How do you factor #x(x-1)^(-1/2) + 2(x-1)^(1/2)#?

1 Answer
May 31, 2017

Answer:

#((3x-2)sqrt(x-1))/(x-1)#

Explanation:

First of all you need to know that #x^-n=1/x^n# and #x^(1/m)=root(m)x#.
so #x(x-1)^(-1/2)+2(x-1)^(1/2)=x/sqrt(x-1)+2sqrt(x-1)=(x+2sqrt(x-1)sqrt(x-1))/sqrt(x-1)=(x+2(x-1))/sqrt(x-1)=(x+2x-2)/sqrt(x-1)=(3x-2)/sqrt(x-1)=(3x-2)(x-1)^(-1/2)#
You can stop here but usually you don't want a square root as a denominator so u multiply for #sqrt(x-1)/sqrt(x-1#:
#(3x-2)/sqrt(x-1)*sqrt(x-1)/sqrt(x-1)=((3x-2)sqrt(x-1))/(x-1)#