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CAT 2020 Slot 2 — QA Question 11

Solving Quadratic EquationsMedium

In how many ways can a pair of integers (x , a) be chosen such that x2 − 2|x| + |a - 2| = 0?

Answer & solution

Correct answer: 7

  • A

    4

  • B

    5

  • C

    6

  • 7

Solution

Medium

Complete the square in ∣x∣|x| to fold the equation into a sum of two non-negative pieces equal to a constant. Since aa is an integer, ∣a−2∣−1|a-2|-1 is an integer, which forces only a couple of cases; enumerate the integer pairs (x,a)(x,a) in each.

1

Complete the square. Treat ∣x∣|x| as the variable and add/subtract 11.

x2−2∣x∣+∣a−2∣=0⇒ (∣x∣−1)2−1+∣a−2∣=0⇒ (∣x∣−1)2+(∣a−2∣−1)=0\begin{aligned} &x^2 - 2|x| + |a-2| = 0\\ &\Rightarrow\ \big(|x|-1\big)^2 - 1 + |a-2| = 0\\ &\Rightarrow\ \big(|x|-1\big)^2 + \big(|a-2|-1\big) = 0 \end{aligned}
2

Restrict the cases. (∣x∣−1)2≥0\big(|x|-1\big)^2\ge 0, and since aa is an integer, ∣a−2∣−1|a-2|-1 is an integer. For the sum to be zero, ∣a−2∣−1|a-2|-1 must be ≤0\le 0, i.e. 00 or −1-1.

3

Case 1: ∣a−2∣−1=0|a-2|-1 = 0. Then (∣x∣−1)2=0\big(|x|-1\big)^2 = 0.

∣a−2∣=1⇒a=1 or 3∣x∣=1⇒x=±1⇒ 2×2=4 pairs\begin{aligned} &|a-2| = 1 \Rightarrow a = 1\ \text{or}\ 3\\ &|x| = 1 \Rightarrow x = \pm 1\\ &\Rightarrow\ 2\times 2 = 4\ \text{pairs} \end{aligned}
4

Case 2: ∣a−2∣−1=−1|a-2|-1 = -1. Then (∣x∣−1)2=1\big(|x|-1\big)^2 = 1.

∣a−2∣=0⇒a=2∣x∣−1=±1⇒∣x∣=0 or 2⇒x=0, ±2⇒ 1×3=3 pairs\begin{aligned} &|a-2| = 0 \Rightarrow a = 2\\ &|x|-1 = \pm 1 \Rightarrow |x| = 0\ \text{or}\ 2 \Rightarrow x = 0,\ \pm 2\\ &\Rightarrow\ 1\times 3 = 3\ \text{pairs} \end{aligned}
5

Total. Add the counts from steps 3 and 4.

4+3=7 pairs4 + 3 = 7\ \text{pairs}
7 ways7\ \text{ways}

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CAT 2020 Slot 2 QA Q11 — Solving Quadratic Equations solution | TheCATExam