How do you solve #sqrt(6x-3) - 1 = x#?

1 Answer
May 24, 2018

#x=2#

Explanation:

#"isolate "sqrt(6x-3)" by adding 1 to both sides"#

#sqrt(6x-3)=x+1#

#color(blue)"square both sides"#

#(sqrt(6x-3))^2=(x+1)^2#

#6x-3=x^2+2x+1#

#"rearrange into "color(blue)"standard form ";ax^2+bx+c=0#

#"subtract "6x-3" from both sides"#

#0=x^2-4x+4#

#0=(x-2)^2rArrx=2#

#color(blue)"As a check"#

Substitute this value into the equation and if both sides are equal then it is the solution.

#"left "=sqrt(12-3)-1=sqrt9-1=3-1=2=" right"#

#x=2" is the solution"#