# Thread: Problem with solving an equation

1. ## 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!

2. ## Re: Problem with solving an equation

First things first:

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

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

So our x is strictly bigger than 2.

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

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

If $\displaystyle x \neq 5$:

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

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

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

So x = 5 is the only solution.

3. ## Re: Problem with solving an equation

Hi Zhirkov.

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

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

5. ## 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?

6. ## Re: Problem with solving an equation

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:

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

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

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

$\displaystyle -\sqrt{x-2}=1$ is impossible because $\displaystyle \sqrt{x-2}>0 \Rightarrow -\sqrt{x-2}<0$ and 1 > 0.