SOLUTION: Hi, I need help on solving extraneous equations. (square root)3x=(squareroot)x+6 (squareroot)3x+2-(SR)2x+7=0 (SR)2x+6-(SR)x-1=2 thank you!

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hi, I need help on solving extraneous equations. (square root)3x=(squareroot)x+6 (squareroot)3x+2-(SR)2x+7=0 (SR)2x+6-(SR)x-1=2 thank you!      Log On


   

Question 711991: Hi, I need help on solving extraneous equations.
(square root)3x=(squareroot)x+6
(squareroot)3x+2-(SR)2x+7=0
(SR)2x+6-(SR)x-1=2
thank you!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
There are a few problems with what you posted:
  • Every equation you posted is ambiguous. None of them make clear what is inside the square root. For example, the first equation could be any of the following:
    sqrt%283%29x=sqrt%28x%29%2B6
    sqrt%283x%29=sqrt%28x%29%2B6
    sqrt%283%29x=sqrt%28x%2B6%29
    sqrt%283x%29=sqrt%28x%2B6%29
    You have parentheses around the words "square root". Putting parentheses around the radicand (the expression within a radical is called a radicand) is much more important.
  • I don't think "extraneous equations" is a meaningful phrase. Solving square root equations like these can result in what are called "extraneous solutions".
Please re-post your question using parentheses to make the equations unambiguous.