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Math Help - Passage of a norm of a vector exercise

  1. #1
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    Passage of a norm of a vector exercise

    Hi,

    I've an example of vector exercise that achieved the following step onwards to resolving. |vector r| is the norm of the vector.

    |vector r|^2 = 100(25t^2-14t+2) = 100((5t-(14/10)^2) - (14/10)^2 +2)

    I don't understand how the passage is made, because for me the values inside the parameter of the first count is a quadratic function who has the consequent result:
    t=(-14+/-squarert(-4))/50.

    Can anybody explain me how the passage of the |vector r|^2 is made?


    Thanks,
    Pedro
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  2. #2
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    Hi Pedro

    (5t-\frac{14}{10})^2\;=\;25t^2-14t+(\frac{14}{10})^2

    Therefore
    25t^2-14t+2\;=\;(5t-\frac{14}{10})^2-(\frac{14}{10})^2+2

    I hope this will help you
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  3. #3
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    Quote Originally Posted by pedrosacosta View Post
    Hi,

    I've an example of vector exercise that achieved the following step onwards to resolving. |vector r| is the norm of the vector.

    |vector r|^2 = 100(25t^2-14t+2) = 100((5t-(14/10))^2 - (14/10)^2 +2)

    I don't understand how the passage is made, because for me the values inside the parameter of the first count is a quadratic function who has the consequent result:
    t=(-14+/-squarert(-4))/50.

    Can anybody explain me how the passage of the |vector r|^2 is made?


    Thanks,
    Pedro
    You've made an insidious typo: See the correction in red.

    Here is nothing done but completing the square:

    |\vec r|^2=100(25t^2-14t+2)=100((5t)^2-14t+\left(\frac{14}{10}\right)^2 - \left(\frac{14}{10}\right)^2+2) = 100\left(\left(5t-\left(\frac{14}{10}\right)\right)^2-\left(\frac{14}{10}\right)^2+2\right)

    You only have to translate the ^2 behind the closing paranthese.
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