# 3x + 1 - x - 1 = 2

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.

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

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?

footprints said:
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.

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

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:
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:

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:
Galileo
Homework Helper
fourier jr said:
how did i do everybody? Looks good to me footprints said:
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?
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:
$$2(x-1)=\sqrt{(3x+1)(x-1)}$$

footprints said:
boaz: Where did the 2 from $$2\sqrt{x-1}+4$$ come from?
according to next formula :
$(a+b)^2=a^2+2ab+b^2$
$a=2; b=\sqrt{x-1}$

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

HallsofIvy
Homework Helper
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, $\sqrt{16}- \sqrt{4}= 4-2= 2$.

boaz said:
according to next formula :
$(a+b)^2=a^2+2ab+b^2$
$a=2; b=\sqrt{x-1}$
Oh so thats how you get it Thank to all who helped. 