Problem with solving an equation

**Zhirkov** · Feb 17th 2013, 11:39 AM

I am stuck with this equation and would be glad if someone could help me out.
So please help me with how to solve this equation.

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

Thanks!

**veileen** · Feb 17th 2013, 12:13 PM

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.

**Paze** · Feb 17th 2013, 12:23 PM

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.

EDIT: Never mind, I see you have received an answer above.

**Zhirkov** · Feb 17th 2013, 12:41 PM

Thank you for your help!

**Zhirkov** · Feb 17th 2013, 01:36 PM

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?

**veileen** · Feb 18th 2013, 07:58 AM

You're welcome! ^^

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.

**LanellePalmer** · Feb 18th 2013, 08:48 PM

Thank you for your help!

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