Question 322195
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The general solution of a quadratic equation in standard form, namely *[tex \LARGE ax^2\ +\ bx\ +\ c\ =\ 0] is:


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


Example:


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


Here, *[tex \Large a\ =\ 1], *[tex \Large b\ = -2], and *[tex \Large c\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-2)\ \pm\ \sqrt{(-2)^2\ -\ 4(1)(-3)}}{2(1)}\ =\ \frac{2\ \pm\ \sqrt{4\ +\ 12}}{2}\ =\ \frac{2\ \pm\ 4}{2}\ =\ 1\ \pm\ 2]


Therefore, *[tex \Large x\ =\ 3] or *[tex \Large x\ =\ -1]


Example 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 1\ =\ 0]


Here, *[tex \Large a\ =\ 1], *[tex \Large b\ = -1], and *[tex \Large c\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-1)\ \pm\ \sqrt{(-1)^2\ -\ 4(1)(-1)}}{2(1)}\ =\ \frac{1\ \pm\ \sqrt{1\ +\ 4}}{2}\ =\ \frac{1\ \pm\ \sqrt{5}}{2}]


Therefore, *[tex \Large x\ =\ \frac{1\ +\ \sqrt{5}}{2}] or *[tex \Large x\ =\ \frac{1\ -\ \sqrt{5}}{2}]


Example 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ +\ 4\ =\ 0]


Here, *[tex \Large a\ =\ 1], *[tex \Large b\ = -1], and *[tex \Large c\ =\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-1)\ \pm\ \sqrt{(-1)^2\ -\ 4(1)(4)}}{2(1)}\ =\ \frac{1\ \pm\ \sqrt{1\ -\ 16}}{2}\ =\ \frac{1\ \pm\ \sqrt{-15}}{2}\ =\ \frac{1\ \pm\ i\sqrt{15}}{2}]


Therefore, *[tex \Large x\ =\ \frac{1\ +\ i\sqrt{15}}{2}] or *[tex \Large x\ =\ \frac{1\ -\ i\sqrt{15}}{2}]


Where *[tex \Large i] is the imaginary number defined by *[tex \Large i^2\ =\ -1]


Where did it all come from?


Consider the general form of the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


Add the additive inverse of the constant term to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ =\ -c]


Multiply both sides by the reciprocal of the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{b}{a}x\ =\ \frac{-c}{a}]


Divide the first degree coefficient by 2, square the result, then add that result to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{b}{a}x\ +\ \frac{b^2}{4a^2}\ =\ \frac{-c}{a}\ +\ \frac{b^2}{4a^2}]


Factor the perfect square LHS and combine the fractions in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ \frac{b}{2a}\right)^2\ =\ \frac{b^2\ -\ 4ac}{4a^2}]


Take the square root of both sides, considering both positive and negative roots:


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


Add *[tex \Large \frac{-b}{2a}] to both sides:


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


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