One more time!If q(x)=5-x^2 and p(q(x))= 4-x^2/x^3 when x≠0, then what is p(1/4) equal to? Can you show me the steps and possibly explain them so I understand?

since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

[tex]\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\ x = \frac{\sqrt{19} }{2}[/tex]

so we conclude that
[tex]q(\frac{\sqrt{19} }{2} ) = 1/4[/tex]

therefore

[tex]p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right) \right)[/tex]

plug [tex] x=\sqrt{19}/2[/tex] into p( q(x) ) to get answer

[tex]p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left( \frac{\sqrt{19} }{2}\right)^2 }{ \left( \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow [/tex]

[tex]\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow -\dfrac{6\sqrt{19} }{361}[/tex]

[tex]p(1/4) = -\dfrac{6\sqrt{19} }{361}[/tex]