Question

If ${(4a+c)}^2\le 4b^2$ then one root of $a{x}^2+bx+c=0$ lies in

• A. $(-2,2)$
• B. $(-1,1)$
• C. $(-\infty ,-2)$
• D. $(2,\infty )$

Answer 1

The correct option is A $(-2,2)$

$\because (4a+c{)}^2⩽4b^2$

$\Rightarrow 4a+c⩽\pm 2b$

$\Rightarrow 4a\pm 2b+c⩽0$

$\therefore 4a+2b+c⩽0$

or , $4a-2b+c⩽0$

let $f\left(x\right)=a{x}^2+bx+c$

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

$f(-2)=4a-2b+c\le 0$

Case 1 :- $a>0$

Case 2 :- $a<0$

$\therefore$ Roots should lie between $(-2,2)$