How do you solve #sqrt(x + 6) + sqrt(2-x) = 4#?

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Answer 1 · 2015-10-07T00:28:40

I found #x=-2#

Let us try squaring both sides:

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

#cancel(x)+6+2sqrt(x+6)sqrt(2-x)+2cancel(-x)=16#

#8+2sqrt((x+6)(2-x))=16# taking #8# to the right:

#2sqrt((x+6)(2-x))=8# taking #2# to the right (dividing):

#sqrt((x+6)(2-x))=4#

let us square again:

#(x+6)(2-x)=16#

#2x-x^2+12-6x-16=0#

#-x^2-4x-4=0# using the Quadratic Formula we get:

#x_(1,2)=(4+-sqrt(16-4(-4*-1)))/-2=(4+-sqrt(0))/-2=-2#

Let us try this solution into our original equation as #x#:
#sqrt(-2+6)+sqrt(2+2)=2+2=4# Yes.