INMO - 1998

Algebra Level 3

Suppose $a,b,c$ are three real numbers such that the quadratic equation

$x^2 - (a+b+c)x + (ab+bc+ca) = 0$

has roots of the form $\alpha + \iota\beta$ ( $\iota^2 = -1$ ), where $\alpha > 0$ & $\beta \neq 0$ are real numbers, then

a,b,c

are positive

a

c

are positive &

b

is negative None of These

a,b,c

are negative

1 solution

David Vaccaro
Jul 26, 2014

$x^{2}-(a+b+c)x+a(b+c)=k$ has real solutions for all $k\geq0$ . (Since there are real roots for $k=0$ at $x=a$ and $x=b+c$ .)

Since the given quadratic has no real roots we have $bc>0$ . Similarly $ac>0$ and $ab>0$ . so $a$ , $b$ and $c$ are all the same sign.

Since $2\alpha=(a+b+c)>0$ we must have that $a$ , $b$ and $c$ are positive.