To solve the quadratic equation without using the quadratic formula, we can use other methods like factoring or completing the square. For the given equation:

$6z^2 + z + 6 = 0$

we will first attempt to factor it. If factoring is not feasible or obvious, completing the square will be the alternative method.

Attempt to Factor

For factoring, we need to express the quadratic equation in the form $(az + b)(cz + d) = 0$. However, factoring directly seems challenging for this equation because there are no obvious integer factors of $6 \times 6 = 36$ that sum up to $1$. Therefore, we move to completing the square.

Completing the Square

To complete the square, we first need to make the coefficient of $z^2$ equal to 1. We do this by dividing every term by 6:

$z^2 + \frac{1}{6}z + 1 = 0$

Next, we move the constant term to the other side of the equation:

$z^2 + \frac{1}{6}z = -1$

Now, we need to add a number to both sides of the equation to make the left side a perfect square trinomial. This number is $\left(\frac{\text{coefficient of } z}{2}\right)^2$:

$\left(\frac{\frac{1}{6}}{2}\right)^2 = \left(\frac{1}{12}\right)^2 = \frac{1}{144}$

Adding $\frac{1}{144}$ to both sides:

$z^2 + \frac{1}{6}z + \frac{1}{144} = -1 + \frac{1}{144}$

Simplify the right side by finding a common denominator:

$-1 + \frac{1}{144} = -\frac{144}{144} + \frac{1}{144} = -\frac{143}{144}$

Now the equation is:

$z^2 + \frac{1}{6}z + \frac{1}{144} = -\frac{143}{144}$

The left-hand side is a perfect square:

$\left(z + \frac{1}{12}\right)^2 = -\frac{143}{144}$

To solve for $z$, we take the square root of both sides:

$z + \frac{1}{12} = \pm \sqrt{-\frac{143}{144}}$

Simplify the square root:

$z + \frac{1}{12} = \pm \frac{\sqrt{143} \, j}{12}$

Here, $j = \sqrt{-1}$ is the imaginary unit. Solving for $z$:

$z = -\frac{1}{12} \pm \frac{\sqrt{143} \, j}{12}$

So, the solutions are:

$z_1 = -\frac{1}{12} + \frac{\sqrt{143} \, j}{12}, \quad z_2 = -\frac{1}{12} - \frac{\sqrt{143} \, j}{12}$

Summary of the Solutions

The zeros of the system $H_1(z) = 6z^2 + z + 6$ are:

$z_1 = \frac{-1 + j\sqrt{143}}{12}, \quad z_2 = \frac{-1 - j\sqrt{143}}{12}$

These solutions match what we found using the quadratic formula, but we obtained them by completing the square.

Would you like further clarification or have any more questions?

Related Questions:

1. What are other methods to solve a quadratic equation?
2. How do you apply completing the square for equations with complex solutions?
3. What is the geometric interpretation of the zeros in the complex plane?
4. How does the presence of complex zeros affect the stability of a system?
5. What are the implications of the magnitude of zeros on the frequency response of an FIR filter?

Tip:

When completing the square, always remember to balance the equation by adding the same value to both sides. This ensures that the equality is maintained and the solution remains valid.