To find the solutions of the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex], we can use factoring by introducing a substitution that simplifies the polynomial.
### Step-by-Step Solution:
1. Substitute [tex]\( y = x^2 \)[/tex]:
This substitution simplifies the fourth-degree equation into a quadratic equation. Rewriting the equation using this substitution:
[tex]\[
x^4 - 5x^2 - 14 = 0 \implies y^2 - 5y - 14 = 0
\][/tex]
2. Solve the quadratic equation [tex]\( y^2 - 5y - 14 = 0 \)[/tex]:
Use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -14 \)[/tex]:
[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]
This gives us the solutions:
[tex]\[
y = \frac{14}{2} = 7 \quad \text{and} \quad y = \frac{-4}{2} = -2
\][/tex]
3. Back-substitute [tex]\( y = x^2 \)[/tex]:
Now, substituting back [tex]\( y = x^2 \)[/tex]:
[tex]\[
x^2 = 7 \quad \text{and} \quad x^2 = -2
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
- For [tex]\( x^2 = 7 \)[/tex]:
[tex]\[
x = \pm \sqrt{7}
\][/tex]
- For [tex]\( x^2 = -2 \)[/tex]:
[tex]\[
x = \pm \sqrt{-2} = \pm \sqrt{2}i
\][/tex]
### Conclusion:
The solutions to the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[
x = \pm \sqrt{7}, \pm \sqrt{2}i
\][/tex]
Therefore, matching with the given options, the correct one is:
[tex]\[
x = \pm \sqrt{7} \text{ and } x = \pm i \sqrt{2}
\][/tex]
Thus, the solution is:
[tex]\[
x = \pm \sqrt{7} \text{ and } x = \pm i \sqrt{2}
\][/tex]
