document.write( "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). \n" ); document.write( "
Algebra.Com's Answer #56766 by ptaylor(2198)\"\" \"About 
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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).\r
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\n" ); document.write( "\n" ); document.write( "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:\r
\n" ); document.write( "\n" ); document.write( "\"%28sqrt%285w-1%29%29%5E2=%28w-3%29%5E2\" or\r
\n" ); document.write( "\n" ); document.write( "\"%285w-1%29=w%5E2-6w%2B9\" subtract (5w-1) from both sides\r
\n" ); document.write( "\n" ); document.write( "\"0=w%5E2-6w-5w%2B9%2B1\" collect like terms and rearrange\r
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\n" ); document.write( "\n" ); document.write( "\"w%5E2-11w%2B10=0\" quadratic in standard form and it can be factored. The factors are:
\n" ); document.write( "\"%28x-10%29%28x-1%29=0\"
\n" ); document.write( "\"x=10\"
\n" ); document.write( "and\r
\n" ); document.write( "\n" ); document.write( "\"x=1\"\r
\n" ); document.write( "\n" ); document.write( "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. \r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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