Question

Find the value of $a$ which satisfies the equation $a{x}^2+4x+1=0$ for some real value of $x$.

• A. $a\le 0$
• B. $a\le 2$
• C. $a\le 4$
• D. None

Answer 1

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

The correct options are
A $a\le 0$
B $a\le 2$
C $a\le 4$

$D=$ Discriminant

$D\ge 0$

Comparing with $a{x}^2+bx+c=0$, we get
$a=a,b=4,c=1$

$b^2-4ac\ge 0$
$\Rightarrow 4^2-4a\ge 0$
$\Rightarrow 16-4a\ge 0$
$\Rightarrow 16\ge 4a$
$\Rightarrow 4\ge a$
$\Rightarrow a\le 4$

Hence, all options are satisfied.