Question

How do you solve `sqrt[x+2]=sqrt[4-x]` and find any extraneous solutions?

Answer 1

`x=1`

Explanation:

`"to access the contents of the radicals"`

`color(blue)"square both sides"`

`•color(white)(x)sqrtaxxsqrta=(sqrta)^2=a`

`(sqrt(x+2))^2=(sqrt(4-x))^2`

`rArrx+2=4-x`

`"add x to both sides"`

`x+x+2=4cancel(-x)cancel(+x)`

`rArr2x+2=4`

`"subtract 2 from both sides"`

`2xcancel(+2)cancel(-2)=4-2`

`rArr2x=2`

`"divide both sides by 2"`

`(cancel(2) x)/cancel(2)=2/2`

`rArrx=1`

`color(blue)"As a check"`

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

`"left side "=sqrt(1+2)=sqrt3`

`"right side "=sqrt(4-1)=sqrt3`

`rArrx=1" is the solution"`

There are no other solutions hence no extraneous solutions.