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Equations with Modulus or Squares

Introduction

Modulus and square equations often appear in aptitude tests to test your understanding of absolute values and the concept of distance from zero. These equations require splitting the equation into separate cases based on the definition of the modulus or using the squaring property carefully.

This pattern is important because modulus and square-based equations commonly appear in reasoning and algebra sections, often with trick-based simplifications.

Pattern: Equations with Modulus or Squares

Pattern

For modulus equations, always split into two cases depending on whether the expression inside | | is positive or negative.

  • If |x| = a → then x = a or x = -a.
  • If |x - k| = m → then x - k = m or x - k = -m → giving x = k ± m.

For square equations, use x² = a → x = ±√a.

When both sides involve modulus or squares, convert them into equivalent linear or quadratic forms before solving.

Step-by-Step Example

Question

Solve: |x - 3| = 5

Solution

  1. Step 1: Split into two cases based on modulus definition

    Case 1: x - 3 = 5 ⇒ x = 8.

    Case 2: x - 3 = -5 ⇒ x = -2.

  2. Step 2: Combine both valid results

    Both values satisfy the modulus equation.

  3. Final Answer:

    x = 8 or x = -2.

  4. Quick Check:

    |8 - 3| = |5| = 5 ✅, |-2 - 3| = |-5| = 5 ✅

Quick Variations

1. Equations involving squares: e.g., (x - 2)² = 9 → x - 2 = ±3 → x = 5 or -1.

2. Nested modulus equations: e.g., |2x - 3| = 1 → two cases → solve both linear equations.

3. Equations where both sides have modulus: e.g., |x - 2| = |x - 5| → check symmetry and midpoints.

4. Squared modulus: e.g., |x|² = a² ⇒ same as x² = a².

Trick to Always Use

  • Step 1: Remove modulus by creating two linear cases (positive and negative).
  • Step 2: For squares, take square roots carefully (± signs are essential).
  • Step 3: Check every solution by substitution-sometimes squaring introduces extraneous roots.

Summary

Summary

In the Equations with Modulus or Squares pattern:

  • Always create two cases when solving modulus or squared forms.
  • For modulus, use the definition: |x - a| = b ⇒ x = a ± b.
  • For squares, use x² = a ⇒ x = ±√a.
  • Verify all solutions to eliminate invalid or extraneous roots.

Practice

(1/5)
1. Solve: |x| = 7
easy
A. x = 7 or −7
B. x = 7
C. x = −7
D. x = 0

Solution

  1. Step 1: Use modulus definition

    |x| = 7 ⇒ x = 7 or x = -7.

  2. Final Answer:

    x = 7 or -7 → Option A.

  3. Quick Check:

    |7| = 7 and |-7| = 7 ✅

Hint: For |x| = a, write x = ±a.
Common Mistakes: Omitting the negative root or treating |x| as 'just x'.
2. Solve: |x - 4| = 2
easy
A. x = 2 or 6
B. x = 2 only
C. x = −2 or 6
D. x = 4 or 6

Solution

  1. Step 1: Split into cases

    x - 4 = 2 or x - 4 = -2.

  2. Step 2: Solve each case

    Case 1: x = 6. Case 2: x = 2.

  3. Final Answer:

    x = 2 or 6 → Option A.

  4. Quick Check:

    |2 - 4| = 2 and |6 - 4| = 2 ✅

Hint: For |x - a| = b, use x = a ± b.
Common Mistakes: Missing one case or sign errors when isolating x.
3. Solve: (x - 3)² = 9
easy
A. x = 3 or 9
B. x = 0 only
C. x = 6 or 0
D. x = −6 or 9

Solution

  1. Step 1: Take square root

    x - 3 = ±3.

  2. Step 2: Solve both cases

    Case +: x - 3 = 3 ⇒ x = 6. Case -: x - 3 = -3 ⇒ x = 0.

  3. Final Answer:

    x = 6 or 0 → Option C.

  4. Quick Check:

    (6 - 3)² = 9 and (0 - 3)² = 9 ✅

Hint: For (x - a)² = b, set x - a = ±√b.
Common Mistakes: Forgetting the ± when taking square roots.
4. Solve: |2x - 3| = 5
medium
A. x = 4 or 1
B. x = 4 or −1
C. x = 3 or −4
D. x = 3 or −3

Solution

  1. Step 1: Split into two linear cases

    2x - 3 = 5 or 2x - 3 = -5.

  2. Step 2: Solve each case

    Case 1: 2x = 8 ⇒ x = 4. Case 2: 2x = -2 ⇒ x = -1.

  3. Final Answer:

    x = 4 or -1 → Option B.

  4. Quick Check:

    |2·4 - 3| = |5| = 5 and |2·(-1) - 3| = |-5| = 5 ✅

Hint: For |ax + b| = c, solve ax + b = ±c then divide by a.
Common Mistakes: Arithmetic errors when solving the negative-case equation.
5. Solve: |x + 1| = |x - 5|
medium
A. x = 1
B. x = 3
C. x = 4
D. x = 2

Solution

  1. Step 1: Use symmetry / midpoint idea

    |x + 1| = |x - 5| means distance from x to -1 equals distance from x to 5, so x is the midpoint of -1 and 5.

  2. Step 2: Compute midpoint

    x = (-1 + 5)/2 = 4/2 = 2.

    3. Final Answer:

    x = 2 → Option D.

  3. Quick Check:

    |2 + 1| = 3 and |2 - 5| = 3 ✅

Hint: For |x - a| = |x - b|, x = (a + b)/2 (midpoint).
Common Mistakes: Overcomplicating instead of recognizing the midpoint property.

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