Question 295727
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The first thing you did that I wouldn't have done is choose to solve this quadratic using the quadratic formula.  Not that it is actually wrong to do so, it is just that factoring this one is so much easier.  However, since you chose to use the quadratic formula, let's do it properly.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-3)\ \pm\ \sqrt{(-3)^2\ -\ 4(1)(-28)}}{2(1)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{3\ \pm\ \sqrt{9\ -\ (-112)}}{2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{3\ \pm\ \sqrt{121}}{2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{3\ \pm\ 11}{2} ]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 7] or *[tex \LARGE x = -4]


Now, had you simply recognized that *[tex \Large -7\ \times\ 4\ =\ -28] and *[tex \Large -7\ +\ 4\ =\ -3], you could have written:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ -\ 28\ =\ (x\ -\ 7)(x\ +\ 4)\ =\ 0]


and therefore *[tex \Large x\ =\ 7] or *[tex \Large x\ =\ -4]



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