How do you solve #1= \sqrt{- 6- 2k} - \sqrt{- 4- k}#?

1 Answer
George C.
Sep 22, 2016

#k = -5#

Explanation:

Given:

#1 = sqrt(-6-2k)-sqrt(-4-k)#

Add #sqrt(-4-k)# to both sides to get:

#1+sqrt(-4-k) = sqrt(-6-2k)#

Square both sides (noting that this may introduce extraneous solutions) to get:

#1+2sqrt(-4-k)+(-4-k) = -6-2k#

which simplifies to:

#2sqrt(-4-k)-3-k = -6-2k#

Add #3+k# to both sides to get:

#2sqrt(-4-k) = -3-k = -(3+k)#

Square both sides (noting that this may introduce extraneous solutions) to get:

#4(-4-k) = 9+6k+k^2#

which simplifies to:

#-16-4k = k^2+6k+9#

Add #16+4k# to both sides to get:

#0 = k^2+10k+25 = (k+5)^2#

So the only possible solution is #k=-5#

Since we squared the equation (twice) we need to check this actually works:

#sqrt(-6-2k)-sqrt(-4-k) = sqrt(-6-2(color(blue)(-5)))-sqrt(-4-(color(blue)(-5)))#

#color(white)(sqrt(-6-2k)-sqrt(-4-k)) = sqrt(-6+10)-sqrt(-4+5)#

#color(white)(sqrt(-6-2k)-sqrt(-4-k)) = sqrt(4)-sqrt(1)#

#color(white)(sqrt(-6-2k)-sqrt(-4-k)) = 2-1#

#color(white)(sqrt(-6-2k)-sqrt(-4-k)) = 1#

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