To find the solutions of the equation [tex]\(x^4 + 3x^2 + 2 = 0\)[/tex] using [tex]\(u\)[/tex] substitution, follow these steps:
1. Identify the substitution: Let [tex]\(u = x^2\)[/tex].
2. Transform the original equation: Substituting [tex]\(u\)[/tex] for [tex]\(x^2\)[/tex] in the original equation gives us the quadratic equation:
[tex]\[u^2 + 3u + 2 = 0.\][/tex]
3. Solve the quadratic equation: We solve the quadratic [tex]\(u^2 + 3u + 2 = 0\)[/tex] for [tex]\(u\)[/tex]. This can be done using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 2\)[/tex].
4. Calculate the discriminant:
[tex]\[\Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1.\][/tex]
5. Find the roots of the quadratic equation:
[tex]\[u = \frac{-3 \pm \sqrt{1}}{2 \cdot 1} = \frac{-3 \pm 1}{2}.\][/tex]
This gives us the roots:
[tex]\[u_1 = \frac{-3 + 1}{2} = -1 \quad \text{and} \quad u_2 = \frac{-3 - 1}{2} = -2.\][/tex]
6. Convert back to [tex]\(x\)[/tex]: Recall that [tex]\(u = x^2\)[/tex]. We now set up two separate equations for [tex]\(x\)[/tex]:
[tex]\[
x^2 = -1 \quad \text{and} \quad x^2 = -2.
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
- For [tex]\(x^2 = -1\)[/tex]:
[tex]\[x = \pm i.\][/tex]
- For [tex]\(x^2 = -2\)[/tex]:
[tex]\[x = \pm \sqrt{-2} = \pm i \sqrt{2}.\][/tex]
8. Combine the solutions: The solutions for [tex]\(x\)[/tex] are:
[tex]\[
x = \pm i \quad \text{and} \quad x = \pm i \sqrt{2}.
\][/tex]
So, the solutions of the equation [tex]\(x^4 + 3x^2 + 2 = 0\)[/tex] are [tex]\(x = \pm i\)[/tex] and [tex]\(x = \pm i \sqrt{2}\)[/tex]. Therefore, the correct answer is:
[tex]\[x = \pm i \quad \text{and} \quad x = \pm i \sqrt{2}.\][/tex]
