Question

If $f\left(x\right)=a{x}^2+bx+c$ has no real zeros and $a+b+c<0$, then

• A. $c=0$
• B. $c>0$
• C. $c<0$
• D. None of these

Answer 1

The correct option is C $c<0$

Given: $f\left(x\right)=a{x}^2+bx+c$ has no real zeroes and $a+b+c<0$

To find : $c=0$ or $c>0$ or $c<0$

$c\ne 0$

$\because$ If $c=0$, then $f\left(x\right)$ will have real roots which is not true $(\because x=-\frac{-b\pm \sqrt{b^2-4a\left(0\right)}}{2a})$

$\because a+b+c<0$

$\Rightarrow f\left(1\right)<0$

$\Rightarrow f\left(1\right)$ lies below $x-$axis $----\left(I\right)$

$\because f\left(x\right)$ has no real roots.

$\therefore f\left(x\right)\ne 0$ for any real values of $x$

$\Rightarrow$ graoh of $f\left(x\right)$ does not cut $x-$axis$----\left(II\right)$

$\Rightarrow f\left(0\right)=c<0$ (From $\left(I\right)$ & $\left(II\right)$)