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14 January, 14:34

How many and what type of solutions does the equation have? 2k² = 9 + 3k

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  1. 14 January, 16:42
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    2k² = 9 + 3k

    ⇒ 2k² - 9 - 3k = 0

    ⇒ 2k² - 3k - 9 = 0

    b² - 4ac = 3² - 4*2 * (-9) = 81>0

    It has 2 real roots.
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