What are the solutions of the equation [tex]x^4 - 5x^2 - 14 = 0[/tex]? Use factoring to solve.

A. [tex]x = \pm \sqrt{7}[/tex] and [tex]x = \pm \sqrt{2}[/tex]

B. [tex]x = \pm i \sqrt{7}[/tex] and [tex]x = \pm i \sqrt{2}[/tex]

C. [tex]x = \pm i \sqrt{7}[/tex] and [tex]x = \pm \sqrt{2}[/tex]

D. [tex]x = \pm \sqrt{7}[/tex] and [tex]x = \pm i \sqrt{2}[/tex]



Answer :

To solve the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] using factoring, let's proceed step-by-step.

1. Introduce a Substitution: Let [tex]\( y = x^2 \)[/tex]. Then the original equation becomes
[tex]\[ y^2 - 5y - 14 = 0. \][/tex]
This is a quadratic equation in terms of [tex]\( y \)[/tex].

2. Solve the Quadratic Equation: We need to solve
[tex]\[ y^2 - 5y - 14 = 0. \][/tex]
To factor this quadratic equation, look for two numbers that multiply to [tex]\(-14\)[/tex] and add to [tex]\(-5\)[/tex]. These numbers are [tex]\(-7\)[/tex] and [tex]\(2\)[/tex]. Thus, the quadratic factors as
[tex]\[ (y - 7)(y + 2) = 0. \][/tex]

3. Solve for [tex]\( y \)[/tex]:
[tex]\[ y - 7 = 0 \quad \text{or} \quad y + 2 = 0 \][/tex]
So,
[tex]\[ y = 7 \quad \text{or} \quad y = -2. \][/tex]

4. Back-Substitute [tex]\( y = x^2 \)[/tex]:
[tex]\[ x^2 = 7 \quad \text{or} \quad x^2 = -2. \][/tex]

5. Solve for [tex]\( x \)[/tex]:
- If [tex]\( x^2 = 7 \)[/tex],
[tex]\[ x = \pm \sqrt{7}. \][/tex]
- If [tex]\( x^2 = -2 \)[/tex],
[tex]\[ x = \pm \sqrt{-2} = \pm i \sqrt{2}. \][/tex]

Thus, the solutions to the original equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[ x = \pm \sqrt{7} \quad \text{and} \quad x = \pm i \sqrt{2}. \][/tex]

Therefore, the correct answer is:

[tex]\[ x = \pm \sqrt{7} \quad \text{and} \quad x = \pm i \sqrt{2}. \][/tex]