3x + 1 - x - 1 = 2

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

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  • #2
StatusX
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square both sides, then rearrange it so the radical thats left is on one side by itself, and then square again.
 
  • #3
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Ok so I rearrange it and I get [tex]\sqrt{3x+1}=2+\sqrt{x-1}[/tex]
When you say square it, won't I get 3x +1 = 2 + x - 1?
 
  • #4
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footprints said:
How do you solve this? [tex]\sqrt{3x+1} - \sqrt{x-1} = 2[/tex]
I know it isn't 3x + 1 - x - 1 = 2.
This isn't homework. I just saw it somewhere else.

start with:
[tex] \sqrt{3x+1} - \sqrt{x-1} = 2[/tex]

square both sides & simplify:
[tex] 2(x-1) = \sqrt{(3x+1)(x-1)} [/tex]

square both sides again:
[tex] 4(x-1)^2 = (3x+1)(x-1) [/tex]

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

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

from the original equation, [tex] 1 \leq x [/tex] (so no sqrt of negative #s)


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

how did i do everybody? :blushing:
 
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  • #5
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another way to do it but similar to yours is the following :

[itex]\sqrt{3x+1}-\sqrt{x-1}=2 \rightarrow \sqrt{3x+1}=\sqrt{x-1}+2[/itex]
[itex]\sqrt{3x+1}^2=(\sqrt{x-1}+2)^2 \rightarrow 3x+1=x-1+2\cdot2\sqrt{x-1}+4[/itex]
[itex]3x+1=x-1+2\cdot2\sqrt{x-1}+4 \rightarrow 2x-2=2\cdot2\sqrt{x-1}[/itex]
[itex]x-1=2\sqrt{x-1} \rightarrow (x-1)^2=4\sqrt{x-1}^2[/itex]
[itex]x^2-2x+1=4(x-1)[/itex]

note : after finding the solution, you have to check if it is the answer by positioning it in the first equation. :shy:
 
  • #6
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fourier jr: How did you go from [tex] \sqrt{3x+1} - \sqrt{x-1} = 2[/tex] to [tex] 2(x-1) = \sqrt{(3x+1)(x-1)} [/tex]? Did you skip any steps? Cuz I'm a bit slow.

boaz: Where did the 2 from [tex]2\sqrt{x-1}+4[/tex] come from?
 
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  • #7
Galileo
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fourier jr said:
how did i do everybody? :blushing:
Looks good to me :biggrin:

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

boaz: Where did the 2 from [tex]2\sqrt{x-1}+4[/tex] come from?
Square both sides of:

[tex]\sqrt{3x+1} - \sqrt{x-1} = 2[/tex]

You'll get:
[tex](3x+1)+(x-1)-2\sqrt{(3x+1)(x-1)}=4[/tex]

which can be rearranged to:
[tex]2(x-1)=\sqrt{(3x+1)(x-1)}[/tex]
 
  • #8
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footprints said:
boaz: Where did the 2 from [tex]2\sqrt{x-1}+4[/tex] come from?
according to next formula :
[itex](a+b)^2=a^2+2ab+b^2[/itex]
however, in your example :
[itex]a=2; b=\sqrt{x-1}[/itex]
 
  • #9
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footprints said:
fourier jr: How did you go from [tex] \sqrt{3x+1} - \sqrt{x-1} = 2[/tex] to [tex] 2(x-1) = \sqrt{(3x+1)(x-1)} [/tex]? 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
 
  • #10
HallsofIvy
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I think it is slightly easier to follow if you shift [itex]\sqrt{x-1}[/itex] to the other side first: [itex]\sqrt{3x+1}= \sqrt{x-1}+ 2[/itex] . Squaring both sides gives
[itex]3x+ 1= x-1+ 4\sqrt{x-1}+ 4[/itex].

Now subtract x+ 3 from both sides: [itex]2x- 2= 4\sqrt{x-1}[/itex] and divide by 2.
[itex]x- 1= 2\sqrt{x-1}[/itex] 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 [itex]\sqrt{4}-\sqrt{0}= 2[/itex].
If x= 5, then 3x+1= 16 and x- 1= 4. Yes, [itex]\sqrt{16}- \sqrt{4}= 4-2= 2[/itex].
 
  • #11
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boaz said:
according to next formula :
[itex](a+b)^2=a^2+2ab+b^2[/itex]
however, in your example :
[itex]a=2; b=\sqrt{x-1}[/itex]
Oh so thats how you get it :rolleyes:
Thank to all who helped. :smile:
 
  • #12
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u guys makes it too complicated. why dont u just substitute 'x' by 1
 
  • #13
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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.
 
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