anonymous
  • anonymous
Solve for x. [(SQRT.x+1)+(SQRT.x-1)]=(SQRT.2x+1)
Mathematics
katieb
  • katieb
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phi
  • phi
it's not clear what is inside the square root signs. can you use the equation editor?
anonymous
  • anonymous
\[\sqrt{x+1} + \sqrt{x-1} = \sqrt{2x+1}\]
anonymous
  • anonymous
I know that I need to square everything to get rid of my square roots

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phi
  • phi
if you had (a+b) and squared that: (a+b)(a+b) what do you get ?
anonymous
  • anonymous
\[a^{2}+2ab+b ^{2}\]
phi
  • phi
now we assume \( a= \sqrt{x+1} \) and \( b= \sqrt{x-1}\) in \[ a^{2}+2ab+b ^{2} \] what do you get ?
anonymous
  • anonymous
\[x ^{2}-1 \] ??
phi
  • phi
don't skip steps, just replace a^2, a and b, and b^2 in \[ a^{2}+2ab+b ^{2} \]
anonymous
  • anonymous
\[x ^{2}-x+x-1 \]
phi
  • phi
a^2 is x+1 b^2 is x-1 2*a*b is \(2 \sqrt{x+1}\sqrt{x-1} \) so you should get \[ x+1 + 2 \sqrt{x+1}\sqrt{x-1} + x-1 \]
phi
  • phi
we are doing \[ \left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2 \]
anonymous
  • anonymous
Okay....So I can't square \[\sqrt{x+1}\] and \[\sqrt{x-1}\] individually? I have to foil them together right?
phi
  • phi
I am not following you. yes, you can square \( \sqrt{x+1}\) to get x+1 but when we square \[ \left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2 \] we follow the same pattern as \[ (a+b)^2 = a^2 +2ab+b^2\] in other words \[ \left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2 = x+1 + 2 \sqrt{x+1}\sqrt{x-1} + x-1 \] that simplifies a little: \[ 2x+ 2 \sqrt{x+1}\sqrt{x-1} \]
phi
  • phi
meanwhile, we also square the right-hand side \[ \left( \sqrt{2x+1}\right)^2 \]
anonymous
  • anonymous
\[4x ^{2}+4x+1 \]??
anonymous
  • anonymous
Or would that just actually be 2x+1 ??
phi
  • phi
you did (2x+1)^2 it is easier than that square of a square root: drop the square root sign
phi
  • phi
yes, just 2x+1
phi
  • phi
so now we have \[ 2x+ 2 \sqrt{x+1}\sqrt{x-1} = 2x+1 \] I would add -2x to both sides as the next step.
anonymous
  • anonymous
Okay so \[2\sqrt{x+1}\sqrt{x-1} =1\]
phi
  • phi
yes. any idea what would be a good step now?
anonymous
  • anonymous
I know I need to do something with my square roots on the left side, but I don't know what..
phi
  • phi
use this idea: \[ (a\cdot b)^2 = a^2 \cdot b^2 \]
anonymous
  • anonymous
\[2\sqrt{x+1}^{2}\sqrt{x-1}^{2}\] ??
phi
  • phi
and also the 2 what you are doing is squaring the whole left side \[ ( 2 \sqrt{x+1} \sqrt{x-1} )^2 \] and that means multiply that mess times itself \[ 2 \sqrt{x+1} \sqrt{x-1} \cdot 2 \sqrt{x+1} \sqrt{x-1} \] when you multiply, you can chaange the order \[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1} \]
phi
  • phi
the "rule" is if you square (a*b) you square each term inside: a^2 * b^2 (this extends to (a*b*c)^2 or any number of factors: square each one)
anonymous
  • anonymous
Is this another instance where I can drop my square root sign??
phi
  • phi
yes, \[ \sqrt{x}\cdot \sqrt{x} =x \] we can use that rule twice in \[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1} \]
anonymous
  • anonymous
So, \[4\times(x+1)\times(x-1)\]
anonymous
  • anonymous
Would I distribute my four to the (x+1) and (x-1) and then foil??
phi
  • phi
yes, but we should not use \( \times\) in algebra (it looks like an x) we also have to square the right-hand side and write the whole equation
anonymous
  • anonymous
\[4[(x+1)*(x-1)] = 1^{2}\] \[(4x+4)(4x-4) = 1\] Am I good so far?
phi
  • phi
you only multiply by 4 once the easiest way is to first do (x+1)(x-1) you get x^2 -1 (difference of squares) now multiply by 4 4(x^2-1) = 4x^2 -4 that is all equal to the right side 4x^2-4 = 1
phi
  • phi
FYI, notice when you got (4x+4) (4x-4) that is the same as 4*(x+1) * 4 * (x-1) but we only have 4*(x+1)*(x-1)
anonymous
  • anonymous
I got \[\sqrt{\frac{ 1 }{ 2 }}\]
phi
  • phi
starting with \[ 4x^2-4 = 1 \] add +4 to both sides
anonymous
  • anonymous
like \[x = \sqrt{\frac{ 5 }{ 2 }}\]
phi
  • phi
closing in on it, but still wrong
anonymous
  • anonymous
oh my goodness. \[x = \frac{ \sqrt{5} }{ 4 }\]
phi
  • phi
4x^2 = 5 x^2 = 5/4 x= sqrt(5/4) which is also \[ \frac{\sqrt{5}}{\sqrt{4}}= \frac{\sqrt{5}}{2} \]
anonymous
  • anonymous
Okay got it. Thank you so much!!!
phi
  • phi
we should also check x= - sqrt(5) /2 but we can throw out the negative square root because in the original problem we have \( \sqrt{x-1}\) and if x is negative, we will have a negative number inside a square root and we don't allow that. in other words, it's a good idea to make a note that x= - sqrt(5)/2 is not allowed.
anonymous
  • anonymous
I really appreciate your help and patience with me. I am working on a review for a test I have tomorrow and was really struggling with this one. I really appreciate the help!!

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