Question 235514
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If you have a quadratic equation of the form *[tex \LARGE ax^2\ +\ bx\ + c\ =\ 0], then the two roots can be calculated using the coefficients of the quadratic trinomial thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a}]


In your case, the quadratic is in *[tex \Large h] and the coefficients are *[tex \Large a\ =\ 9], *[tex \Large b\ =\ -6], and *[tex \Large c\ =\ 7].  Just plug in the numbers and do the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-6) \pm sqrt{(-6)^2 - 4(9)(7)}}{2(9)}]


In this problem, the quantity under the radical, that is to say *[tex \Large b^\ -\ 4ac] is less than zero.  That means that your two roots will be a conjugate pair of complex roots of the form *[tex \Large a\ \pm\ bi] where *[tex \Large i] is the imaginary number defined by *[tex \Large i^2\ =\ -1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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