# Solve equation

• Oct 12th 2013, 02:21 AM
Paze
Solve equation
I'm having severe difficulties solving this equation:

$\displaystyle \frac{\sqrt{x^2-50x+1525}}{x-25}+\frac{\sqrt{x^2+400}}{x}=0$

I tried most conventional algebra methods but none seem to work! Thanks.
• Oct 12th 2013, 06:06 AM
Paze
Re: Solve equation
Really need help on this one. It's a string in a series of long calculus computations that are due for monday :)
• Oct 12th 2013, 06:31 AM
Shakarri
Re: Solve equation
You can use conventional algebra methods

$\displaystyle \frac{\sqrt{x^2-50x+1525}}{x-25}=-\frac{\sqrt{x^2+400}}{x}$

Cross-multiply

$\displaystyle (x)\sqrt{x^2-50x+1525}=-(x-25)\sqrt{x^2+400}$

Then square both sides. It starts as a nasty equation but simplifies to something quite nice.
• Oct 12th 2013, 06:38 AM
Paze
Re: Solve equation
Quote:

Originally Posted by Shakarri
You can use conventional algebra methods

$\displaystyle \frac{\sqrt{x^2-50x+1525}}{x-25}=-\frac{\sqrt{x^2+400}}{x}$

Cross-multiply

$\displaystyle (x)\sqrt{x^2-50x+1525}=-(x-25)\sqrt{x^2+400}$

Then square both sides. It starts as a nasty equation but simplifies to something quite nice.

Thanks. I realized later that $\displaystyle \left((a+b)(\sqrt{c+d})\right)^2$ actually simplifies to $\displaystyle \left((a+b)^2(c+d)\right)$

I hadn't realized this before and it made the calculation that much easier! Thanks!