Question 1122911
<font face="Times New Roman" size="+2">


The way you wrote the equation, it cannot be solved, at least in the sense that I think you mean.  That is because *[tex \Large X] and *[tex \Large x] are two different things.  That means you have a single equation in two different variables.  You can solve for *[tex \Large X] in terms of *[tex \Large x], or for *[tex \Large x] in terms of *[tex \Large X].  However, I'm going to go out on a limb and assume that you were just too lazy to proofread and correct your post before you submitted it and that you actually meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ -\ 14x\ +\ 49\ =\ 0]


Factor out an *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\(x^2\ -\ 14x\ +\ 49)\ =\ 0]


Then factor the trinomial


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\(x\ -\ 7\)\(x\ -\ 7\)\ =\ 0]


Since at least one of the three factors must be zero so that the entire product is zero, either:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 0]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 7]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 7]


So you have to roots, 0 and 7, one of which has a multiplicity of two.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

</font>