Question

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

• A. $\left(-2,4\right)$
• B. $\left(1,2\right)$
• C. $\left(-\sqrt{2},\sqrt{2}\right)$
• D. $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
• E. $\left(-4,2\right)$

Answer 1

The correct option is D

$\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

Explanation for the correct option:

Solving the quadratic inequality:

In $x^2+4ax+2$, we have $a=1,b=4a,\mathrm{and}c=2$.

As $x^2+4ax+2>0$, so $b^2-4ac<0$.

$$\begin{gathered} \Rightarrow 16a^2-8<0 \\ \Rightarrow 16a^2<8 \\ \Rightarrow \left(a-\frac{1}{\sqrt{2}}\right)\left(a+\frac{1}{\sqrt{2}}\right)<0 \end{gathered}$$

So, $-\frac{1}{\sqrt{2}}<a<\frac{1}{\sqrt{2}}$

Hence, option D is correct.