Question 715877
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Step 1:  Divide through by the lead coefficient.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{1}{4}x\ +\ \frac{3}{4}\ =\ 0]


Step 2: Add the opposite of the constant term to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{1}{4}x\ =\ -\frac{3}{4}]


Step 3: Divide the coefficient on the first degree term by 2, square the result, and then add that result to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{1}{4}x\ +\ \frac{1}{64}\ =\ -\frac{3}{4}\ +\ \frac{1}{64}]


Step 4: The result of step 3 is a perfect square trinomial in the LHS.  Factor it.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ \frac{1}{8}\right)^2\ =\ -\frac{47}{64}]


Step 5: Take the square root of both sides.  Remember to consider both positive and negative roots.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{1}{8}\ =\ \pm\frac{i\sqrt{47}}{8}]


Step 6: Add the opposite of the constant term in the LHS to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\  \frac{-1\ \pm{i\sqrt{47}}}{8}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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