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

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

$RHS=x-5=10-5=5$

$LHS=RHS$

$QED$

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