Question

Can someone help me on this please

Answer 1

Answer:

x=0,1,±2

Step-by-step explanation:

x^3(x^2-5)=-4x

x^2(x^2-5)=-4

Take x^2=a, we have

a*(a-5)=-4, a^2-5a+4=0, (a-1)(a-4)=0, a=1,4

x^2=0, 1, 4. x=0,1,±2

Answer 2

Answer:

x = 1 and x = 2

Step-by-step explanation:

${x}^{5} - 5 {x}^{3} = - 4x \\ {x}^{4} - 5 {x}^{2} + 4 = 0 \\ ( {x}^{2} - 4)( {x}^{2} - 1)=0$

x² = 4 or x² = 1

1. x = ±√4 = -2 or 2
2. x = ±√1 = -1 or 1

Since x > 0, the negative integers are rejected and the solutions to this equations are x = 1 and x = 2