SOLUTION: 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).

Algebra ->  Radicals -> SOLUTION: 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).      Log On


   

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).
Found 2 solutions by Earlsdon, ptaylor:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for w:
sqrt%285w-1%29+=+w-3 Square both sides.
5w-1+=+w%5E2-6w%2B9 Subtract 5w from both sides.
-1+=+w%5E2-11w%2B9 Add 1 to both sides.
0+=+w%5E2-11w%2B10 Factor this quadratic equation.
0+=+%28w-1%29%28w-10%29 Apply the zero products principle:
w-1+=+0 or w-10+=+0 so...
w+=+1 or w+=+10

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
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%285w-1%29 although it is a legal operation. You want to square both sides, in other words:
%28sqrt%285w-1%29%29%5E2=%28w-3%29%5E2 or
%285w-1%29=w%5E2-6w%2B9 subtract (5w-1) from both sides
0=w%5E2-6w-5w%2B9%2B1 collect like terms and rearrange

w%5E2-11w%2B10=0 quadratic in standard form and it can be factored. The factors are:
%28x-10%29%28x-1%29=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