Question 998153
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As you wrote the problem, it cannot be solved.  That is because K and k are two distinctly different things.  If I presume that you meant: *[tex \Large kx^2\ +\ (k\ +\ 2)x\ -\ 3\ =\ 0], then in order for the roots to be real, *[tex \Large (k\ +\ 2)^2\ -\ 4k(-3)\ \geq\ 0].  In order for both roots to be positive, *[tex \Large \frac{-(k\ +\ 2)\ +\ \sqrt{(k\ +\ 2)^2\ -\ 4k(-3)}}{2k}\ \geq\ 0] and *[tex \Large \frac{-(k\ +\ 2)\ -\ \sqrt{(k\ +\ 2)^2\ -\ 4k(-3)}}{2k}\ \geq\ 0] must both hold.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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