# 3x + 1 - x - 1 = 2

1. Dec 6, 2004

### footprints

How do you solve this? $$\sqrt{3x+1} - \sqrt{x-1} = 2$$
I know it isn't 3x + 1 - x - 1 = 2.
This isn't homework. I just saw it somewhere else.

2. Dec 6, 2004

### StatusX

square both sides, then rearrange it so the radical thats left is on one side by itself, and then square again.

3. Dec 6, 2004

### footprints

Ok so I rearrange it and I get $$\sqrt{3x+1}=2+\sqrt{x-1}$$
When you say square it, won't I get 3x +1 = 2 + x - 1?

4. Dec 6, 2004

### fourier jr

$$\sqrt{3x+1} - \sqrt{x-1} = 2$$

square both sides & simplify:
$$2(x-1) = \sqrt{(3x+1)(x-1)}$$

square both sides again:
$$4(x-1)^2 = (3x+1)(x-1)$$

move everything over to the left-hand side:
$$4(x-1)^2 - (3x+1)(x-1) = 0$$

simplify:
$$4(x-1)^2 - (3x+1)(x-1) = (x-1)[4(x-1)-3x-1] = (x-1)(x-5) = 0$$

from the original equation, $$1 \leq x$$ (so no sqrt of negative #s)

so it looks like the solution is x=1 & x=5

how did i do everybody?

Last edited: Dec 6, 2004
5. Dec 6, 2004

### boaz

another way to do it but similar to yours is the following :

$\sqrt{3x+1}-\sqrt{x-1}=2 \rightarrow \sqrt{3x+1}=\sqrt{x-1}+2$
$\sqrt{3x+1}^2=(\sqrt{x-1}+2)^2 \rightarrow 3x+1=x-1+2\cdot2\sqrt{x-1}+4$
$3x+1=x-1+2\cdot2\sqrt{x-1}+4 \rightarrow 2x-2=2\cdot2\sqrt{x-1}$
$x-1=2\sqrt{x-1} \rightarrow (x-1)^2=4\sqrt{x-1}^2$
$x^2-2x+1=4(x-1)$

note : after finding the solution, you have to check if it is the answer by positioning it in the first equation. :shy:

6. Dec 7, 2004

### footprints

fourier jr: How did you go from $$\sqrt{3x+1} - \sqrt{x-1} = 2$$ to $$2(x-1) = \sqrt{(3x+1)(x-1)}$$? Did you skip any steps? Cuz I'm a bit slow.

boaz: Where did the 2 from $$2\sqrt{x-1}+4$$ come from?

Last edited: Dec 7, 2004
7. Dec 7, 2004

### Galileo

Looks good to me

Square both sides of:

$$\sqrt{3x+1} - \sqrt{x-1} = 2$$

You'll get:
$$(3x+1)+(x-1)-2\sqrt{(3x+1)(x-1)}=4$$

which can be rearranged to:
$$2(x-1)=\sqrt{(3x+1)(x-1)}$$

8. Dec 7, 2004

### boaz

according to next formula :
$(a+b)^2=a^2+2ab+b^2$
$a=2; b=\sqrt{x-1}$

9. Dec 7, 2004

### fourier jr

ya i squared, simplified & put the square-root stuff one side in one step

10. Dec 7, 2004

### HallsofIvy

I think it is slightly easier to follow if you shift $\sqrt{x-1}$ to the other side first: $\sqrt{3x+1}= \sqrt{x-1}+ 2$ . Squaring both sides gives
$3x+ 1= x-1+ 4\sqrt{x-1}+ 4$.

Now subtract x+ 3 from both sides: $2x- 2= 4\sqrt{x-1}$ and divide by 2.
$x- 1= 2\sqrt{x-1}$ and square again.

(x-1)2= x2- 2x+ 1= 4(x- 1) so x2- 6x+ 5= 0.
That factors as (x- 5)(x- 1)= 0 which has x= 1 and x= 5 as solutions.

Checking: if x= 1 then 3x+1= 4 and x-1= 0. Yes $\sqrt{4}-\sqrt{0}= 2$.
If x= 5, then 3x+1= 16 and x- 1= 4. Yes, $\sqrt{16}- \sqrt{4}= 4-2= 2$.

11. Dec 8, 2004

### footprints

Oh so thats how you get it
Thank to all who helped.

12. Dec 13, 2004

### taurinne

u guys makes it too complicated. why dont u just substitute 'x' by 1

13. Dec 13, 2004

### fourier jr

because although x=1 works, it may not be the only solution. after fiddling with the original equation you get a quadratic which has 2 solutions.

Last edited: Dec 13, 2004