Question

In how many ways can a pair of integers \((x , a)\) be chosen such that \(x^2-2|x|+|a-2|=0\) ?

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The Correct Option is D

We are given the equation:

\[ x^2 - 2|x| + |a - 2| = 0 \]

Let’s simplify by substituting \( |x| = t \). Then the equation becomes:

\[ t^2 - 2t + |a - 2| = 0 \]

Solving the quadratic in \( t \):

\[ t = 1 \pm \sqrt{1 - |a - 2|} \]

Case 1: \( a > 2 \Rightarrow |a - 2| = a - 2 \)

\[ |x| = 1 \pm \sqrt{1 - (a - 2)} = 1 \pm \sqrt{3 - a} \]

For \( |x| \) to be real and integer, \( 3 - a \geq 0 \Rightarrow a \leq 3 \)

Also since \( a > 2 \), only \( a = 3 \) is valid. 
For \( a = 3 \Rightarrow |x| = 1 \Rightarrow x = \pm 1 \) ⇒ 2 solutions

Case 2: \( a = 2 \Rightarrow |a - 2| = 0 \)

\[ |x| = 1 \pm \sqrt{1 - 0} = 1 \pm 1 \Rightarrow |x| = 0, 2 \]

\[ \Rightarrow x = 0, \pm 2 \Rightarrow 3 \text{ solutions} \]

Case 3: \( a < 2 \Rightarrow |a - 2| = 2 - a \)

\[ |x| = 1 \pm \sqrt{1 - (2 - a)} = 1 \pm \sqrt{a - 1} \]

To keep values real and integer, \( a \geq 1 \) and \( a < 2 \) ⇒ Only possible value is \( a = 1 \)

\[ \Rightarrow |x| = 1 \Rightarrow x = \pm 1 \Rightarrow 2 \text{ more solutions} \]

Summary of Valid (x, a) Pairs:

• (-1, 3)
• (1, 3)
• (-2, 2)
• (0, 2)
• (2, 2)
• (-1, 1)
• (1, 1)

Total number of integer solutions: \( \boxed{7} \)

Answer 2

We are given the equation:

\[ |x|^2 - 2|x| + |a - 2| = 0 \]

Step 1: Analyze the Quadratic in \( |x| \)

The equation is a quadratic in \( |x| \):

\[ |x|^2 - 2|x| + |a - 2| = 0 \]

We know that for real solutions of \( |x| \), the discriminant must be a perfect square. Also, since \( |x| \geq 0 \), the roots must be non-negative.

Step 2: Try Values of \( |a - 2| \)

• If \( |a - 2| = 0 \), the equation becomes:
• If \( |a - 2| = 1 \), the equation becomes:

Step 3: Total Possible Integer Pairs \((x, a)\)

Combining all valid combinations:

• \((x = 0, a = 2)\)
• \((x = 2, a = 2)\)
• \((x = -2, a = 2)\)
• \((x = 1, a = 1)\)
• \((x = -1, a = 1)\)
• \((x = 1, a = 3)\)
• \((x = -1, a = 3)\)

✅ Therefore, the total number of valid integer pairs \( (x, a) \) is: \[ \boxed{7} \]