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5 November, 15:13

Solve

9x^2 + 42x + 49 = 0

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  1. 5 November, 18:18
    0
    Because of the relatively large coefficients {9, 42, 49}, applying the quadratic formula would be a bit messy. Instead, I've chosen to "complete the square:"

    9x^2 + 42x + 49 = 0 can be re-written as 9 [ x^2 + (42/9) x ] = - 49

    Dividing both sides by 9, we get [ x^2 + (42/9) x ] = - 49/9

    Completing the square: [ x^2 + (42/9) x + (21/9) ^2 - (21/9) ^2 ] = - 49/9

    [ x + 21/9 ]^2 = 441/81 - 441/81 = 0

    Then [ x + 21/9 ] = 0, and x = - 21/9 (this is a double root).
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