$f(x) = x^2 +(a+3)\lvert x \rvert + 4 = 0$
This is the quadratic equation They have given condition that find the range of $a$ for which the roots are real
So what I did was
$D\geq 0$
Solved the condition and got
$(a+7)(a-1) \geq 0$
So $a \in (-\infty , -7] \cup [1,\infty)$
But the given key is only $a \in (-\infty , -7]$
Could someone explain where I went wrong so $[1,\infty)$ does not come in the solution
$f(x) = x^2 +(a+3)\lvert x \rvert + 4 = 0$
If $\alpha$ and $\beta$ are roots of the equation,
$\alpha + \beta = - (a+3)$
$\alpha \beta = 4$
If $a \gt - 3, $ the quadratic has both negative roots but $|x|$ cannot be negative.
Hence the only solution that works is $a \in (-\infty , -7)$