Question

For $x^2-(a+3)|x|+4=0$ to have real solutions, the range of $a$ is

• A. $(-\infty ,-7]\cup [1,\infty )$
• B. $(-3,\infty )$
• C. $(-\infty ,-7]$
• D. $[1,\infty )$

Answer 1

The correct option is D $[1,\infty )$

$x^2-(a+3)|x|+4=0$
Let $|x|=t>0$
Then, $t^2-(a+3)t+4=0\cdots \left(1\right)$
For real solutions,
$D\ge 0$
$\Rightarrow (a+3{)}^2-4^2\ge 0$
$\Rightarrow (a+3+4)(a+3-4)\ge 0$
$\Rightarrow (a+7)(a-1)\ge 0$
$\Rightarrow a\in (-\infty ,-7]\cup [1,\infty )$

Also, since eqn $\left(1\right)$ has positive roots, sum of roots is positive.
i.e., $a+3>0$
$\Rightarrow a>-3\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$a\in [1,\infty )$

$\text{Alternate:}$
$a=\frac{x^2+4}{|x|}-3$
$=|x|+\frac{4}{|x|}-3$
$\ge 2\sqrt{|x|\cdot \frac{4}{|x|}}-3(\because \text{A.M.}\ge \text{G.M.})$
$=4-3=1$
$\Rightarrow a\ge 1$