getting some trouble

with dis

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

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- Jul 23rd 2010, 05:49 PMmath321rational expression
getting some trouble

with dis

square root of x^2+6x=x+square root of 2x - Jul 23rd 2010, 06:14 PMskeeter
- Jul 23rd 2010, 06:27 PMArchie Meade
$\displaystyle \sqrt{x^2+6x}=x+\sqrt{2x}$

$\displaystyle \sqrt{x^2+6x}=\sqrt{\left(x+\sqrt{2x}\right)^2}$

$\displaystyle \sqrt{x^2+6x}=\sqrt{x^2+2x\sqrt{2x}+2x}$

$\displaystyle 6x=2x\sqrt{2x}+2x$

$\displaystyle 6x-2x=4x=2x\sqrt{2x}$

Do not do the following....

$\displaystyle \frac{4x}{2x}=2=\sqrt{2x}$

$\displaystyle 2=\sqrt{4}=\sqrt{2x}$

as you lose a solution.

Instead, factor...

$\displaystyle 4x-2x\sqrt{2x}=0$

$\displaystyle x\left(4-2\sqrt{2x}\right)=0$ - Jul 23rd 2010, 08:11 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 23rd 2010, 11:25 PMCaptainBlack
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, 04:26 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.

$\displaystyle 2x=x\sqrt{2x}$

$\displaystyle x\left(2-\sqrt{2x}\right)=0$

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.