a) find the zeros of t(x) =2x^2 + ix + 3.

b) The zeros are not complex conjugates. Explain why this does not contradict with the Conjugate Zeros Theorem

Use the quadratic equation for the roots.

x = (1/4)[-i +/- isqrt23]
(check my work)

The Conjugate zeros theorem only applies to polynomials with real coefficients.

To find the zeros of a quadratic equation, we need to set the equation equal to zero and solve for x.

a) For the equation t(x) = 2x^2 + ix + 3, we set it equal to zero:

2x^2 + ix + 3 = 0

Now, to solve this equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = i, and c = 3. Plugging in these values, we get:

x = (-i ± √(i^2 - 4(2)(3))) / (2(2))

Simplifying further:

x = (-i ± √(-1 - 24)) / 4
x = (-i ± √(-25)) / 4
x = (-i ± 5i) / 4

Now, we can simplify each expression separately:

x1 = (-i + 5i) / 4 = 4i / 4 = i
x2 = (-i - 5i) / 4 = -6i / 4 = -3i / 2

Therefore, the zeros of t(x) = 2x^2 + ix + 3 are x = i and x = -3i / 2.

b) The fact that the zeros of t(x) = 2x^2 + ix + 3 are not complex conjugates does not contradict the Conjugate Zeros Theorem.

The Conjugate Zeros Theorem states that if a polynomial equation has complex zeros, then their conjugates are also zeros. In this case, the zeros of the equation are i and -3i / 2, which are not conjugates of each other.

However, the Conjugate Zeros Theorem is specifically applicable to polynomials with real coefficients. In this case, the coefficient of the linear term in the quadratic equation, ix, is not purely real since it contains the imaginary unit i. Therefore, the Conjugate Zeros Theorem does not apply to this equation.

In summary, the fact that the zeros of t(x) = 2x^2 + ix + 3 are not complex conjugates does not contradict the Conjugate Zeros Theorem because the theorem is only applicable to polynomials with real coefficients.