Number of roots (part -1)

We note that the function $f(x) = x^2+a|x|+1$ is an even function. That is $f(x) = f(-x)$ and is symmetrical about the y-axis( $x=0$ ). Therefore, for $f(x)$ to have two real roots, one of them must be positive, say $\alpha$ , and the other must be $-\alpha$ . There is only one root for $x>0$ .

For $x>0$ , we have $f(x) = x^2 + ax + 1$ and for $x^2 + ax + 1 = 0$ $\Rightarrow x = \dfrac{-a \pm \sqrt{a^2-4}}{2}$ . For $f(x)$ to have only one positive root $\Rightarrow a^2 = 4$ $\Rightarrow a = \boxed{-2}$ because if $a=2$ , $f(x)>0$ for $x>0$ .

The graph of the $f(x) = x^2-2|x|+1$ is shown below: