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anonymous
 one year ago
Solve for x.
[(SQRT.x+1)+(SQRT.x1)]=(SQRT.2x+1)
anonymous
 one year ago
Solve for x. [(SQRT.x+1)+(SQRT.x1)]=(SQRT.2x+1)

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phi
 one year ago
Best ResponseYou've already chosen the best response.1it's not clear what is inside the square root signs. can you use the equation editor?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x+1} + \sqrt{x1} = \sqrt{2x+1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know that I need to square everything to get rid of my square roots

phi
 one year ago
Best ResponseYou've already chosen the best response.1if you had (a+b) and squared that: (a+b)(a+b) what do you get ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a^{2}+2ab+b ^{2}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1now we assume \( a= \sqrt{x+1} \) and \( b= \sqrt{x1}\) in \[ a^{2}+2ab+b ^{2} \] what do you get ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1don't skip steps, just replace a^2, a and b, and b^2 in \[ a^{2}+2ab+b ^{2} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1a^2 is x+1 b^2 is x1 2*a*b is \(2 \sqrt{x+1}\sqrt{x1} \) so you should get \[ x+1 + 2 \sqrt{x+1}\sqrt{x1} + x1 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1we are doing \[ \left( \sqrt{x+1}+\sqrt{x1} \ \right)^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay....So I can't square \[\sqrt{x+1}\] and \[\sqrt{x1}\] individually? I have to foil them together right?

phi
 one year ago
Best ResponseYou've already chosen the best response.1I am not following you. yes, you can square \( \sqrt{x+1}\) to get x+1 but when we square \[ \left( \sqrt{x+1}+\sqrt{x1} \ \right)^2 \] we follow the same pattern as \[ (a+b)^2 = a^2 +2ab+b^2\] in other words \[ \left( \sqrt{x+1}+\sqrt{x1} \ \right)^2 = x+1 + 2 \sqrt{x+1}\sqrt{x1} + x1 \] that simplifies a little: \[ 2x+ 2 \sqrt{x+1}\sqrt{x1} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1meanwhile, we also square the righthand side \[ \left( \sqrt{2x+1}\right)^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or would that just actually be 2x+1 ??

phi
 one year ago
Best ResponseYou've already chosen the best response.1you did (2x+1)^2 it is easier than that square of a square root: drop the square root sign

phi
 one year ago
Best ResponseYou've already chosen the best response.1so now we have \[ 2x+ 2 \sqrt{x+1}\sqrt{x1} = 2x+1 \] I would add 2x to both sides as the next step.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so \[2\sqrt{x+1}\sqrt{x1} =1\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes. any idea what would be a good step now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know I need to do something with my square roots on the left side, but I don't know what..

phi
 one year ago
Best ResponseYou've already chosen the best response.1use this idea: \[ (a\cdot b)^2 = a^2 \cdot b^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2\sqrt{x+1}^{2}\sqrt{x1}^{2}\] ??

phi
 one year ago
Best ResponseYou've already chosen the best response.1and also the 2 what you are doing is squaring the whole left side \[ ( 2 \sqrt{x+1} \sqrt{x1} )^2 \] and that means multiply that mess times itself \[ 2 \sqrt{x+1} \sqrt{x1} \cdot 2 \sqrt{x+1} \sqrt{x1} \] when you multiply, you can chaange the order \[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x1} \cdot \sqrt{x1} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1the "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
 one year ago
Best ResponseYou've already chosen the best response.0Is this another instance where I can drop my square root sign??

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, \[ \sqrt{x}\cdot \sqrt{x} =x \] we can use that rule twice in \[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x1} \cdot \sqrt{x1} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, \[4\times(x+1)\times(x1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would I distribute my four to the (x+1) and (x1) and then foil??

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, but we should not use \( \times\) in algebra (it looks like an x) we also have to square the righthand side and write the whole equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[4[(x+1)*(x1)] = 1^{2}\] \[(4x+4)(4x4) = 1\] Am I good so far?

phi
 one year ago
Best ResponseYou've already chosen the best response.1you only multiply by 4 once the easiest way is to first do (x+1)(x1) you get x^2 1 (difference of squares) now multiply by 4 4(x^21) = 4x^2 4 that is all equal to the right side 4x^24 = 1

phi
 one year ago
Best ResponseYou've already chosen the best response.1FYI, notice when you got (4x+4) (4x4) that is the same as 4*(x+1) * 4 * (x1) but we only have 4*(x+1)*(x1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got \[\sqrt{\frac{ 1 }{ 2 }}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1starting with \[ 4x^24 = 1 \] add +4 to both sides

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like \[x = \sqrt{\frac{ 5 }{ 2 }}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1closing in on it, but still wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh my goodness. \[x = \frac{ \sqrt{5} }{ 4 }\]

phi
 one year ago
Best ResponseYou've already chosen the best response.14x^2 = 5 x^2 = 5/4 x= sqrt(5/4) which is also \[ \frac{\sqrt{5}}{\sqrt{4}}= \frac{\sqrt{5}}{2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay got it. Thank you so much!!!

phi
 one year ago
Best ResponseYou've already chosen the best response.1we should also check x=  sqrt(5) /2 but we can throw out the negative square root because in the original problem we have \( \sqrt{x1}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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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