Question

If X > 0, what is one possible solution to this equation x^3 (x^2-5) = -4x

Answer 1

Answer:

$x {}^{3} ( {x}^{2} - 5) = - 4x$

distribute x³ :

$x {}^{5} - 5 {x}^{3} = - 4x$

move variable to the left-hand side:

${x}^{5} - 5 {x}^{3} + 4x = 0$

factorize out x :

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

write -5x as difference:

$x( {x}^{4} - {x}^{2} - 4 {x}^{2} + 4) = 0$

factorize x² from the equation:

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

factorize -4 from the equation:

$x( {x}^{2} ( {x}^{2} - 1) - 4( {x}^{2} - 1) = 0$

factorize (x²-1) from the equation:

$x( {x}^{2} - 1)( {x}^{2} - 4) = 0$

products:

1) x = 0

*ignore this as your question wants x>0

2) x²-1 = 0

x² = 1

x = √1

x=1

3) x²-4 = 0

x² = 4

x = √4

x=2

thus, x=1, x=2