Question

If $a{x}^2+bx+c=0$ has no real roots and $a,b,c,\in R$ such that $a+c>0$, then

• A. $a+b+c>0$
• B. $a-b+c>0$
• C. $a+c=b$
• D. $a+b+c=0$

Answer 1

Answer: (A), (B)

The correct options are
A $a+b+c>0$
B $a-b+c>0$

Since roots of $a{x}^2+bx+c=0$ are not real
The graph of $y=a{x}^2+bx+C$ either lies above the x-axis or below the x-axis
This will depend up on the sign of $a$
Now $y\left(1\right)=a+b+c$ and $y(-1)=a-b+c$
Thus $y\left(1\right)+y(-1)=2(a+c)>0$ given
This mean at least one of $y\left(1\right)$ or $y(-1)$ is +ve
But we know graph will lie above or below the x-axis
Which means both $y\left(1\right)$ and $y(-1)$ are +ve
Hence option A and B are correct.