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Equations with Modulus or Squares

Introduction

Modulus और square वाली equations aptitude tests में इसलिए दी जाती हैं ताकि आप absolute values और zero से distance की समझ को apply कर सकें। ऐसी equations को हल करने के लिए modulus की definition के आधार पर cases में split करना पड़ता है या फिर squaring property को ध्यान से इस्तेमाल करना होता है।

यह pattern महत्वपूर्ण है क्योंकि modulus और square-based equations reasoning और algebra sections में अक्सर आती हैं, जिनमें कई बार trick-based simplifications भी होते हैं।

Pattern: Equations with Modulus or Squares

Pattern

Modulus equations में हमेशा दो cases बनाएं-यह इस पर निर्भर करता है कि | | के अंदर वाला expression positive है या negative।

  • अगर |x| = a हो → तो x = a या x = -a होगा।
  • अगर |x - k| = m हो → तो x - k = m या x - k = -m → यानी x = k ± m।

Square equations में x² = a हो → तो x = ±√a।

अगर दोनों sides में modulus या squares हों, तो उन्हें पहले linear या quadratic form में बदलकर solve करें।

Step-by-Step Example

Question

Solve: |x - 3| = 5

Solution

  1. Step 1: Modulus definition के आधार पर दो cases बनाएं

    Case 1: x - 3 = 5 ⇒ x = 8.

    Case 2: x - 3 = -5 ⇒ x = -2.

  2. Step 2: दोनों valid results लिखें

    दोनों values modulus equation को satisfy करती हैं।

  3. Final Answer:

    x = 8 या x = -2.

  4. Quick Check:

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

Quick Variations

1. Squares वाली equations: जैसे (x - 2)² = 9 → x - 2 = ±3 → x = 5 या -1.

2. Nested modulus: जैसे |2x - 3| = 1 → दो linear cases → दोनों solve करें।

3. दोनों sides में modulus: जैसे |x - 2| = |x - 5| → symmetry या midpoint logic से solve करें।

4. Squared modulus: जैसे |x|² = a² ⇒ वही form x² = a²।

Trick to Always Use

  • Step 1: Modulus हटाकर दो linear cases बनाएं (positive और negative दोनों)।
  • Step 2: Squares के लिए square root लेते समय ± sign ज़रूर लगाएं।
  • Step 3: हर solution को substitute करके check करें-कई बार squaring से extra roots आ जाते हैं।

Summary

Summary

Equations with Modulus or Squares pattern में:

  • Modulus या square forms में हमेशा दो cases बनते हैं।
  • Modulus के लिए: |x - a| = b ⇒ x = a ± b।
  • Squares के लिए: x² = a ⇒ x = ±√a।
  • हर solution verify करें ताकि invalid या extraneous roots हट सकें।

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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