Question 330251
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Use the quadratic formula for all 4 of these.


For a quadratic equation in standard form:  *[tex \LARGE ax^2\ +\ bx +\ c\ =\ 0] the two roots are given by:


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


For your first one:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -2\ \ ], and *[tex \LARGE c\ =\ -13]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-2)\ \pm\ \sqrt{(-2)^2\ -\ 4(1)(-13)}}{2(1)}]


Just do the arithmetic and simplify.  Do the other three exactly the same way.  Remember, if the discriminant, that is the part of the formula that is under the radical, is negative, then you have 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].  Hint:  This is the situation in the 2nd of your problems.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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