How do you solve x=sqrt(x+6) and find any extraneous solutions?
1 Answer
Apr 30, 2018
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"
