Real Numbers?

Question

If p and q are real, prove that the expression

x^2 + x(p + q) + p^2 -pq +q^2

cannot be negative

If p and q are real, prove that the expression

x^2 + x(p + q) + p^2 -pq +q^2

cannot be negative

*Not Homework*

Answer 1

Consider the discriminant of this quadratic in x:
(p+q)^2 - 4p^2 + 4pq - 4q^2
= p^2 + 2pq + q^2 - 4p^2 + 4pq - 4q^2
= -3p^2 + 6pq - 3q^2
= -3(p^2 - 2pq + q^2)
= -3(p-q)^2.

Since p-q is real, and a real number squared is positive, this discriminant is always negative; thus our quadratic in x has no real roots, and therefore is always positive (or always be negative, but the x^2 term is positive, so this is not the case).

Answer 2

Well, if the expression is ever negative, then it has to have real roots. Solving for x, we find that the roots contain the expression

√(-(p-q)²)

In other words, for any real p and q, it's always a complex number, and therefore x cannot have any real roots nor is ever negative.

Answer 3

It's a parabola..right?
And it's max/min value is got by differentiating
> y'=2x+(p+q) and y'=0 where x=-(p+q)/2 (min/max TV)
subst this in2 ur orig equation..
> y=3(p-q)^2/4 always =/>0. for p,q={real}

Answer 4

Complete the square, and you should see that the expression is the sum of two squares, and thus cannot be negative.

Answer 5

is this your homework?