Question

Solving this equation in comments.

Answer 1

\[(x+1)^\frac{ -1 }{2} +\frac{ (x-1)(x+1)^\frac{ 1 }{ 2 } }{ \sqrt{x^3 +1} }=0\]

Answer 2

1. to get rid of the fractional exponents \[\frac{ 1 }{x+1 }+\frac{ (x-1)\sqrt{x+1} }{ \sqrt{x^3 + 1} }=0\]

Answer 3

nope

Answer 4

\[\frac{ 1 }{\sqrt{x+1} }+\frac{ (x-1)\sqrt{x+1} }{ \sqrt{x^3 + 1} }=0\]

Answer 5

ah yest i forgot. then next i would do this \[\frac{ (x-1)\sqrt{x+1} }{ \sqrt{x^3 +1} }=-\frac{ 1 }{ \sqrt{x+1}}\]

Answer 6

and then do this\[ \sqrt{x+1} [(x-1)\sqrt{x+1}] =-\sqrt{x^3 +1}\] which is\[\sqrt{x+1}[x \sqrt{x+1}-\sqrt{x+1}]=-\sqrt{x^3 -1}\]

Answer 7

so it becomes \[x(x-1)-(x+1)=-\sqrt{x^3+1}\]
which is\[x^2-2x-1=\sqrt{x^3+1}\]

Answer 8

and then i would do this \[(x^2-2x-1)^2=(\sqrt{x^3+1})^2\] and that turns into this\[x^4-4x^3-2x^2+8x+1=x^3+1\] which simplifies to\[x^4-5x^3-2x^2+8x=0\]

Answer 9

Now i can factor out an x making it \[x(x^3-5x^2-2x+8)=0\] but where would i go from here?

Answer 10

nevermind...i messed a bit up. I figured it out.

Answer 11

\[\sqrt{x+1} [(x-1)\sqrt{x+1}] =-\sqrt{x^3 +1}\]\[(x+1)(x-1)=-\sqrt[3]{x^2+1}\]

Answer 12

it should be \[x^2-1=-\sqrt{x^3+1}\] which i would square both sides

Answer 13

where did you get the cube root? @satellite73 ?