Answer
The solutions to this equation are $x=-1,$ $x=0$ and $x=3$
Work Step by Step
$4(x+1)^{1/2}-5(x+1)^{3/2}+(x+1)^{5/2}=0$
Take out common factor $(x+1)^{1/2}$:
$(x+1)^{1/2}[4-5(x+1)+(x+1)^{2}]=0$
Rearrange the expression inside brackets:
$(x+1)^{1/2}[(x+1)^{2}-5(x+1)+4]=0$
Let $x+1$ be equal to $u$:
$u^{1/2}(u^{2}-5u+4)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u^{1/2}=0$
$u=0$
$u^{2}-5u+4=0$ (Solve this second equation by factoring):
$(u-1)(u-4)=0$
Again, set both factors equal to $0$ and solve each individual equation for $u$:
$u-1=0$
$u=1$
$u-4=0$
$u=4$
Substitute $u$ back to $x+1$ into each solution and solve for $x$:
$u=0$
$x+1=0$
$x=-1$
$u=1$
$x+1=1$
$x=1-1$
$x=0$
$u=4$
$x+1=4$
$x=4-1$
$x=3$
The original equation is true for all values of $x$ found.
The solutions to this equation are $x=-1,$ $x=0$ and $x=3$
