How do you solve #sqrt(x+4 )= 1- sqrt(3x+13)#?
1 Answer
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) = 2x+10#
Divide both sides by
#sqrt(3x+13) = x+5#
Square both sides to get:
#3x+13 = x^2+10x+25#
Subtract
#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