Radical-Radical equation

**dhiab** · June 26th 2009, 06:46 AM

Solve in R this equation :

**earboth** · June 26th 2009, 07:16 AM

Originally Posted by dhiab

Solve in R this equation :

1. Substitute $u = \sqrt{x-1}$

then $u^2 = x-1$

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

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

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

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

3. Now resubstitute:

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

and solve for x

4. This part is for you.

**dhiab** · June 26th 2009, 08:14 AM

Hello : Thank you , but I'have one question :
why condition? 2=<u=<3

**earboth** · June 26th 2009, 10:23 AM

Originally Posted by dhiab

Hello : Thank you , but I'have one question :
why condition? 2=<u=<3

That isn't a condition but the solution of the equation!

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

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

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

3. You get three different domains of the equation:

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

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

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

Now collect all partial solutions and you'll get: $2 \leq u \leq 3$

**dhiab** · June 26th 2009, 11:49 AM

Originally Posted by earboth

That isn't a condition but the solution of the equation!

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

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

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

3. You get three different domains of the equation:

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

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

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

Now collect all partial solutions and you'll get: $2 \leq u \leq 3$

Hello, THANK YOU for resolution
Now I'have :

**earboth** · June 27th 2009, 12:35 AM

Originally Posted by dhiab

Hello, THANK YOU for resolution
Now I'have :

I'm not sure what exactly you have done ...

As I wrote in my previous post:

3. Now resubstitute:

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

and solve for x

you only have to get rid of the square-root. So square the complete inequality and you'll get:

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

I've attached the graph of the function.

**dhiab** · June 27th 2009, 03:38 AM

Originally Posted by earboth

I'm not sure what exactly you have done ...

As I wrote in my previous post:

you only have to get rid of the square-root. So square the complete inequality and you'll get:

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

I've attached the graph of the function.

THANK YOU

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