if b^2 - 3ac

Question

if b^2 - 3ac<0, and a is not 0, then show that ax^3+bx^2+cx+d=0 has exactly 1 real root.....   @MIT 18.02 Multiva…

myininaya · Answer

any thoughts satellite?

anonymous · Answer

yes

anonymous · Answer

your cubic polynomial is 
$$ax^3+bx^2+cx+d$$ and we know it has at least one real solution.  
the derivative is 
$$3ax^2+2bc+c$$ 
and if that has not real zeros then it is always positive (negative), implying your cubic polynomial is always increasing (decreasing) , meaning it can have only one zero.  

your condition above says exactly that the discriminant is negative, meaning the derivative has no zeros, so it is alway positive (negative) and so your function is always increasing (decreasing)

anonymous · Answer

ok a couple typos as usual.  derivative is 
$$3ax^2+2bx+c$$

myininaya · Answer

oh satellite i bet you wish you could go type-o free 
yeah i wasn't sure what knowledge w could use
but cal is awesome

myininaya · Answer

goodnight

anonymous · Answer

discriminant of that thing is 
$$4b^2-12ac$$ and if 
$$b^2-3ac<0\implies 4b^2-12ac<0$$ meaning the derivative is alway either above or below the x - axis aka always positive or negative

anonymous · Answer

good night to you as well.

anonymous · Answer

get some rest so you can torture your early morning class sufficiently

anonymous · Answer

*teach