getting some trouble

with dis

square root of x^2+6x=x+square root of 2x

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- Jul 23rd 2010, 06:49 PMmath321rational expression
getting some trouble

with dis

square root of x^2+6x=x+square root of 2x - Jul 23rd 2010, 07:14 PMskeeter
- Jul 23rd 2010, 07:27 PMArchie Meade
- Jul 23rd 2010, 09:11 PMhotspice
wondering what will be wrong in solving algebraically to find x which eventually equals 2? as problem is neither a quadratic equation nor an inequality?

- Jul 24th 2010, 12:25 AMCaptainBlack
Please use brackets to make you equation unambiguous, the above could plausibly mean either:

square_root(x^2+6x)=x+square_root(2x)

or

square_root(x^2)+6x=x+square_root(2x)

I know that the first of these is the most likely, but someone will always fail to make that assumption and interpret it differently.

CB - Jul 24th 2010, 05:26 AMArchie Meade
It finds the x=2 solution, but not the x=0 solution.

Quadratic equations may be solved by factoring, but factoring when possible, finds all solutions and is not confined to quadratics.

zero multiplied by another value is zero, so the factored version shows that x=0 is also a solution.

If we divide both sides by x, we lose that particular solution, though it finds the x=2 answer.

Also remember that we really should not divide by x because we cannot divide by zero anyway,

so when we do that, we are finding only non-zero solutions.

Do check CaptainBlack's post also.