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What are the solutions of the equation [tex]\(x^4 + 3x^2 + 2 = 0\)[/tex]? Use [tex]\(u\)[/tex] substitution to solve.

A. [tex]\(x = \pm i \sqrt{2}\)[/tex] and [tex]\(x = \pm 1\)[/tex]
B. [tex]\(x = \pm i \sqrt{2}\)[/tex] and [tex]\(x = \pm i\)[/tex]
C. [tex]\(x = \pm \sqrt{2}\)[/tex] and [tex]\(x = \pm i\)[/tex]
D. [tex]\(x = \pm \sqrt{2}\)[/tex] and [tex]\(x = \pm 1\)[/tex]

Sagot :

To solve the equation [tex]\(x^4 + 3x^2 + 2 = 0\)[/tex] using the [tex]\(u\)[/tex] substitution method, follow these steps:

1. Substitute [tex]\(u\)[/tex] for [tex]\(x^2\)[/tex]:
Let [tex]\(u = x^2\)[/tex]. Then, [tex]\(u^2 = (x^2)^2 = x^4\)[/tex].

Substituting [tex]\(u\)[/tex] into the original equation gives us:
[tex]\[u^2 + 3u + 2 = 0\][/tex]

2. Solve the quadratic equation:
The equation [tex]\(u^2 + 3u + 2 = 0\)[/tex] is a standard quadratic equation. It can be factored as:
[tex]\[(u + 1)(u + 2) = 0\][/tex]

This gives us two solutions for [tex]\(u\)[/tex]:
[tex]\[u + 1 = 0 \quad \text{or} \quad u + 2 = 0\][/tex]
[tex]\[u = -1 \quad \text{or} \quad u = -2\][/tex]

3. Substitute back [tex]\(x^2\)[/tex] for [tex]\(u\)[/tex]:
Remember that [tex]\(u = x^2\)[/tex]. So, we need to solve for [tex]\(x\)[/tex] in the equations [tex]\(x^2 = -1\)[/tex] and [tex]\(x^2 = -2\)[/tex].

4. Solve [tex]\(x^2 = -1\)[/tex]:
We know that [tex]\(x^2 = -1\)[/tex] has complex solutions:
[tex]\[x = \pm i\][/tex]

5. Solve [tex]\(x^2 = -2\)[/tex]:
Similarly, [tex]\(x^2 = -2\)[/tex] also has complex solutions:
[tex]\[x = \pm \sqrt{2} i\][/tex]

Hence, the solutions of the equation [tex]\(x^4 + 3x^2 + 2 = 0\)[/tex] are:
[tex]\[x = \pm i\][/tex]
[tex]\[x = \pm \sqrt{2} i\][/tex]

So, the correct answer is:
[tex]\[x = \pm i \text{ and } x = \pm i \sqrt{2}\][/tex]

This corresponds to the answer:
[tex]\[x= \pm i \text{ and } x= \pm i \sqrt{2}\][/tex]