How do you solve #sqrt(x+3)-sqrt x=sqrt(4x-5)#?

This is a tricky equation, so you have first to determine the dominion of it:

#x+3>=0 and x>0 and 4x-5>=0#

#x>=-3 and x>0 and x>=5/4 => x>=5/4#

The standard way to solve this type of equations is to square the parcels, admiting that:

#color(red)(if a=b => a^2=b^2)#

However this brings false solutions, because

#color(red)(if a=-b => a^2=b^2)#

So we have to check the solutions after we obtain the results.

So now let's start:

#sqrt(x+3)-sqrt(x)=sqrt(4x-5)#

#(sqrt(x+3)-sqrt(x))^2=(sqrt(4x-5))^2#

#x+3-2sqrt((x+3)x)+x=4x-5#

Now, you continue to have a "sqrt" in the equation, so you have to square it again. Rearrange the equation in order to isolate the root:

#2sqrt(x^2+3x)=4x-5-x-3-x#

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

#sqrt(x^2+3x)=x-4#

squaring:

#x^2+3x=x^2-8x+16#

Which gives:

#x=16/11#

First #16/11>5/4?#(the dominion determined above)

Put them in the same denominator:

#(16/11)xx(4/4)>(5/4)xx(11/11) ?#

#64/44>55/44, true#

Now, is the solution true?

#sqrt(16/11+3)-sqrt(16/11)=sqrt(4xx16/11-5)#

#sqrt(49/11)-sqrt(16/11)=sqrt(9/11)#

#(sqrt(49)-sqrt(16))/sqrt(11)=sqrt(9/11)#

#(7-4)/sqrt(11)=3/sqrt(11), true#

#1#. When dealing with radicals, try to eliminate them first. Thus, start by squaring both sides of the equation.

#sqrt(x+3)-sqrt(x)=sqrt(4x-5)#

#(sqrt(x+3)-sqrt(x))^2=(sqrt(4x-5))^2#

#2#. Simplify.

#(sqrt(x+3)-sqrt(x))(sqrt(x+3)-sqrt(x))=4x-5#

#x+3-sqrt(x(x+3))-sqrt(x(x+3))+x=4x-5#

#2x+3-sqrt(x^2+3x)-sqrt(x^2+3x)=4x-5#

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

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

#sqrt(x^2+3x)=-x+4#

#3#. Since the left side contains a radical, square the whole equation again.

#(sqrt(x^2+3x))^2=(-x+4)^2#

#4#. Simplify.

#(sqrt(x^2+3x))(sqrt(x^2+3x))=(-x+4)(-x+4)#

#x^2+3x=x^2-4x-4x+16#

#color(red)cancelcolor(black)(x^2)+3x=color(red)cancelcolor(black)(x^2)-8x+16#

#3x=-8x+16#

#5#. Solve for #x#.

#11x=16#

#color(green)(x=16/11)#

#:.#, #x# is #16/11#.