Question 497133
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Let's follow the process and maybe you can figure out where you went astray.


Step 1:  Constant term to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ 16x\ =\ -13]


Step 2:  Divide by the lead coefficient:


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


Step 3:  Divide the 1st degree term coefficient by 2.  Add the square of that result to both sides:  -4 divided by 2 is -2.  -2 squared is 4.  Add 4 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ +\ 4\ =\ -\frac{13}{4}\ +\ 4\ =\ -\frac{13}{4}\ +\ \frac{16}{4}]


Step 4:  Factor the perfect square trinomial in the LHS and do the arithmetic in the RHS:


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


Step 5:  Take the root of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 2\ =\ \pm\ \frac{\sqrt{3}}{2}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2\ \pm\ \frac{\sqrt{3}}{2}]


Which you could render, if you were so inclined:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{4\ \pm\ \sqrt{3}}{2}]


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