- #1

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I know it isn't 3x + 1 - x - 1 = 2.

This isn't homework. I just saw it somewhere else.

- Thread starter footprints
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- #1

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I know it isn't 3x + 1 - x - 1 = 2.

This isn't homework. I just saw it somewhere else.

- #2

StatusX

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- #3

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When you say square it, won't I get 3x +1 = 2 + x - 1?

- #4

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footprints said:

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?

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- #5

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[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]

- #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?

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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Looks good to mefourier jr said:how did i do everybody?

Square both sides of: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?

[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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according to next formula :footprints said:boaz: Where did the 2 from [tex]2\sqrt{x-1}+4[/tex] come from?

[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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ya i squared, simplified & put the square-root stuff one side in one stepfootprints 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.

- #10

HallsofIvy

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[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)

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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Oh so thats how you get itboaz 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]

Thank to all who helped.

- #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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