How do you solve #x=sqrt(x+6)# and find any extraneous solutions?

1 Answer
Apr 30, 2018

Answer:

#x=3,x=-2" is extraneous"#

Explanation:

#color(blue)"square both sides"#

#"note that "sqrtaxxsqrta=(sqrta)^2=a#

#x^2=(sqrt(x+6))^2#

#rArrx^2=x+6#

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

#rArrx^2-x-6=0#

#"the factors of - 6 which sum to - 1 are - 3 and + 2"#

#rArr(x-3)(x+2)=0#

#"equate each factor to zero and solve for x"#

#x+2=0rArrx=-2#

#x-3=0rArrx=3#

#color(blue)"As a check"#

Substitute these values into the equation and if both sides equate then they are the solutions.

#x=-2to" left "=-2" right "=sqrt4=2#

#-2!=2rArrx=-2" is extraneous"#

#x=3to"left "=3" right "=sqrt9=3#

#rArrx=3" is the solution"#