What are the solutions of the equation [tex]$x^4+6x^2+5=0$[/tex]? Use [tex]$u$[/tex] substitution to solve.

A. [tex]x = i[/tex] and [tex]x = i \sqrt{5}[/tex]
B. [tex]x = \pm i[/tex] and [tex]x = \pm i \sqrt{5}[/tex]
C. [tex]x = \pm \sqrt{-1}[/tex] and [tex]x = \pm \sqrt{-5}[/tex]
D. [tex]x = \pm-1[/tex] and [tex]x = \pm-\sqrt{5}[/tex]



Answer :

To solve the equation [tex]\(x^4 + 6x^2 + 5 = 0\)[/tex] using [tex]\(u\)[/tex] substitution, we follow these steps:

1. Substitute [tex]\(u = x^2\)[/tex].

This transforms the original equation into:
[tex]\[ (x^2)^2 + 6(x^2) + 5 = u^2 + 6u + 5 = 0. \][/tex]

2. Solve the quadratic equation [tex]\(u^2 + 6u + 5 = 0\)[/tex].

This standard form quadratic equation can be solved using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a=1\)[/tex], [tex]\(b=6\)[/tex], and [tex]\(c=5\)[/tex].

[tex]\[ u = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{-6 \pm \sqrt{36 - 20}}{2} = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2}. \][/tex]

This gives us two solutions for [tex]\(u\)[/tex]:

[tex]\[ u = \frac{-6 + 4}{2} = \frac{-2}{2} = -1, \quad \text{and} \quad u = \frac{-6 - 4}{2} = \frac{-10}{2} = -5. \][/tex]

So, the two values for [tex]\(u\)[/tex] are:
[tex]\[ u = -1 \quad \text{and} \quad u = -5. \][/tex]

3. Convert back to [tex]\(x\)[/tex] by solving for [tex]\(x\)[/tex] from [tex]\(u = x^2\)[/tex].

For [tex]\(u = -1\)[/tex]:
[tex]\[ x^2 = -1 \implies x = \pm \sqrt{-1} = \pm i. \][/tex]

For [tex]\(u = -5\)[/tex]:
[tex]\[ x^2 = -5 \implies x = \pm \sqrt{-5} = \pm \sqrt{5}i. \][/tex]

4. Combine all solutions.

Therefore, the solutions to the equation [tex]\(x^4 + 6x^2 + 5 = 0\)[/tex] are:
[tex]\[ x = \pm i \quad \text{and} \quad x = \pm \sqrt{5}i. \][/tex]

So, the correct answers are:

[tex]\[ x = \pm i \quad \text{and} \quad x = \pm \sqrt{5}i. \][/tex]