If $x^{2} + 4ax + 2 0$ for all values of $x$, then $a$ lies in the interval

Question

If $x^{2} + 4ax + 2 > 0$ for all values of $x$, then $a$ lies in the interval

Dr. Abubacker · Accepted Answer

(d)
$x^{2} + 4ax + 20$
$	herefore\ \ (4a)^{2} - 4 	imes 2 < 0$
[$\because$ if $f(x) > 0$, then]
$\Rightarrow \ \ 16a^{2} < 8\  \Rightarrow a^{2} < \frac{1}{2}$
$\Rightarrow \  - \frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}}$

Q. If $x^{2} + 4ax + 2 > 0$ for all values of $x$, then $a$ lies in the interval

Explanation

Reviewed by Dr. Abubacker

Want more practice?