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Math Help - how to solve by square root principle

  1. #1
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    how to solve by square root principle

    Solve by using square root principle

    [ (x+1)/2 ]^2 - (x/2) = 2

    Answer key: √7

    I tried the following:

    [ (x+1)/2 ]^2 - (x/2) = 2

    (x^2/4) + (1/4) - (x/2) = 2

    -bring the (1/4) to RHS

    (x^2/4) - (x/2) = (7/4)

    -LCD 4

    x^2 - 2x = 7

    I'm stuck now as I have a middle term (-2x), otherwise the square root principle would work out nice. How to solve this using the square root principle?

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shenton View Post
    Solve by using square root principle

    [ (x+1)/2 ]^2 - (x/2) = 2

    Answer key: √7

    I tried the following:

    [ (x+1)/2 ]^2 - (x/2) = 2

    (x^2/4) + (1/4) - (x/2) = 2

    -bring the (1/4) to RHS

    (x^2/4) - (x/2) = (7/4)

    -LCD 4

    x^2 - 2x = 7

    I'm stuck now as I have a middle term (-2x), otherwise the square root principle would work out nice. How to solve this using the square root principle?

    Thanks.
    [ (x+1)/2 ]^2 - (x/2) = 2

    expand the square:

    [ x^2/4+x/2+1/4 ]- (x/2) = 2

    x^2+1=8

    x^2=7.

    RonL
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  3. #3
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    Thanks!

    This part is confusing:

    [ (x/2) + (1/2) ]^2

    I worked it out as:

    (x^2/4) + (x/4) + (x/4) + (1/4)

    Add (x/4) with (x/4), this gives me (2x/4):

    (x^2/4) + (2x/4) + (1/4)

    Simplify (2x/4) by dividing by 2, this gives me (x/2):

    (x^2/4) + (x/2) + (1/4)

    Are the above workings right? Is that how I expand the binomial fraction?

    Thanks.
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  4. #4
    Senior Member OReilly's Avatar
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    Quote Originally Posted by shenton View Post
    Thanks!

    This part is confusing:

    [ (x/2) + (1/2) ]^2

    I worked it out as:

    (x^2/4) + (x/4) + (x/4) + (1/4)

    Add (x/4) with (x/4), this gives me (2x/4):

    (x^2/4) + (2x/4) + (1/4)

    Simplify (2x/4) by dividing by 2, this gives me (x/2):

    (x^2/4) + (x/2) + (1/4)

    Are the above workings right? Is that how I expand the binomial fraction?

    Thanks.
    Yes, you have expand it ok.
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  5. #5
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    Thanks for confirming the workings.
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