Solve inRthisequation: (Rofl)

http://www.mathramz.com/xyz/latexren...79548e6986.png (Headbang)

Printable View

- Jun 26th 2009, 06:46 AMdhiabRadical-Radical equation
**Solve in**R**this**equation**: (Rofl)**

**http://www.mathramz.com/xyz/latexren...79548e6986.png (Headbang)** - Jun 26th 2009, 07:16 AMearboth
1. Substitute $\displaystyle u = \sqrt{x-1}$

then $\displaystyle u^2 = x-1$

2. $\displaystyle \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ becomes:

$\displaystyle \sqrt{u^2+4-4u}+\sqrt{u^2+9-6u}=1$ which is:

$\displaystyle \sqrt{(u-2)^2}+\sqrt{(u-3)^2}=1$ and you get:

$\displaystyle |u-2|+|u-3|=1~\implies~2\leq u \leq 3$

3. Now resubstitute:

$\displaystyle 2\leq \sqrt{x-1} \leq 3$

and solve for x

4. This part is for you. - Jun 26th 2009, 08:14 AMdhiab
**Hello : Thank you , but I'have one question :**

**why condition? 2=<u=<3** - Jun 26th 2009, 10:23 AMearboth
That isn't a condition but the solution of the equation!

Use the definition of the absolute value: $\displaystyle |a|=\left\{\begin{array}{r}a, a \geq 0\\-a, a < 0\end{array}\right.$

1. $\displaystyle |u-2|= \left\{\begin{array}{r}u-2, u \geq 2\\-(u-2), u < 2\end{array}\right.$

2. $\displaystyle |u-3|= \left\{\begin{array}{r}u-3, u \geq 3\\-(u-3), u < 3\end{array}\right.$

3. You get three different domains of the equation:

$\displaystyle u\geq 3 : u-2+u-3=1~\implies~2u=6~\implies~u=3$

$\displaystyle 2 \leq u < 3 : u-2-u+3=1~\implies~0=0~\implies~2 \leq u < 3$

$\displaystyle u < 2 : -u+2-u+3=1~\implies~-2u=-4~\implies~u=2$

Now collect all partial solutions and you'll get: $\displaystyle 2 \leq u \leq 3$ - Jun 26th 2009, 11:49 AMdhiab
**Hello, THANK YOU for resolution**

**Now I'have**:

http://www.mathramz.com/xyz/latexren...9321421221.png

http://www.mathramz.com/xyz/latexren...7d0a11f294.png

http://www.mathramz.com/xyz/latexren...f61cf2eaa4.png

http://www.mathramz.com/xyz/latexren...dee623e757.png(Wait) - Jun 27th 2009, 12:35 AMearboth
I'm not sure what exactly you have done ... (Wondering)

As I wrote in my previous post:

Quote:

3. Now resubstitute:

$\displaystyle 2\leq \sqrt{x-1} \leq 3$

and solve for x

$\displaystyle 2\leq \sqrt{x-1} \leq 3~\implies~4\leq x-1 \leq 9~\implies~\boxed{5\leq x \leq 10}$

I've attached the graph of the function. - Jun 27th 2009, 03:38 AMdhiab