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

2 Answers
Jul 10, 2017

Answer:

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

Answer:

#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