How do you solve #sqrt(x+4 )= 1- sqrt(3x+13)#?

1 Answer
Mar 11, 2018

Answer:

#x=-4#

Explanation:

Given:

#sqrt(x+4) = 1-sqrt(3x+13)#

We can square both sides to get:

#x+4=1-2sqrt(3x+13)+(3x+13) = 3x+14-2sqrt(3x+13)#

Add #2sqrt(3x+13)-x-4# to both ends to get:

#2sqrt(3x+13) = 2x+10#

Divide both sides by #2# to get:

#sqrt(3x+13) = x+5#

Square both sides to get:

#3x+13 = x^2+10x+25#

Subtract #3x+13# from both sides to get:

#0 = x^2+7x+12 = (x+3)(x+4)#

So:

#x = -3" "# or #x = -4#

We now need to check for extraneous roots, since these may have been introduced when we squared the equation:

#sqrt((color(blue)(-3))+4) = 1 != -1 = 1-2 = 1-sqrt(4) = 1-sqrt(3(color(blue)(-3))+4)#

#sqrt((color(blue)(-4))+4) = 0 = 1-1 = 1-sqrt(1) = 1-sqrt(3(color(blue)(-4))+13)#

So #x=-3# is an extraneous solution caused by squaring #1 != -1#, while #x=-4# is a solution of the original equation.