Question 282020
<pre> <font size = 4 color = "indigo"><b>
{{{sqrt(4x+5)-sqrt(3x-6)=8}}}

Let {{{A=sqrt(4x+5)}}} and {{{B=sqrt(3x-6)}}}

{{{A-B=8}}}

Solve for one of the letters, say A:

{{{A=8+B}}}

Square both sides:

{{{A^2=(8+B)^2}}}

{{{A^2=64+16B+B^2}}}

Replace {{{A^2}}} by {{{A=4x+5}}} and {{{B^2}}} by {{{3x-6}}}

{{{4x+5=64+16B+3x-6}}}

Solve for 16B:

{{{4x+5=58+16B+3x}}}

{{{x+5=58+16B}}}

{{{x-53=16B}}}

Square both sides:

{{{x^2-106x+2809=256B^2}}}

Replace {{{B^2}}} by {{{3x-6}}}

{{{x^2-106x+2809=256(3x-6)}}}

{{{x^2-106x+2809=768x-1536}}}

{{{x^2-874x+4345=0}}}

Factor the left side:

{{{(x-869)(x-5)=0}}}

Solutions are {{{x=869}}} and {{{x=5}}}

We must check even root radical equations for extraneous solutions:

Substitute {{{x=869}}} into original equation:

{{{sqrt(4x+5)-sqrt(3x-6)=8}}}

{{{sqrt(4(869)+5)-sqrt(3(869)-6)=8}}}

{{{sqrt(3481)-sqrt(2601)=8}}}

{{{59-51=8}}}

{{{8=8}}}

That is true so {{{x=869}}} is a solution to the
original equation.

Substitute {{{x=5}}} into original equation:

{{{sqrt(4x+5)-sqrt(3x-6)=8}}}

{{{sqrt(4(5)+5)-sqrt(3(5)-6)=8}}}

{{{sqrt(25)-sqrt(9)=8}}}

{{{5-3=8}}}

{{{2=8}}}

That is false so {{{x=5}}} is not a solution to the
original equation.

So there is one solution {{{x=869}}}

Edwin</pre>