How do you solve #sqrt(x-8)+sqrt(x+3)=1#?

1 Answer
Jan 30, 2016

Answer:

#sqrt(x-8)+sqrt(x+3) = 1# has no solutions

Explanation:

#sqrt(x-8)+sqrt(x+3) = 1#

#=> (sqrt(x-8)+sqrt(x+3))^2 = 1^2#

#=> (x-8) + 2sqrt((x-8)(x+3)) + (x+3) = 1#

#=> 2sqrt(x^2-5x-24) = -2x+6#

#=> sqrt(x^2-5x-24) = -x+3#

#=>(sqrt(x^2-5x-24))^2 = (-x+3)^2#

#=>x^2-5x-24 = x^2-6x+9#

#=> x = 33#

In the process of squaring, we may have generated extraneous solutions, and so we must check to see if our possible solution is valid.

#sqrt(33-8)+sqrt(33+3) = sqrt(25)+sqrt(36) = 5+6 = 11#

As the only possible value for #x# we found is not a solution, there is no solution. As another way of seeing this, note that the graph of
#sqrt(x-8)+sqrt(x+3)-1#
never intersects the #x# axis, and thus the equivalent equation
#sqrt(x-8)+sqrt(x+3)-1 = 0#
has no solutions.

graph{sqrt(x-8)+sqrt(x+3)-1 [-3.92, 76.08, -4, 36]}