How do you solve #sqrt(2x+2) - sqrt(x+2) = 1#?

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Answer 1 · 2015-11-02T03:02:13

X = 7

Start off by shifting;
#sqrt(2x+2) = sqrt(x+2) + 1 #

Square both sides;

#2x + 2 = (x+2) + 1 + 2sqrt(x+2)#

Simplify as much as possible;

#x -1 = 2sqrt(x+2)#

Square once again;

#x^2 + 1 -2x = 4(x+2)#

#x^2 + 1 -2x = 4x+ 8#
Simplify;
#x^2 - 6x - 7 = 0#

Factor;

#(x-7)(x + 1) = 0#

So #x = 7 or -1#

But now we can be hasty in saying it has 2 solutions;

Make sute both solutions are real

Lets check for 7

#sqrt(2x+2) - sqrt(x+2)#

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Lets check for -1

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We end up with x = 7