Question

How do you solve `[sqrt(4 - x)] - [sqrt(x + 6)] = 2` and find any extraneous solutions?

Answer 1

Since the arguments of square roots may not be negative, we already know that `x<=4andx>=-6`

Explanation:

Or: `-6<=x<=+4`

And now for the difference of `2`:
It may be clear that `x` has to be negative, or the difference (`2`) will never be positive.
We try whole values, and `x=-5` works out, as
`sqrt(4-(-5))-sqrt(-5+6)=sqrt9-sqrt1=3-1=2`
graph{sqrt(4-x)-sqrt(x+6) [-11.25, 11.25, -5.625, 5.625]}

Answer 2

`color(blue)(x=4 or x=-6` extraneous solution

Explanation:

`[sqrt(4-x)]-[sqrt(x+6)]=2`

`:.sqrt(4-x)=2+sqrt(x+6)`

square both sides

`:.(sqrt(4-x))^2=(2+sqrt(x+6))^2`

`sqrt(4-x)*sqrt(4-x)=4-x`

`:.4-x=10+4sqrt(x+6)+x`

`:.4-10-x-x=4sqrt(x+6)`

`:.-6-2x=4sqrt(x+6)`

square both sides

`:.(-6-2x)^2=(4sqrt(x+6))^2`

`:.36+24x+4x^2=16(sqrt(x+6))^2`

`:.36+24x+4x^2=16(x+6)`

`:.36+24x+4x^2=16x+96`

`:.36-96+24x-16x+4x^2=0`

`:.4x^2+8x-96=0`

`:.4(x^2+2x-24=0)`

`:.(x-4)(x+6)=0`

`:.color(blue)(x=4 or x=-6` extraneous solution