Question 312470
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You have two difficulties here.

First is the fact that you are trying to take the square root of a negative number.  Remember, *[tex \Large \sqrt{-1}] is not defined in the real numbers -- no wonder your calculator is pitching a fit!  In order to find an approximation of the square root of -73 you need to factor it into two parts, (-1)(73).  Use the definition of the imaginary number *[tex \LARGE i], that is *[tex \LARGE i^2\ =\ -1], to write *[tex \Large \sqrt{-73}\ =\ i\sqrt{73}\ \approx 8.544i]


But that problem is secondary to the fact that you improperly calculated the value under the radical in the quadratic formula.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ -\ 7x\ +\ 2\ =\ 0]


For this equation, *[tex \Large a\ =\ 3], *[tex \Large b\ =\ -7], and *[tex \Large c\ =\ 2]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-7)\ \pm\ \sqrt{(-7)^2\ -\ 4(3)(2)}}{2(3)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{7\ \pm\ \sqrt{49\ -\ 24}}{6} ]


Note: -7 times -7 is +49.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{7\ \pm\ \sqrt{25}}{6} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{1}{3} ]



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