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Can someone help me solve this quadratic !

  1. May 25, 2016 #1
    1. The problem statement, all variables and given/known data
    I can't determine where the 40x comes from.
    From: 4x + 5 = 8√(1 - x)
    To: 16x^2 +40x + 25 = 64 - 64x^2
    2. Relevant equations
    √1-x^2 = √1-x *√x+1

    3. The attempt at a solution
    4x + 5 = 8√(1-x)
    4x^2 + 5^2 = (8 - 8x)*(8x + 8)
    16x^2 +25 = 64 - 64x^2
     
  2. jcsd
  3. May 25, 2016 #2

    Math_QED

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    What is (a+b)^2 equal to?
     
  4. May 25, 2016 #3
    I am not sure what you mean?
     
  5. May 25, 2016 #4
    (a+b)^2 = a^2 + b^2
     
  6. May 25, 2016 #5

    Math_QED

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    this is a frequently made mistake.

    Well (a+b)^2 = (a+b)(a+b) = ...
    Use distributivity
     
  7. May 25, 2016 #6
    Ahhhh I see, thank you!
     
  8. May 25, 2016 #7

    Mark44

    Staff: Mentor

    What you wrote as a relevant equation doesn't apply here. ##[8\sqrt{1 - x}]^2 \ne (8 - 8x)(8x + 8)##
     
  9. May 27, 2016 #8
    alternatively ##4x+5=8√(1-x)##, can be expressed as, ##(4x+5)^2=64(1-x)##
     
  10. May 27, 2016 #9

    Math_QED

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    One should be careful because those might not be equivalent.
     
  11. May 28, 2016 #10
    ok elaborate please.
     
  12. May 28, 2016 #11

    Math_QED

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    The first equation (the original one) has one solution for x. The second one has 2 solutions for x. We use indeed that => but to make sure you can use <=> you have to say that one solution for x is not valid.
     
  13. May 28, 2016 #12

    Mark44

    Staff: Mentor

    The basic idea is that if you square both sides of an equation, the new equation might have solutions that don't satisfy the original equation. Here's a very simple example:

    (Eqn 1) x = -2
    Square both sides to get
    (Eqn 2) ##x^2 = 4##
    The first equation has -2 as its solution. The second equation has -2 and 2 as its solutions.
     
  14. May 29, 2016 #13
    agreed , we can have the two solutions as stated in post 12 ,but we can go ahead and state that one solution does not satisfy the equation, my take in post number 8, is to indicate that one way of solving such quadratics is by squaring both sides, then determine later which solutions satisfy the original equation.
     
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