A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Solve for x. [(SQRT.x+1)+(SQRT.x-1)]=(SQRT.2x+1)

  • This Question is Closed
  1. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it's not clear what is inside the square root signs. can you use the equation editor?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know that I need to square everything to get rid of my square roots

  4. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if you had (a+b) and squared that: (a+b)(a+b) what do you get ?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[a^{2}+2ab+b ^{2}\]

  6. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now we assume \( a= \sqrt{x+1} \) and \( b= \sqrt{x-1}\) in \[ a^{2}+2ab+b ^{2} \] what do you get ?

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x ^{2}-1 \] ??

  8. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    don't skip steps, just replace a^2, a and b, and b^2 in \[ a^{2}+2ab+b ^{2} \]

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x ^{2}-x+x-1 \]

  10. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[ x+1 + 2 \sqrt{x+1}\sqrt{x-1} + x-1 \]

  11. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we are doing \[ \left( \sqrt{x+1}+\sqrt{x-1} \ \right)^2 \]

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay....So I can't square \[\sqrt{x+1}\] and \[\sqrt{x-1}\] individually? I have to foil them together right?

  13. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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: \[ 2x+ 2 \sqrt{x+1}\sqrt{x-1} \]

  14. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    meanwhile, we also square the right-hand side \[ \left( \sqrt{2x+1}\right)^2 \]

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[4x ^{2}+4x+1 \]??

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Or would that just actually be 2x+1 ??

  17. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you did (2x+1)^2 it is easier than that square of a square root: drop the square root sign

  18. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, just 2x+1

  19. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay so \[2\sqrt{x+1}\sqrt{x-1} =1\]

  21. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes. any idea what would be a good step now?

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know I need to do something with my square roots on the left side, but I don't know what..

  23. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    use this idea: \[ (a\cdot b)^2 = a^2 \cdot b^2 \]

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[2\sqrt{x+1}^{2}\sqrt{x-1}^{2}\] ??

  25. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1} \]

  26. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this another instance where I can drop my square root sign??

  28. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, \[ \sqrt{x}\cdot \sqrt{x} =x \] we can use that rule twice in \[ 2 \cdot 2\cdot \sqrt{x+1} \cdot \sqrt{x+1} \cdot \sqrt{x-1} \cdot \sqrt{x-1} \]

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So, \[4\times(x+1)\times(x-1)\]

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Would I distribute my four to the (x+1) and (x-1) and then foil??

  31. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[4[(x+1)*(x-1)] = 1^{2}\] \[(4x+4)(4x-4) = 1\] Am I good so far?

  33. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  34. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got \[\sqrt{\frac{ 1 }{ 2 }}\]

  36. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    starting with \[ 4x^2-4 = 1 \] add +4 to both sides

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    like \[x = \sqrt{\frac{ 5 }{ 2 }}\]

  38. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    closing in on it, but still wrong

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh my goodness. \[x = \frac{ \sqrt{5} }{ 4 }\]

  40. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    4x^2 = 5 x^2 = 5/4 x= sqrt(5/4) which is also \[ \frac{\sqrt{5}}{\sqrt{4}}= \frac{\sqrt{5}}{2} \]

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay got it. Thank you so much!!!

  42. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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!!

  44. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.