Question 301413
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ 16x\ +\ 3\ =\ 0]


Step 1:  Divide through by the lead coefficient:


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


Step 2:  Add the Additive Inverse of the Constant term to both sides:


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


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


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


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


Step 4: Factor the perfect square in the LHS and simplify the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ 2\right)^2\ =\ \frac{13}{4}]


Step 5: Take the square root of both sides, remembering to consider both the positive and negative root


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 2\ =\ \pm\sqrt{\frac{13}{4}}\ =\ \frac{\pm\sqrt{13}}{2}]


Step 6: Finish solving:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2 +\frac{\pm\sqrt{13}}{2}\ =\ \frac{4\ \pm\ \sqrt{13}}{2}]


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