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

The given equation is:

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

Let \( t = |x| \). The equation transforms to:

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

Solving for \( t \):

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

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

Substituting into the equation for \( t \):

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

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

Given \( a>2 \), the only valid value is \( a = 3 \).

For \( a = 3 \), \( |x| = 1 \), yielding \( x = \pm 1 \). This gives 2 solutions.

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

Substituting into the equation for \( t \):

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

This results in \( |x| = 0 \) or \( |x| = 2 \).

Thus, \( x = 0, \pm 2 \), providing 3 solutions.

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

Substituting into the equation for \( t \):

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

For real and integer values of \( |x| \), we require \( a \geq 1 \) and \( a<2 \). The only possible integer value for \( a \) is \( a = 1 \).

For \( a = 1 \), \( |x| = 1 \), yielding \( x = \pm 1 \). This adds 2 more solutions.

Summary of Valid (x, a) Pairs:

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

The total number of integer solutions is \( \boxed{7} \).