \omega^2 must be a non-negative real number. Therefore, s^2 must be a non-positive real number (i.e., s^2 \le 0).

Transforming the Polynomial Equation

Let's simplify the analysis by substituting y = s^2 into the polynomial equation:

$$y^2+5y+(4+K)=0$$

For the original polynomial p(s) to have all its roots on the imaginary axis, the roots of this quadratic equation in y must be non-positive real numbers. Specifically, y must represent values of -\omega², which are less than or equal to zero.

Conditions for Non-Positive Real Roots

For a quadratic equation $a{y}^2+by+c=0$ to have non-positive real roots, the following three conditions must be met:

- Discriminant (D) Condition: The discriminant must be non-negative for the roots to be real. $$D=b^2-4ac\ge 0$$
- Sum of Roots Condition: The sum of the roots $y_1+y_2=-b/a$ must be non-positive.
- Product of Roots Condition: The product of the roots $y_1{y}_2=c/a$ must be non-negative. This ensures both roots are either positive or negative. Combined with the sum condition, it guarantees both roots are non-positive.

Applying Conditions to Our Polynomial

For the quadratic equation $y^2+5y+(4+K)=0$:

Here, $a=1$, $b=5$, and $c=(4+K)$.

1. Discriminant Condition

$$D=5^2-4(1)(4+K)\ge 0$$

$$25-16-4K\ge 0$$

$$9-4K\ge 0$$

$$9\ge 4K$$

$$K\le \frac{9}{4}$$

2. Sum of Roots Condition

$$y_1+y_2=-\frac{b}{a}=-\frac{5}{1}=-5$$

Since the sum of roots is -5, which is already negative, this condition is satisfied for any real K. This ensures that if real roots exist, at least one is negative. Combined with the product condition, it forces both to be non-positive.

3. Product of Roots Condition

$$y_1{y}_2=\frac{c}{a}=\frac{4+K}{1}=4+K$$

For both roots to be non-positive, their product must be non-negative. If one root is 0, the product is 0. If both are negative, the product is positive.

$$4+K\ge 0$$

$$K\ge -4$$

Combining the Conditions for K

We need to satisfy all three conditions simultaneously to find the complete range of K:

- $K\le \frac{9}{4}$ (from discriminant)
- $K\ge -4$ (from product of roots)

Combining these, the complete range for K is:

$$-4\le K\le \frac{9}{4}$$

Root Verification for Boundary K Values

Let's check the polynomial's behavior at the boundary values of K.

- **Case 1: $K=-4$ ** The polynomial becomes $p(s)=s^4+5s^2+4+(-4)=s^4+5s^2$. Setting $p(s)=0$: $s^4+5s^2=0$ Factoring out $s^2$: $s^2(s^2+5)=0$ This gives roots $s^2=0$ (so $s=0$, which is on the imaginary axis) and $s^2=-5$ (so $s=\pm j\sqrt{5}$, which are also on the imaginary axis). All roots are indeed on the imaginary axis for $K=-4$.
- **Case 2: $K=\frac{9}{4}$ ** The polynomial becomes $p(s)=s^4+5s^2+4+\frac{9}{4}=s^4+5s^2+\frac{16+9}{4}=s^4+5s^2+\frac{25}{4}$. Setting $p(s)=0$: $s^4+5s^2+\frac{25}{4}=0$ Let $y=s^2$: $y^2+5y+\frac{25}{4}=0$ This is a perfect square trinomial: ${\left(y+\frac{5}{2}\right)}^2=0$ So, $y=-\frac{5}{2}$ (a repeated real root). Substituting back $s^2=y$, we get $s^2=-\frac{5}{2}$. This gives roots $s=\pm j\sqrt{\frac{5}{2}}$, which are also on the imaginary axis (these are repeated roots). All roots are on the imaginary axis for $K=\frac{9}{4}$.

Both boundary values satisfy the condition, confirming the inclusion of K = -4 and K = 9/4 in the range.

Therefore, the complete range of K for which p(s) has all its roots on the imaginary axis is **$-4\le K\le \frac{9}{4}$ **.