Question

If a > 0 and the equation $a{x}^2+bx+c=0$ has two real roots $\alpha$ and $\beta$ such that $\left|\alpha \right|\le 1,\left|\beta \right|\le 1$, then which of the following inequalities holds true?

• A. $a+b+c\ge 0$
• B. $a-b+c\ge 0$
• C. $a+\left|b\right|+c\ge 0$
• D. $a-c\ge 0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $a+b+c\ge 0$
B $a-b+c\ge 0$
C $a+\left|b\right|+c\ge 0$
D $a-c\ge 0$

$y=a{x}^2+bx+c$
since both the roots lie between the interval of [-1,1]
$y(-1)\ge 0$ and $y\left(1\right)\ge 0$
Therefore
$a-b+c\ge 0$ and $a+b+c\ge 0$
Therefore
$a+|b|+c\ge 0$
Also $\alpha \beta \le |\alpha \beta |\le |\alpha ||\beta |\le 1$
Therefore
$\frac{c}{a}\le 1$
$c\le a$
$c-a\le 0$

hence all options are correct