Question 68462
{{{sqrt(3x-2)+sqrt(2x+2)+1=0}}}  First thing that you want to do is get rid of the radicals by squaring both sides but before you do that you want to rearrange the terms on either side of the equal sign in a manner that simplifies that process.  Sometimes you will need to square both sides more than once.  Lets move the 1 to the right side of the equal sign by subtracting 1 from both sides:

{{{sqrt(3x-2)+sqrt(2x+2)=-1}}}   Now lets square both sides:

{{{(3x-2)+2sqrt((2x+2)(3x-2))+(2x+2)=+1}}}   Lets collect like terms:

{{{5x+2sqrt((2x+2)(3x-2))=+1}}}    Now we will subtract 5x from both sides

{{{2sqrt((2x+2)(3x-2))=1-5x}}}   Next lets square both sides again

{{{4(2x+2)(3x-2)=1-10x+25x^2}}}  Simplifying some we get

{{{8(x+1)(3x-2)=1-10x+25x^2}}}    Next lets clear parens

{{{24x^2+8x-16=1-10x+25x^2}}} Subtract 1 and 25x^2 from and add 10x to both sides

{{{24x^2-25x^2+10x+8x-16-1=0}}} Collecting like terms:

{{{-x^2+18x-17=0}}}  Multiply through by -1

{{{x^2-18x+17=0}}}  ---------quadratic in standard form

This quadratic can be factored:

{{{(x-1)(x-17)=0}}}

{{{x=1}}}

and

{{{x=17}}}

ck

substitute 1 in the original equation:

{{{sqrt(3x-2)+sqrt(2x+2)+1=0}}}  equals

{{{sqrt(3-2)+sqrt(2+2)+1=0}}} and this equals
{{{sqrt(1)+sqrt(4)+1=0}}}
(+or-1)+(+or-2)+1=0
{{{+1-2+1=0}}}
{{{0=0}}}

verify that it also checks for x=17


Hope this helps---ptaylor