what are the solutions of the equation $x^{4}-5x^{2}-36 = 0$? use factoring to solve.\n$x=pm2$ and $x =…

what are the solutions of the equation $x^{4}-5x^{2}-36 = 0$? use factoring to solve.\n$x=pm2$ and $x = pm3$\n$x=pm2i$ and $x=pm3$\n$x=pm2$ and $x=pm3i$\n$x=pm2i$ and $x=pm3i$

what are the solutions of the equation $x^{4}-5x^{2}-36 = 0$? use factoring to solve.\n$x=pm2$ and $x = pm3$\n$x=pm2i$ and $x=pm3$\n$x=pm2$ and $x=pm3i$\n$x=pm2i$ and $x=pm3i$

Answer

Explanation:

Step1: Let (y = x^{2})

The equation (x^{4}-5x^{2}-36 = 0) becomes (y^{2}-5y - 36=0).

Step2: Factor the quadratic equation

We factor (y^{2}-5y - 36) as ((y - 9)(y+4)=0) since (-9\times4=- 36) and (-9 + 4=-5).

Step3: Solve for (y)

Set each factor equal to zero: (y - 9=0) gives (y = 9); (y + 4=0) gives (y=-4).

Step4: Substitute back (y = x^{2})

When (y = 9), we have (x^{2}=9), so (x=\pm3). When (y=-4), we have (x^{2}=-4), so (x=\pm2i).

Answer:

(x = \pm2i) and (x=\pm3)