We need to find all integer pairs $(x, a)$ that satisfy: $x^2 - 2|x| + |a-2| = 0$ $$\newline$$ Since we have absolute value expressions, we need to consider different cases based on the signs of $x$ and $(a-2)$. $$\newline$$

$$\newline$$ When dealing with absolute values: $\newline$ - $|x| = x$ when $x \geq 0$ $\newline$ - $|x| = -x$ when $x < 0$ $\newline$ - $|a-2| = a-2$ when $a \geq 2$ $\newline$ - $|a-2| = -(a-2) = 2-a$ when $a < 2$ $$\newline$$ This gives us 4 cases to analyze. $$\newline$$

$$\newline$$ For $x \geq 0$ and $a \geq 2$: $$\newline$$ $|x| = x$ and $|a-2| = a-2$ $$\newline$$ Our equation becomes: $x^2 - 2x + (a-2) = 0$ $$\newline$$ For a quadratic $ax^2 + bx + c = 0$ to have real solutions, its discriminant $D = b^2 - 4ac$ must be non-negative. $$\newline$$ Here: $D = (-2)^2 - 4(1)(a-2) = 4 - 4(a-2) = 4 - 4a + 8 = 12 - 4a$ $$\newline$$ For real solutions: $12 - 4a \geq 0$ $$\newline$$ $a \leq 3$ $$\newline$$ Since $a \geq 2$ and $a \leq 3$, we have $a \in \{2, 3\}$. $$\newline$$ When $a = 2$: $x^2 - 2x + 0 = 0$ $$\newline$$ $x(x-2) = 0$ $$\newline$$ $x = 0, 2$ $$\newline$$ Valid pairs: $(0, 2)$ and $(2, 2)$ $$\newline$$ When $a = 3$: $x^2 - 2x + 1 = 0$ $$\newline$$ $(x-1)^2 = 0$ $$\newline$$ $x = 1$ $$\newline$$ Valid pair: $(1, 3)$ $$\newline$$

$$\newline$$ For $x \geq 0$ and $a < 2$: $$\newline$$ $|x| = x$ and $|a-2| = 2-a$ $$\newline$$ Our equation becomes: $x^2 - 2x + (2-a) = 0$ $$\newline$$ Discriminant: $D = 4 - 4(2-a) = 4 - 8 + 4a = 4a - 4$ $$\newline$$ For real solutions: $4a - 4 \geq 0$ $$\newline$$ $a \geq 1$ $$\newline$$ Since $a < 2$ and $a \geq 1$, we have $a = 1$. $$\newline$$ When $a = 1$: $x^2 - 2x + 1 = 0$ $$\newline$$ $(x-1)^2 = 0$ $$\newline$$ $x = 1$ $$\newline$$ Valid pair: $(1, 1)$ $$\newline$$

$$\newline$$ For $x < 0$ and $a \geq 2$: $$\newline$$ $|x| = -x$ and $|a-2| = a-2$ $$\newline$$ Our equation becomes: $x^2 - 2(-x) + (a-2) = 0$ $$\newline$$ $x^2 + 2x + (a-2) = 0$ $$\newline$$ Discriminant: $D = 4 - 4(a-2) = 12 - 4a$ $$\newline$$ For real solutions: $12 - 4a \geq 0$ $$\newline$$ $a \leq 3$ $$\newline$$ Since $a \geq 2$ and $a \leq 3$, we have $a \in \{2, 3\}$. $$\newline$$ When $a = 2$: $x^2 + 2x + 0 = 0$ $$\newline$$ $x(x+2) = 0$ $$\newline$$ $x = 0, -2$ $$\newline$$ Since we need $x < 0$, only $x = -2$ is valid $$\newline$$ Valid pair: $(-2, 2)$ $$\newline$$ When $a = 3$: $x^2 + 2x + 1 = 0$ $$\newline$$ $(x+1)^2 = 0$ $$\newline$$ $x = -1$ $$\newline$$ Valid pair: $(-1, 3)$ $$\newline$$

$$\newline$$ For $x < 0$ and $a < 2$: $$\newline$$ $|x| = -x$ and $|a-2| = 2-a$ $$\newline$$ Our equation becomes: $x^2 + 2x + (2-a) = 0$ $$\newline$$ Discriminant: $D = 4 - 4(2-a) = 4a - 4$ $$\newline$$ For real solutions: $4a - 4 \geq 0$ $$\newline$$ $a \geq 1$ $$\newline$$ Since $a < 2$ and $a \geq 1$, we have $a = 1$. $$\newline$$ When $a = 1$: $x^2 + 2x + 1 = 0$ $$\newline$$ $(x+1)^2 = 0$ $$\newline$$ $x = -1$ $$\newline$$ Valid pair: $(-1, 1)$ $$\newline$$

$$\newline$$ All valid integer pairs $(x, a)$: $$\newline$$ $(0, 2)$ $$\newline$$ $(2, 2)$ $$\newline$$ $(1, 3)$ $$\newline$$ $(1, 1)$ $$\newline$$ $(-2, 2)$ $$\newline$$ $(-1, 3)$ $$\newline$$ $(-1, 1)$ $$\newline$$ Total number of ways = 7 $$\newline$$

$$\newline$$ When solving equations with absolute values, we always: $$\newline$$ Identify the critical points where expressions inside absolute values change sign $$\newline$$ Create cases based on these critical points $$\newline$$ Solve each case separately $$\newline$$ Verify that our solutions satisfy the original conditions for each case

CAT (QA) PYQs

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