Find the Real Solutions

**adkinsjr** · October 7th 2009, 10:01 AM

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

$\sqrt{x+5}=x$

$x+5=x^2$

$x^2-x-5=0$

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

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

**earboth** · October 7th 2009, 10:26 AM

Originally Posted by adkinsjr

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

$\sqrt{x+5}=x$

$x+5=x^2$

$x^2-x-5=0$

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

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

From the first equation you know that $\sqrt{x+5}$ 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.

**PatrickFoster** · October 7th 2009, 10:30 AM

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?

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