Certainly! To solve the equation [tex]\( x^3(x^2 - 5) = -4x \)[/tex] for [tex]\( x > 0 \)[/tex], we will follow these steps:
1. Simplify and rearrange the equation:
[tex]\[
x^3(x^2 - 5) + 4x = 0
\][/tex]
2. Factor out the common term:
[tex]\[
x(x^4 - 5x^2 + 4) = 0
\][/tex]
Since [tex]\( x \neq 0 \)[/tex] (as [tex]\( x > 0 \)[/tex]), we focus on solving the quadratic factor:
[tex]\[
x^4 - 5x^2 + 4 = 0
\][/tex]
3. Substitute a new variable to simplify:
Let [tex]\( y = x^2 \)[/tex]. Hence, the equation becomes:
[tex]\[
y^2 - 5y + 4 = 0
\][/tex]
4. Solve the quadratic equation:
This can be done either by factoring or using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
For the given quadratic equation:
[tex]\[
y^2 - 5y + 4 = 0
\][/tex]
we have [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex].
Calculating the discriminant:
[tex]\[
\Delta = b^2 - 4ac = (-5)^2 - 4(1)(4) = 25 - 16 = 9
\][/tex]
So,
[tex]\[
y = \frac{5 \pm \sqrt{9}}{2} = \frac{5 \pm 3}{2}
\][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{5 + 3}{2} = 4 \quad \text{and} \quad y = \frac{5 - 3}{2} = 1
\][/tex]
5. Back-substitute [tex]\( y = x^2 \)[/tex]:
[tex]\[
x^2 = 4 \quad \Rightarrow \quad x = 2 \quad \text{(since \( x > 0 \))}
\][/tex]
[tex]\[
x^2 = 1 \quad \Rightarrow \quad x = 1 \quad \text{(since \( x > 0 \))}
\][/tex]
Thus, we find that the equation [tex]\( x^3(x^2 - 5) = -4x \)[/tex] has two possible positive solutions: [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex].
Given about finding one possible solution, we conclude that:
[tex]\[
\boxed{1}
\][/tex]
is one possible solution for [tex]\( x > 0 \)[/tex].
