Suppose that $ b^2 - 4ac 0$ for all x.

Question

Suppose that $ b^2 - 4ac < 0$. Show that there are no numbers x satisfying$$x^2 + bx + c = 0$$In fact, $ x^2 + bc + c > 0$ for all x.

jamesj · Answer

where's your a gone?

And in fact, it could be strictly positive or strictly negative.

anonymous · Answer

Imaginary numbers, becuase u have a square root of a negative.

anonymous · Answer

I don't know where the a went :( that's the question out of the book

anonymous · Answer

it says hint: complete the square

anonymous · Answer

is this of diff eqs??? becuase i just took a diff eqs test and it had this on it, so imaginary number are the product of  b2−4ac<0

anonymous · Answer

It's basic calculus

anonymous · Answer

from Spivak

jamesj · Answer

Well if we assume that a = 1, then

$$ x^2 + bx + c = (x + b/2a)^2 + 1/4a^2 ( 4ac - b^2 ) > 0  \ \ \ \ \forall \ x $$

anonymous · Answer

Now Spivak asks to prove that $ x^2 + xy + y^2 > 0$ provided that x and y are not both 0.

anonymous · Answer

I'm curious about that one. We know that x^2 and y^2 will always be positive regardless of the sign of x or y. I wonder how would can verify that x^2 + y^2 > xy, for x or y being negative.

jamesj · Answer

Well, it's fairly obvious what he wants you to do.  

EITHER treat this as a quadratic in x or y and use the relation you just proved.

OR repeat this method of completing the square

anonymous · Answer

Ah ha. $$\left ( x + {y \over 2} \right)^2 + {2y^2 \over 4}$$ All quantities are squared. http://www.wolframalpha.com/input/?i=x%5E2+%2B+xy+%2B+y%5E2

anonymous · Answer

Rather 3/4 in the second term.