Question 799788
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"Sole" the equation?  Is that "sole" as in the fish or as in the bottom part of a shoe?


If you actually want to "solve" the equation by completing the square, read on.


1.  Put the terms with variables in the LHS, and leave the constant term in the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ =\ 3]


2.  Divide the coefficient on the first degree term by 2, square the result, and add that result to both sides of the equation.  2 divided by 2 is 1, 1 squared is 1, add 1 to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ 4]


3.  Factor the perfect square polynomial in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)^2\ =\ 4]


4.  Take the square root of both sides (Remember both positive and negative roots)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ \pm2]


5.  Solve both linear equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -3]


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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