# Task 2. (10 points)

Find the value of the parameter $p$ for which the equation $p x^{2}=|x-1|$ has exactly three solutions.

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# Solution

Notice that for $p>0$.

The given equation will have three roots if the parabola $y=p x^{2}$ intersects the "corner" $y=|x-1|$ at exactly three points. Since the left branch of the "corner" intersects the parabola at exactly two points when $p>0$, the right branch must be tangent to the parabola.

This happens if and only if the equation $p x^{2}=x-1$ has exactly one solution, which means

$$
D=1-4 p=0 \quad \Leftrightarrow \quad p=\frac{1}{4}
$$

Answer: $p=\frac{1}{4}$.