How do you solve #sqrt(3x-5)=x-5# and check your solution?

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Answer 1 · 2017-07-28T20:45:48

The solution is #S={10}#

The equation is

#sqrt(3x-5)=x-5#

The conditions are :

#3x-5>=0#, #x>=5/3#

Squaring the equation

#(sqrt(3x-5))^2=(x-5)^2#

#3x-5=x^2-10x+25#

#x^2-13x+30=0#

Factorising this quadratic equation

#(x+3)(x-10)=0#

Therefore,

#x+3=0#, #=>#, #x=-3#, this solution is not valid since #x>=5/3#

#x-10=0#, #=>#, #x=10#

Verification

#LHS=sqrt(3x-5)=sqrt(30-5)=sqrt25=5#

#RHS=x-5=10-5=5#

#LHS=RHS#

#QED#