Question 631890
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Multiply the equation by the reciprocal of the lead coefficient:


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


Add the opposite of the constant term to both sides:


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

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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{3}{2}x\ +\ \frac{9}{16}\ =\ -\frac{1}{2}\ +\ \frac{9}{16}]


Factor the perfect square trinomial in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \frac{3}{4}\right)^2\ =\ \frac{1}{16}]


Take the square root of both sides, remembering to consider both the positive and negative roots


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{3}{4}\ =\ \pm\sqrt{\frac{1}{16}}]


Simplify and specify both roots:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{1}{2}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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