Confused??

[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.

[StatusX]

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

[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?

[fourier jr]

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.


start with:
\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? :blushing:

[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:

[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?

[Galileo]

how did i do everybody? :blushing:
Looks good to me :biggrin:

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)}

[boaz]

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
however, in your example :
a=2; b=\sqrt{x-1}

[fourier jr]

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]

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.

[footprints]

according to next formula :
(a+b)^2=a^2+2ab+b^2
however, in your example :
a=2; b=\sqrt{x-1}
Oh so thats how you get it :rolleyes:
Thank to all who helped. :smile:

[taurinne]

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

[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.