# Problem with solving an equation

• Feb 17th 2013, 11:39 AM
Zhirkov
Problem with solving an equation
I am stuck with this equation and would be glad if someone could help me out.

(x-5)/((x+1)^(1/2))=(-x^2+7*x-10)/((x^2-x-2)^(1/2))

Thanks!
• Feb 17th 2013, 12:13 PM
veileen
Re: Problem with solving an equation
First things first:

$x+1> 0\Rightarrow x > -1$

$x^2-x-2>0\Rightarrow (x+1)(x-2)>0\Rightarrow x\in (-\infty , -1) \cup (2, \infty)$

So our x is strictly bigger than 2.

$\frac{x-5}{\sqrt{x+1}}=\frac{-x^2+7x-10}{\sqrt{x^2-x-2}} \Leftrightarrow \frac{x-5}{\sqrt{x+1}}=\frac{-(x-2)(x-5)}{\sqrt{(x+1)(x-2)}}$

If x = 5, we obtain 0 = 0 which is true, so 5 is one of the solutions.

If $x \neq 5$:

$\frac{x-5}{\sqrt{x+1}}=\frac{-(x-2)(x-5)}{\sqrt{(x+1)(x-2)}}\Leftrightarrow$

$\frac{x-5}{\sqrt{x+1}}=\frac{(x-5)}{\sqrt{x+1}}\cdot \frac{-(x-2)}{\sqrt{x-2}}$

$\Leftrightarrow \frac{-(x-2)}{\sqrt{x-2}}=1\Leftrightarrow -\sqrt{x-2}=1$ Impossible.

So x = 5 is the only solution.
• Feb 17th 2013, 12:23 PM
Paze
Re: Problem with solving an equation
Hi Zhirkov.

You have the equation: $\frac{x-5}{\left({x+1}\right)^{1/2}}=\frac{-x^2+7x-10}{\left({x^2-x-2}\right)^{1/2}}$?

Because it is an unnecessary long computation as far as I know.

• Feb 17th 2013, 12:41 PM
Zhirkov
Re: Problem with solving an equation
• Feb 17th 2013, 01:36 PM
Zhirkov
Re: Problem with solving an equation
A couple of question though if it is okay. How do go from the second last step to the last one and what do you mean with impossible? You can still solve the last step of the equation and get a value of X out of that. And is it possible to just compute the equation until you get x=5?
• Feb 18th 2013, 07:58 AM
veileen
Re: Problem with solving an equation
You're welcome! ^^

Quote:

Originally Posted by Zhirkov
A couple of question though if it is okay. How do go from the second last step to the last one and what do you mean with impossible? You can still solve the last step of the equation and get a value of X out of that. And is it possible to just compute the equation until you get x=5?

Starting from here:

$\frac{x-5}{\sqrt{x+1}}=\frac{(x-5)}{\sqrt{x+1}}\cdot \frac{-(x-2)}{\sqrt{x-2}}$

I multiplied the relation with $\frac{\sqrt{x+1}}{x-5} \neq 0$ and obtained:

$1=\frac{-(x-2)}{\sqrt{x-2}} \Leftrightarrow 1=\frac{-(\sqrt{x-2})^2}{\sqrt{x-2}} \Leftrightarrow 1=-\sqrt{x-2}$

$-\sqrt{x-2}=1$ is impossible because $\sqrt{x-2}>0 \Rightarrow -\sqrt{x-2}<0$ and 1 > 0.
• Feb 18th 2013, 08:48 PM
LanellePalmer
Re: Problem with solving an equation