what are the solutions of the equation $x^{4}+95x^{2}-500 = 0$? use factoring to solve.\n$x=pmsqrt{5}$ and…

what are the solutions of the equation $x^{4}+95x^{2}-500 = 0$? use factoring to solve.\n$x=pmsqrt{5}$ and $x = pm10$\n$x=pm isqrt{5}$ and $x=pm10i$\n$x=pmsqrt{5}$ and $x=pm10i$\n$x=pm isqrt{5}$ and $x=pm10$
Answer
Explanation:
Step1: Let (y = x^{2})
The equation (x^{4}+95x^{2}-500 = 0) becomes (y^{2}+95y - 500=0).
Step2: Factor the quadratic equation
We need to find two numbers that multiply to (-500) and add up to (95). The numbers are (100) and (- 5). So, (y^{2}+95y - 500=(y + 100)(y - 5)=0).
Step3: Solve for (y)
Set each factor equal to zero: If (y + 100=0), then (y=-100). If (y - 5=0), then (y = 5).
Step4: Substitute back (y=x^{2})
When (x^{2}=5), then (x=\pm\sqrt{5}). When (x^{2}=-100), then (x=\pm\sqrt{- 100}=\pm10i).
Answer:
(x=\pm\sqrt{5}) and (x=\pm10i)