Question

The equation $a{x}^2+bx+c=0$ does not have real roots and $c<0$. Which of these is true?

• A. $b>a+c$
• B. Both $\left(a\right)$ & $\left(b\right)$
• C. $4a+2b+c<0$
• D. $a+b+c>0$

Answer 1

The correct option is B

Both $\left(a\right)$ & $\left(b\right)$

Given equations does not have any real roots.

$D<0\Rightarrow b^2-4ac<0$

$b^2<4ac$

$4ac>b^2$

$4ac>0$ $[$As $b^2>0]$

$ac>0$

As $c<0$, $a$ must be $-ve$ for $ac>0$.

So, $a<0$ & $D<0$ $\Rightarrow$ Graph is downward & does not touch $x$-axis

So, $f\left(x\right)<0$

$f(-1)=a-b+c<0\Rightarrow b>a+c$

$f\left(2\right)=4a+2b+c<0$

So, both options $a$ & $b$ are true.