Equation, anyone?

The equation $f(x) = x|x|+a+bx = \begin{cases} x^2+bx+a & x \ge 0 \\ -x^2+bx+a & x < 0 \end{cases}$ .

We know that for $x \ge 0$ , $f(x)$ the number of roots decrease from $2$ to $0$ as $a$ increases. And that for $x<0$ , $f(x)$ is always decreasing, therefore even when there is no root for positive $x$ , there is always a root on the negative $x$ . Therefore, there is always one real root $(Q)$ and there is at most three roots $(P)$ . Therefore the answer is $\boxed{(R)}$ .

The answer can be easily visible with a graph. Notice that $a$ shifts the curves up and $b$ shifts the curves to the right.

If $x \geq 0$ , then the equation becomes $x^{2} + bx + a = 0$ ; its roots are $\frac{-b \pm \sqrt{b^{2} - 4a}}{2}$ . Else, if $x \leq 0$ , then the equation becomes $-x^{2} + bx + a = 0$ ; its roots are $\frac{b \pm \sqrt{b^{2} + 4a}}{2}$ . Now, let's consider the discriminants; if $a > 0$ , then the discriminant of the second equation must also be greater than zero due to $b^{2}$ always being a non-negative number, and thus this equation must have real roots. Conversely, if $a < 0$ , the discriminant of the first equation must be a non-negative number, and thus this equation must also have real roots; if $a = 0$ then we have three real roots: $x = 0, \pm b$ . Thus, we've proven proposition $Q$ .

Proving proposition $P$ is a little harder, but it's easy to see why it is true. Both equations represent a set of parabolas in the cartesian plane; however, we are only interested on the parts which represent the graph of the solution. Assume both equations have real, distinct roots; we should, theoretically, have four different solutions. However, these parabolas have a single point in common at $x = 0$ ; thus, they make a shape on the cartesian plane which resemble a sideways S near $x = 0$ . You can try making a line cross the "S" more than three times to realize it's impossible.

Calculus tells us that the function $x*|x| + bx + a = 0$ has the following properties:

• Two inflection points (One for each parabola; can be a degenerate case if b = 0.)

• The function increases from $-\infty$ until the first inflection point, then (if not degenerate case) it decreases from the first inflection point up to zero, at which point we begin to travel on the second parabola, still decreasing up to the second inflection point, and then the function begins to grow again. up to $\infty$ . If we're working with the degenerate case then the function is always increasing.

So, while it's increasing for the first time, it can at most cross the x axis once; then, if it has crossed the x axis, while it decreases it can at most cross the x axis once; finally, if it crossed the x axis twice, then the function can cross the x axis only one more time as it grows to infinity. Thus, both $P$ and $Q$ are correct and thus $R$ is also correct.