# rational expression

• Jul 23rd 2010, 06:49 PM
math321
rational expression
getting some trouble
with dis

square root of x^2+6x=x+square root of 2x
• Jul 23rd 2010, 07:14 PM
skeeter
Quote:

Originally Posted by math321
getting some trouble
with dis

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

square both sides ...

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

combine like terms ...

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

factor ...

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

set each factor equal to 0 and solve ... check solutions in the original equation.
• Jul 23rd 2010, 07:27 PM
Quote:

Originally Posted by math321
getting some trouble
with dis

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

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

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

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

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

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

Do not do the following....

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

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

as you lose a solution.

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

$x\left(4-2\sqrt{2x}\right)=0$
• Jul 23rd 2010, 09:11 PM
hotspice
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 AM
CaptainBlack
Quote:

Originally Posted by math321
getting some trouble
with dis

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

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 AM
Quote:

Originally Posted by hotspice
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?

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.

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

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

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.