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

Answer: (D)

7

Answer 2

The correct answer is (D): $7$

$x^2-2|x|+|a-2|=0$

$|x|=\frac{2±\sqrt{4-4(|a-2|})}{2}$

$|x|=1±\sqrt{1-|a-2|}$

If $a>2;|a-2|=a-2$

$|x|=1±\sqrt{1-(a-2)}$

= $1±\sqrt{3-a}$

since $x$ is integer $3-a≥0$

$a≤3$

The possible values of $a$ is = $3$

Then $x = ±1$;

If $a=2,|x|=|1±1|,⇒x=±2,0$

If $a<2,|a-2|=2-a$

$|x|=1±\sqrt{1-(2-a)}$

$|x|=1±\sqrt{a-1}$

Since $x$ is integer $a-1≥0⇒a≥1$

$∴$ The possible values of $a$ is $1$

If $a=1,|x|=1⇒x=±1$

$∴$ The possible pairs =$(-1,3),(1,3),(1,1),(-1,1),(2,2),(-2,2),(0,2)$i.e.,$7$