To find the solutions of the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex], let's go through the factoring process step-by-step.
1. Substitute: Begin by setting [tex]\( y = x^2 \)[/tex]. This substitution transforms our original quartic equation into a quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y^2 - 5y - 14 = 0 \][/tex]
2. Quadratic Equation: The equation [tex]\( y^2 - 5y - 14 = 0 \)[/tex] is a standard quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex]. We can solve for [tex]\( y \)[/tex] using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our particular equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -14 \)[/tex]. Plugging these values into the quadratic formula gives:
[tex]\[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{25 + 56}}{2} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ y = \frac{5 \pm 9}{2} \][/tex]
3. Solve for y: Now we solve for the two values of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{5 + 9}{2} = \frac{14}{2} = 7 \][/tex]
[tex]\[ y = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \][/tex]
So, we have [tex]\( y = 7 \)[/tex] and [tex]\( y = -2 \)[/tex].
4. Back Substitute [tex]\( x \)[/tex]: Recall that [tex]\( y = x^2 \)[/tex]. Thus, we need to solve for [tex]\( x \)[/tex] in the equations [tex]\( x^2 = 7 \)[/tex] and [tex]\( x^2 = -2 \)[/tex].
- For [tex]\( x^2 = 7 \)[/tex]:
[tex]\[ x = \pm \sqrt{7} \][/tex]
- For [tex]\( x^2 = -2 \)[/tex]:
Since square roots of negative numbers involve imaginary units,
[tex]\[ x = \pm \sqrt{-2} = \pm \sqrt{2}i \][/tex]
5. Summarize Solutions:
The solutions of the original equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[
x = \pm \sqrt{7} \quad \text{and} \quad x = \pm \sqrt{2}i
\][/tex]
Therefore, the correct choice is:
[tex]\[ x = \pm \sqrt{7} \text{ and } x = \pm i \sqrt{2} \][/tex]
