How do you solve the equation #sqrt(5r-1)=r-5#?

1 Answer
Oct 26, 2017

#r=13#

Explanation:

As we have #sqrt(5r-1)=r-5#, domain of #r# is #[1/5,oo)#

Squaring both sides in #sqrt(5r-1)=r-5# we get

#5r-1=(r-5)^2#

or #5r-1=r^2-10r+25#

or #r^2-15r+26=0#

or #r^2-13r-2r+26=0#

or #r(r-13)-2(r-13)=0#

or #(r-2)(r-13)=0#

i.e. #r=2# or #13#

Observe that both values are within domain but #r=2# results in #sqrt(5r-1)=sqrt9=3!=2-5#, as it takes negative square root of #sqrt(5r-1)# for equality to hold.

Hence, answer is #r=13#.