Question

If both the roots of $a{x}^2+bx+c=0$ are negative, then

• A. $\Delta >0,ab>0,bc<0$
• B. $\Delta >0,a,b,c,$ have the same signs
• C. $\Delta <0,ab>0,ac<0$
• D. $\Delta <0,ab>0,bc>0$

Answer 1

The correct option is A $\Delta >0,ab>0,bc<0$

$a{x}^2+bx+c=0,$ where $a\ne 0.$

Condition for real roots of a quadratic equation is $b^2\ge 4ac.$ This condition has

to be true. As for the second condition to be true, all coefficient should be

positive.

The proof is easy. If you know a little calculus then you can find that a quadratic

function reaches its extremum when $x=\frac{-b}{2a}.$ This result can be derived via

rearranging the terms in the form of $a(x+p{)}^2+q.$ Also we know that the

extremum is always halfway between the two roots. So when both of the roots

are negative then the extremum should also be negative.

$-\frac{b}{2a}<0$

or,$\frac{b}{a}>0$

or, $\frac{ab}{a^2}>0$

or, $ab>0.$

So both a and b should have same sign. Without loss of generality it would be

safe to assume that both a and b is positive (if they were negative then multiply

the quadratic by (-1)). The general from of the roots are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now we have two cases to consider.

First one is when $b^2-4ac=0$: The roots would become automatically zero as

both a and b is zero.

Second one is when $b^2-4ac>0$: We need to the behavior of the root

nearer to zero. If both of the roots are less than zero then so should be the nearer

one. As both a and b are positive,so the root nearer to zero would be

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . If this is less than zero then,

$-b+\sqrt{b^2-4ac}<0$

or, $b>\sqrt{b^2-4ac}$

or, $b^2>b^2-4ac$

or, $ac>0.$

So all three of a,b, c have the same sign. This is the condition on the coefficient.