Question

Consider the quadratic equation $a{x}^2+bx+(a^2+b^2+c^2-ab-bc-ca)=0$ (where $a,b,c$ are distinct real numbers).

If the graph of the quadratic polynomial is mouth opening downwards, then the product of roots of the corresponding quadratic equation is always

• A. positive
• B. negative
• C. depends on the sign of b
• D. can't determine

Answer 1

The correct option is B negative

$a{x}^2+bx+a^2+b^2+c^2-ab-bc-ca=0$
$a{x}^2+bx+\frac{{(a+b)}^2+{(b+c)}^2+(c+a{)}^2}{2}=0$
now,given that graph is concave downwards, so , $a<0$
and product of roots =$\frac{{(a+b)}^2+{(b+c)}^2+(c+a{)}^2}{2a}$
which is always negative