Question 773597
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*[tex \LARGE \ \ \ \ \ \ \ \ \ k\,\cdot\,2^x\ +\ 2^{-x}\ =\ 3]


Multiply by *[tex \LARGE 2^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ k\,\cdot\,2^{2x}\ +\ 1\ =\ 3\,\cdot\,2^x]


Let *[tex \LARGE u\ =\ 2^x], substitute, and put the resulting quadratic in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ ku^2\ -\ 3u\ +\ 1\ =\ 0]


Start the process to complete the square:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ u^2\ -\ \frac{3}{k}u\ +\ \frac{9}{4k^2}\ =\ -\frac{1}{k}\ +\ \frac{9}{4k^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \left(u\ -\ \frac{3}{2k}\right)^2 =\ -\frac{1}{k}\ +\ \frac{9}{4k^2}]


In order for *[tex \LARGE ku^2\ -\ 3u\ +\ 1\ =\ 0] to have two identical roots (or one root with a multiplicity of 2, if you prefer) then *[tex \LARGE  -\frac{1}{k}\ +\ \frac{9}{4k^2}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \frac{1}{k}\ =\ \frac{9}{4k^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ k(4k\ -\ 9)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ k\ =\ 2.25] or *[tex \LARGE k\ =\ 0]


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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