Question

Find the set of values of $a$ for which the inequality $(a^2+3)x^2+(a+2)x+4<-2$ holds true for all $x$.

• A. $a>0$
• B. $a<0$
• C. $-\infty$ $<a<\infty$
• D. $a\in \varphi$

Answer 1

Answer: (D)

$a\in \varphi$

We have,
$(a^2+3)x^2+(a+2)x+6<0$
$\Delta =(a+2{)}^2-4(a^2+3\left)\right(6)$
$\Delta =(a^2+4a+4)-(24a^2-72)$
$\Delta =-23a^2+4a-68$
The ${\Delta }_1$ for this expression in $a$ is given by,
${\Delta }_1=4^2-\left(4\right)(-68)(-23)<0$
Hence $\Delta$ is always negative, and the coefficient of $x^2$ is always positive.
Hence, $(a^2+3)x^2+(a+2)x+6$ , is always positive.
So, no real value of $a$ satisfies the inequality for all values of $x$ given in the question.