Question 674484
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The only way to do it, since it won't factor and you don't want to use the quadratic formula, is to complete the square.


Divide the coefficient on *[tex \LARGE x] by 2, then square the result:  *[tex \LARGE \left(\frac{1}{4}\right)^2\ =\ \frac{1}{16}].  Add to both sides.


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


Factor the perfect square trinomial in the LHS and simplify the RHS.


*[tex \LARGE \left(x\ -\ \frac{1}{4}\right)^2\ =\ \frac{19}{48}]


Take the square root of both sides:


*[tex \LARGE x\ -\ \frac{1}{4}\ =\ \pm\ \sqrt{\frac{19}{48}}]


Add the additive inverse of the constant in the LHS to both sides:


*[tex \LARGE x\ =\ \frac{1}{4}\ \pm\ \sqrt{\frac{19}{48}}]


Now all you have to do is to rationalize the denominator and apply the LCD to combine the two fractions in the RHS.


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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