Question 563703
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Use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-15\ \pm\ \sqrt{225\ +\ 40}}{2(5)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{-15\ +\ \sqrt{265}}{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \frac{-15\ -\ \sqrt{265}}{10}]


Flip the fractions and rationalize the new denominators (see <a href="http://www.algebra.com/algebra/homework/Radicals/rationalizingdenominators1.lesson">Rationalizing Denominators</a> if you need a refresher here)


To obtain:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{1r}\ =\ \frac{-15\ +\ \sqrt{265}}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{2r}\ =\ \frac{-15\ -\ \sqrt{265}}{4}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \left(\frac{-15\ +\ \sqrt{265}}{4}\right)\right)\ =\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \left(\frac{-15\ -\ \sqrt{265}}{4}\right)\right)\ =\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \left(\frac{-15\ +\ \sqrt{265}}{4}\right)\right)\left(x\ -\ \left(\frac{-15\ -\ \sqrt{265}}{4}\right)\right)\ =\ 0]


First, simplfy the signs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ \left(\frac{15\ -\ \sqrt{265}}{4}\right)\right)\left(x\ +\ \left(\frac{15\ +\ \sqrt{265}}{4}\right)\right)\ =\ 0]


Then use FOIL to multiply the two binomials.  Note that the second terms are a conjugate pair so their product is the difference of two squares.


Finally, multiply through by the denominator of the first degree and constant coefficients to write the equation with integer coefficients.


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