Question

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

Answer 1

`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"`