Respuesta :
Answer:
[tex]\{2\} \cup \{3\}[/tex]
Explanation:
[tex]x^3(x^2 - 13) = -36x; \\ x^3(x^2 - 13) + 36x = 0; \\ x(x^2(x^2 - 13) + 36)=0; \\ x(x^4 - 13x^2 + 36) = 0 \\ \left \ [ {{x = 0} \atop {x^4 - 13x^2 + 36 = 0}} \right.[/tex]
We have two equalities.
Let us analyze the latter.
Let [tex]x^2 = t[/tex] on condition that [tex]t \geq 0[/tex], then:
[tex]t^2 - 13t + 36 = 0; \\ a = 1, b = -13, c = 36; \\ D = b^2 - 4ac = (-13)^2 - 4 * 1 * 36 = 169 - 144 = 25 = 5^2, > 0; \\ t_{1, 2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-13) \pm \sqrt{5^2}}{2 * 1} = \frac{13 \pm 5}{2} = \ [ {{\frac{13 - 5}{2} = \frac{8}{2} = 4 } \atop {\frac{13 + 5}{2} = \frac{18}{2} = 9 }} \right.[/tex]
Substitute [tex]t[/tex] back for [tex]x^2[/tex], keep in mind that [tex]t \geq 0[/tex].
[tex]\left \ [ {{x^2 = 4} \atop {x^2 = 9}} \right. \Leftrightarrow \left \ [ {{x = \pm 2} \atop {x = \pm 3}} \right.[/tex]
Therefore, [tex]\{-3\} \cup \{-2\} \cup \{0\} \cup \{2\} \cup \{3\}[/tex] is our set of solutions. Given the condition that we need our variable to be more than zero, our answer is [tex]\{2\} \cup \{3\}[/tex].
