Question 372140
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You can't, in general.  That is because you need two solutions to create a quadratic.  However, in some cases, the character of the solution that you do have allows you to deduce the solution that you don't have.


For example, if you have a solution that looks like


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ +\ \sqrt{3}}{2}]


Then, because of symmetry, the other solution is guaranteed to be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1\ -\ \sqrt{3}}{2}].


Likewise, if you have a single complex number solution, then the other solution is the conjugate of that complex number.


On the other hand, if you only have one rational number solution, you are at a dead end, unless you know that your single rational solution has a multiplicity of two.


Given that you do have two solutions, let's call them *[tex \LARGE \alpha] and *[tex \LARGE \beta], all you have to do is calculate the product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ \alpha)(x\ -\ \beta)]


And set that product equal to zero.

In the case where you have one solution with a multiplicity of two, meaning you have two identical solutions, you can calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ \alpha)^2]


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