Question 7 (1 point) State the total number of complex roots, the possible number of real and imaginary roots and the possible rational roots for each equation. Then finc all roots $$ x^{6}-x^{4}-x^{2}+1=0 $$ a \# of complex roots: 6 Possible \# of real roots: $$ 6,4,2 $$, or 0 Possible \# of imaginary roots: $$ 6,4,2 $$, or 0 Possible rational roots: $$ \pm 1 $$ Roots: $$ \{-1, i,-i\} $$ b \# of complex roots: 6 Possible \# of real roots: $$ 6,4,2 $$, or 0 Possible \# of imaginary roots: $$ 6,4,2 $$, or 0 Possible rational roots: $$ \pm 1 $$ Roots: $$ \{1 $$ mult. $$ 2,-1 $$ mult. $$ 2, i,-i\} $$ c \# of complex roots: 6 Possible \# of real roots: $$ 6,4,2 $$, or 0 Possible \# of imaginary roots: $$ 6,4,2 $$, or 0 Possible rational roots: $$ \pm 1 $$ Roots: $$ \{1 $$ mult. 2, -1 mult. 2\} d \# of complex roots: 6 Possible \# of real roots: $$ 6,4,2 $$, or 0 Possible \# of imaginary roots: $$ 6,4,2 $$, or 0 Possible rational roots: $$ \pm 1 $$ Roots: $$ \{1,-1,-2,2, i,-i\} $$ Review Answers

Question

UpStudy AI · Accepted Answer

We start with the equation

$$
x^6 - x^4 - x^2 + 1 = 0.
$$

**Step 1. Change of Variable**

Let

$$
u = x^2.
$$

Then the equation becomes

$$
u^3 - u^2 - u + 1 = 0.
$$

**Step 2. Factor the Cubic**

Notice that we can factor by grouping:

$$
u^3 - u^2 - u + 1 = u^2(u - 1) - 1(u - 1) = (u - 1)(u^2 - 1).
$$

Since

$$
u^2 - 1 = (u-1)(u+1),
$$

we obtain

$$
(u-1)^2 (u+1) = 0.
$$

So the solutions for $$ u $$ are

$$
u = 1 \quad (\text{double root}) \quad \text{and} \quad u = -1.
$$

**Step 3. Return to $$ x $$ by Undoing the Substitution**

Recall that $$ u = x^2 $$.

- For $$ u = 1 $$:

$$
  x^2 = 1 \quad \Longrightarrow \quad x = 1 \quad \text{or} \quad x = -1.
  $$

Since $$ u = 1 $$ is a double root, each solution $$ x = 1 $$ and $$ x = -1 $$ has multiplicity 2.

- For $$ u = -1 $$:

$$
  x^2 = -1 \quad \Longrightarrow \quad x = i \quad \text{or} \quad x = -i.
  $$

Here, the roots $$ x = i $$ and $$ x = -i $$ are simple (multiplicity 1).

Thus the factorization in terms of $$ x $$ is

$$
x^6 - x^4 - x^2 + 1 = (x^2 - 1)^2 (x^2 + 1).
$$

**Step 4. Summarize the Information**

- Total number of complex roots (counting multiplicities): $$6$$
- Possible numbers of real roots: $$6, 4, 2,$$ or $$0$$
- Possible numbers of imaginary (non-real) roots: $$6, 4, 2,$$ or $$0$$
- Possible rational roots (by the Rational Root Theorem): $$\pm 1$$

**Final Roots:**

$$\{1, -1, i, -i\}$$ with multiplicities as follows:
- $$x = 1$$ has multiplicity $$2$$
- $$x = -1$$ has multiplicity $$2$$
- $$x = i$$ has multiplicity $$1$$
- $$x = -i$$ has multiplicity $$1$$

This corresponds to option (b).
The equation $$ x^6 - x^4 - x^2 + 1 = 0 $$ has the following roots: - $$ x = 1 $$ (with multiplicity 2) - $$ x = -1 $$ (with multiplicity 2) - $$ x = i $$ (with multiplicity 1) - $$ x = -i $$ (with multiplicity 1) So, the roots are $$ \{1, -1, i, -i\} $$.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions