what are the solutions of the equation $x^{4}-5x^{2}-14 = 0$? use factoring to solve.\n$x=pmsqrt{7}$ and $x…

what are the solutions of the equation $x^{4}-5x^{2}-14 = 0$? use factoring to solve.\n$x=pmsqrt{7}$ and $x = pmsqrt{2}$\n$x=pm isqrt{7}$ and $x=pm isqrt{2}$\n$x=pm isqrt{7}$ and $x=pmsqrt{2}$\n$x=pmsqrt{7}$ and $x=pm isqrt{2}$

what are the solutions of the equation $x^{4}-5x^{2}-14 = 0$? use factoring to solve.\n$x=pmsqrt{7}$ and $x = pmsqrt{2}$\n$x=pm isqrt{7}$ and $x=pm isqrt{2}$\n$x=pm isqrt{7}$ and $x=pmsqrt{2}$\n$x=pmsqrt{7}$ and $x=pm isqrt{2}$

Answer

Answer:

D. $x = \pm\sqrt{7}$ and $x=\pm i\sqrt{2}$

Explanation:

Step1: Let $y = x^{2}$

The equation $x^{4}-5x^{2}-14 = 0$ becomes $y^{2}-5y - 14=0$.

Step2: Factor the quadratic equation

We factor $y^{2}-5y - 14$ as $(y - 7)(y+2)=0$.

Step3: Solve for $y$

Set each factor equal to zero: $y - 7=0$ gives $y = 7$; $y + 2=0$ gives $y=-2$.

Step4: Substitute back $y = x^{2}$

When $y = 7$, we have $x^{2}=7$, so $x=\pm\sqrt{7}$. When $y=-2$, we have $x^{2}=-2$, so $x=\pm i\sqrt{2}$.