Question 554682
√(x-3) = √(x+15)-2 
<pre>
{{{sqrt(x-3)}}} = {{{sqrt(x+15)}}} - 2

Square both sides:


[{{{sqrt(x-3)}}}]² = [{{{sqrt(x+15)}}} - 2]²

[{{{sqrt(x-3)}}}]² = [{{{sqrt(x+15)}}} - 2][{{{sqrt(x+15)}}} - 2]

Use FOIL on the right

[{{{sqrt(x-3)}}}]² = [{{{sqrt(x+15)}}}]² - 2{{{sqrt(x+15)}}} - 2{{{sqrt(x+15)}}} + 4

[{{{sqrt(x-3)}}}]² = [{{{sqrt(x+15)}}}]² - 4{{{sqrt(x+15)}}} + 4


x-3 = x+15 - 4{{{sqrt(x+15)}}} + 4

x-3 = x+19 - 4{{{sqrt(x+15)}}} 

-22 = -4{{{sqrt(x+15)}}}

Square both sides again:

(-22)² = [-4{{{sqrt(x+15)}}}]²

484 = 16(x+15)

484 = 16x + 240

244 = 16x 

{{{244/16}}} = x

{{{61/4}}} = x

Now we have to check because sometimes there are 
extraneous (phony) solutions to radical equations:

{{{sqrt(x-3)}}} = {{{sqrt(x+15)}}} - 2

{{{sqrt(61/4-3)}}} = {{{sqrt(61/4+15)}}} - 2

Get LCD of 4

{{{sqrt(61/4-12/4)}}} = {{{sqrt(61/4+60/4)}}} - 2

{{{sqrt(49/4)}}} = {{{sqrt(121/4)}}} - 2

{{{7/2}}} = {{{11/2}}} - 2

{{{7/2}}} = {{{11/2}}} - {{{4/2}}}

{{{7/2}}} = {{{7/2}}}

Edwin</pre>