Question 328801
<font face="Garamond" size="+2">


Solve 


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


by completing the square


Step 1: Move the constant term to the right hand side.


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


Step 2: Multiply both sides by *[tex \Large \frac{1}{a}]


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


Step 3: Divide the coefficient on the first order term by 2, square the result, and add that result to both sides:


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


Step 4: Determine and apply the LCD in the RHS:


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


Step 5: Factor the perfect square trinomial in the LHS:


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


Step 6: Take the square root of both sides. Consider both positive and negative roots:


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


Step 7: Add the opposite of the constant term in the LHS to both sides:


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


Simplify if possible and you are done.  You can find your own example to use.


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