Question 79106
I need to solve for w and find the solution set for the following: Sqrt (5w-1) = w-3. I multiplied both sides by sqrt (5w-1) and got (5w-1) = w-3(sqrt (5w-1).


At least you gave it a shot but you don't want to multiply both sides by the {{{sqrt(5w-1)}}} although it is a legal operation.  You want to square both sides, in other words:

{{{(sqrt(5w-1))^2=(w-3)^2}}} or

{{{(5w-1)=w^2-6w+9}}} subtract (5w-1) from both sides

{{{0=w^2-6w-5w+9+1}}} collect like terms and rearrange


{{{w^2-11w+10=0}}}  quadratic in standard form and it can be factored.  The factors are:
{{{(x-10)(x-1)=0}}} 
{{{x=10}}}
and

{{{x=1}}}

When you are dealing with a quadratic and the A coefficient is 1, then the B coefficient will be the sum of the factors of the C coefficient, if the quadratic is factorable. 




Hope this helps---ptaylor