# Find the Real Solutions

• Oct 7th 2009, 10:01 AM
Find the Real Solutions
I need to find the real solutions to a simple equation:

$\displaystyle \sqrt{x+5}=x$

$\displaystyle x+5=x^2$

$\displaystyle x^2-x-5=0$

$\displaystyle x=\frac{1\pm \sqrt{21}}{2}$

It seems that there should be two real solutions to the equation. However, I am told that the equation has only 1 real solution. (Headbang)
• Oct 7th 2009, 10:26 AM
earboth
Quote:

I need to find the real solutions to a simple equation:

$\displaystyle \sqrt{x+5}=x$

$\displaystyle x+5=x^2$

$\displaystyle x^2-x-5=0$

$\displaystyle x=\frac{1\pm \sqrt{21}}{2}$

It seems that there should be two real solutions to the equation. However, I am told that the equation has only 1 real solution. (Headbang)

From the first equation you know that $\displaystyle \sqrt{x+5}$ is a positive real number. Therefore x must be a positive real number.
But your last equation has also a negative number which can't be a solution of the original equation.
• Oct 7th 2009, 10:30 AM
PatrickFoster
Only one solution
There is only one real solution: 2.79129.

In your first step, you squared both sides. You forgot that it is "plus or minus" though, when you do that. Make sense?