what are all the roots, both real and nonreal, of the equation $x^4 + x^2 = 4x^2 + 4$?

$-\frac{1}{2}, \frac{1}{2}, -2i, 2i$

$-\frac{1}{2}, \frac{1}{2}, -i, i$

$-1, 1, -2i, 2i$

$-2, 2, -i, i$

Question

what are all the roots, both real and nonreal, of the equation $x^4 + x^2 = 4x^2 + 4$?

$-\frac{1}{2}, \frac{1}{2}, -2i, 2i$

$-\frac{1}{2}, \frac{1}{2}, -i, i$

$-1, 1, -2i, 2i$

$-2, 2, -i, i$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Roots of Polynomials", "Complex Roots of Polynomials" ], "new_concepts": [], "current_concepts": [ "Roots of Polynomials", "Complex Roots of Polynomials", "Polynomial Factoring" ] } ## Rearrange the polynomial equation $$ \begin{aligned} x^4 + x^2 &= 4x^2 + 4 \ x^4 - 3x^2 - 4 &= 0 \end{aligned} $$ ## Factor the quadratic in terms of $x^2$ $$ \begin{aligned} (x^2 - 4)(x^2 + 1) &= 0 \end{aligned} $$ ## Solve for all real and nonreal roots $$ \begin{aligned} x^2 - 4 = 0 &\implies x = \pm 2 \ x^2 + 1 = 0 &\implies x = \pm i \ \text{Roots: } &-2, 2, -i, i \end{aligned} $$

(A) $-\frac{1}{2}, \frac{1}{2}, -2i, 2i$

(B) $-\frac{1}{2}, \frac{1}{2}, -i, i$

(C) $-1, 1, -2i, 2i$

(D) $-2, 2, -i, i$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Complex Roots of Polynomials" ] }

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QUESTION IMAGE

Question

Explanation:

Rearrange the polynomial equation

Factor the quadratic in terms of \(x^2\)

Solve for all real and nonreal roots

Answer: