Given: 2x^4=128x Move 128x to left,getting
2x^4-128x=0 ··· Eq.1 Remove the common factor 2x, getting
2x(x³-64)=0 Since 64=4³, then the left side can be rewritten:
2x(x³-4³)=0 Factor the trinomial using a difference of cubes formula,getting
2x(x-4)(x²+4x+16)=0 So,2x=0, x-4=0, and x²+4x+16=0 will satisfy the equation. So, we have:
x+4x+16=0 Solve the equation using the quadratic formula. We have:
x=-2±2(√3)i Thus, Eq.1 can be rewritten as follows:
2x^4-128x=2x(x-4){x-(-2+2(√3)i}{x-(-2-2(√3)i}=0
The answer is: x∈{ 0,4,-2+2(√3)i,-2+2(√3)i }