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Math Help - u substitution

  1. #1
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    u substitution

    i hope im doing it right :P

    Solve:

     \frac{1}{(x+1)^2} + \frac{4}{x+1} +4=0

    if u = x+1 then

     \frac{1}{(u)^2} + \frac {4}{u}+4=0

    Look for LCD =  u^2

     (u)^2 *\frac{1}{(u)^2} + (u)^2 *\frac {4}{u}+ (u)^2*4= 0

     \frac{1u^2}{(u)^2} + \frac {4u^2}{u}+ 4u^2 = 0

     1+ 4u+ 4u^2 = 0

     4u^2+ 4u+1  = 0

     4u^2+ 4u  = -1

    I dont really know if im doing this right... please help..
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  2. #2
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    You don't multiply 1/u^2 by u^2, you multiply the other two terms by u^2.
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  3. #3
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    like this?

    <br />
\frac{1}{(u)^2} + (u)^2 *\frac {4}{u}+ (u)^2*4= 0<br />
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  4. #4
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    Quote Originally Posted by Anemori View Post
    i hope im doing it right :P

    Solve:

     \frac{1}{(x+1)^2} + \frac{4}{x+1} +4=0

    if u = x+1 then

     \frac{1}{(u)^2} + \frac {4}{u}+4=0

    Look for LCD =  u^2

     (u)^2 *\frac{1}{(u)^2} + (u)^2 *\frac {4}{u}+ (u)^2*4= 0

     \frac{1u^2}{(u)^2} + \frac {4u^2}{u}+ 4u^2 = 0

     1+ 4u+ 4u^2 = 0

     4u^2+ 4u+1 = 0

     4u^2+ 4u = -1

    I dont really know if im doing this right... please help..
    \frac{1}{u^2} + \frac{4}{u} + 4 = 0.


    Multiply both sides of the equation by u^2.


    u^2\left(\frac{1}{u^2} + \frac{4}{u} + 4\right) = 0u

    1 + 4u + 4u^2 = 0.


    Now you have a Quadratic equation - use the Quadratic Formula or some other method to solve for u.
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  5. #5
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    using quadratic formula:

    i hope this is right way...

     x = \frac {-4 +- \sqrt4^2- 4(4)(1)}{2(4)}


     x = \frac {-4 +- \sqrt16- (16)}{8}

     x = \frac {-4 +- \sqrt0}{8}

     x = \frac {-4 +- 0}{8}

     x = \frac {-4 +- 0}{8}

     x = -\frac{1}{2}

    Is that right?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Anemori View Post
    using quadratic formula:

    i hope this is right way...

     x = \frac {-4 +- \sqrt4^2- 4(4)(1)}{2(4)}


     x = \frac {-4 +- \sqrt16- (16)}{8}

     x = \frac {-4 +- \sqrt0}{8}

     x = \frac {-4 +- 0}{8}

     x = \frac {-4 +- 0}{8}

     x = -\frac{1}{2}

    Is that right?
    If you have done what Prove it told you to do, then it is correct.
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  7. #7
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    How do you check if it is right?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Anemori View Post
    How do you check if it is right?
    Plug it back in.
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  9. #9
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    1+ 4u+ 4u^2 factors rather easily as (1+ 2u)^2. That gives u= -\frac{1}{2}, not x! Since you took u= x+ 1 you now know that x+1= -\frac{1}{2}.
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