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

2 Answers
Jul 10, 2017

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]}

Jul 10, 2017

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