Last problem for awhile. here is the question

umm. not sure how to put square root into latex but here is the equation.

Square root of 3x+1 - square root of x-1 =2

I know that there will be squaring going on twice because of the 2 radicals but thats where I get stuck.
Help would be great here. Thanks again.

2. Originally Posted by Geomatt
Last problem for awhile. here is the question

umm. not sure how to put square root into latex but here is the equation.

Square root of 3x+1 - square root of x-1 =2

I know that there will be squaring going on twice because of the 2 radicals but thats where I get stuck.
Help would be great here. Thanks again.
$\displaystyle \sqrt {3x + 1} - \sqrt {x - 1} = 2$

$\displaystyle \Rightarrow \left( \sqrt {3x + 1} - \sqrt {x - 1} \right)^2 = 2^2$

$\displaystyle \Rightarrow 3x + 1 - 2 \sqrt {(3x + 1)(x - 1)} + x - 1 = 4$

$\displaystyle \Rightarrow -2 \sqrt {(3x + 1)(x - 1)} = 4 - 4x$

$\displaystyle \Rightarrow \left( -2 \sqrt {(3x + 1)(x - 1)} \right)^2 = (4 - 4x)^2$

$\displaystyle \Rightarrow 4(3x + 1)(x - 1) = 16 - 32x + 16x^2$

$\displaystyle \Rightarrow 16x^2 - 32x + 16 - 4(3x^2 - 2x - 1) = 0$

$\displaystyle \Rightarrow 4x^2 - 24x + 20 = 0$

$\displaystyle \Rightarrow x^2 - 6x + 5 = 0$

$\displaystyle \Rightarrow (x - 5)(x - 1) = 0$

$\displaystyle \Rightarrow x = 5 \mbox { or } x = 1$

Checking both these solutions into the original equation, we find that both work

3. thanks very much! One question though,
I dont understand how the square root of (3x+1- the square root of x-1)^2
results in the next step. I cant figure out basically how step 2 results in step 3

4. Originally Posted by Geomatt
thanks very much! One question though,
I dont understand how the square root of (3x+1- the square root of x-1)^2
results in the next step. I cant figure out basically how step 2 results in step 3
You know that
$\displaystyle (a - b)^2 = a^2 - 2ab + b^2$

Let
$\displaystyle a = \sqrt{3x + 1}$
$\displaystyle b = \sqrt{x - 1}$

Then
$\displaystyle a^2 = (\sqrt{3x + 1})^2 = 3x + 1$
$\displaystyle 2ab = 2\sqrt{3x + 1}\sqrt{x - 1}$
$\displaystyle b^2 = (\sqrt{x - 1})^2 = x - 1$

So
$\displaystyle (\sqrt{3x + 1} - \sqrt{x - 1})^2 = (3x + 1) - 2\sqrt{3x + 1}\sqrt{x - 1} + (x - 1)$

-Dan

5. Originally Posted by Geomatt
thanks very much! One question though,
I dont understand how the square root of (3x+1- the square root of x-1)^2
results in the next step. I cant figure out basically how step 2 results in step 3
Hi,

use the formula:

$\displaystyle (a - b)^2 = a^2 - 2ab + b^2$
Now plug in the first root instaed of a and the second root instead of b and you will get the next step.