Absolute value equations and inequalities — Hinglish notes

2.1.24 · Maths › Algebra — Introduction & Intermediate

Overview

Absolute value number line par zero se kisi bhi number ki distance represent karta hai, jo hamesha non-negative hoti hai. Absolute values wale equations aur inequalities solve karne ke liye is geometric meaning ko samajhna aur use algebraic cases mein translate karna zaroori hai.

Figure — Absolute value equations and inequalities

Inequalities jaise $∣ x ∣ < 5$ ke liye, hum poochte hain: "Kaun se numbers zero se 5 units se kam door hain?" Jawab: $- 5$ aur $5$ ke beech ki sab cheez. $∣ x ∣ > 5$ ke liye: "Kaun si cheez 5 units se zyada door hai?" Jawab: $- 5$ ke left mein sab kuch OR $5$ ke right mein sab kuch.

Definition and Foundation

Ye definition kyun? Hum ek aisa function chahte hain jo distance measure kare, jo hamesha positive hoti hai. Positive numbers ke liye, zero se distance khud woh number hi hai. Negative numbers ke liye, hum sign flip karte hain use positive banane ke liye.

Key property: $∣ x ∣ = x^{2}$ sabhi real $x$ ke liye.

Kyun? Kisi bhi number ko square karne se woh positive ho jaata hai, phir square root lene se positive root milta hai. Yeh absolute value ko Pythagorean distance concept se connect karta hai.

Absolute Value Equations

Case Analysis Method

$∣ A ∣ = b$ solve karne ke liye jahan $b \geq 0$ :

First principles se derivation:

Definition se shuru karo: $∣ A ∣$ equals $A$ jab $A \geq 0$ ho, aur equals $- A$ jab $A < 0$ ho
Toh $∣ A ∣ = b$ ka matlab ya toh:
- Case 1: $A = b$ (jab expression andar positive ya zero ho)
- Case 2: $- A = b$ , jisse milta hai $A = - b$ (jab expression andar negative ho)

Do cases kyun? Absolute value sign ko "strip away" kar deta hai. Dono $5$ aur $- 5$ , absolute value apply karne ke baad $5$ ban jaate hain. Backwards kaam karte hue, agar result $5$ hai, toh original ya toh kuch bhi raha ho sakta tha.

Special case: Agar $b < 0$ ho, toh koi solution nahi hai kyunki absolute values kabhi negative nahi hote.

Solution:

Case 1: $x - 3 = 7 ⟹ x = 10$
- Ye step kyun? Hum assume karte hain ki andar positive hai, toh absolute value use change nahi karta
Case 2: $x - 3 = - 7 ⟹ x = - 4$
- Ye step kyun? Hum assume karte hain ki andar negative hai, toh absolute value uska sign flip kar deta hai (isliye hum use 7 ke negative ke equal set karte hain)

Verification:

$∣10 - 3∣ = ∣7∣ = 7$ ✓
$∣ - 4 - 3∣ = ∣ - 7∣ = 7$ ✓

Geometric interpretation: Humne woh saare numbers dhundh liye jo number line par $3$ se exactly 7 units door hain: $3 + 7 = 10$ aur $3 - 7 = - 4$ .

Solution: Pehle check karo jab right side non-negative ho: humein chahiye $3 x - 1 \geq 0 ⟹ x \geq \frac{1}{3}$

Case 1: $2 x + 5 = 3 x - 1$
- Kyun? Assume karo $2 x + 5 \geq 0$ , toh absolute value use unchanged chhod deta hai
- $5 + 1 = 3 x - 2 x$
- $x = 6$
- Sign assumption check karo: $2 (6) + 5 = 17 \geq 0$ ✓ (andar wakayi non-negative hai)
- RHS non-negative check karo: $x = 6 \geq \frac{1}{3}$ ✓
- Verify: $∣2 (6) + 5∣ = 17$ aur $3 (6) - 1 = 17$ ✓
Case 2: $2 x + 5 = - (3 x - 1)$
- Kyun? Assume karo $2 x + 5 < 0$ , toh absolute value sign flip kar deta hai
- $2 x + 5 = - 3 x + 1$
- $5 x = - 4$
- $x = - \frac{4}{5}$
- Sign assumption check karo: $2 (- \frac{4}{5}) + 5 = - \frac{8}{5} + 5 = \frac{17}{5} > 0$ ✗ — lekin humne assume kiya tha ki $2 x + 5 < 0$ hai! Assumption violate ho gayi, toh ye branch yahan invalid hai.
- RHS non-negative check karo: $x = - \frac{4}{5} < \frac{1}{3}$ ✗ ( $3 x - 1 = - \frac{17}{5} < 0$ )
- Ye solution dono counts par extraneous hai!

Answer: Sirf $x = 6$

Case 2 kyun fail hua? Do independent reasons: (1) sign assumption " $2 x + 5 < 0$ " contradict ho gayi kyunki $2 (- \frac{4}{5}) + 5 = \frac{17}{5} > 0$ hai, aur (2) right side $3 x - 1 = - \frac{17}{5}$ negative hai, lekin absolute values kabhi negative number ke equal nahi ho sakte. Hamesha check karo inner expression ka sign assumption AUR RHS ka sign dono.

Absolute Value Inequalities

Less Than Inequalities

$∣ A ∣ < b$ solve karne ke liye jahan $b > 0$ :

Derivation:

$∣ A ∣ < b$ ka matlab hai " $A$ ki zero se distance $b$ se kam hai"
Ye tab hota hai jab $A$ , $- b$ aur $b$ ke beech ho
Algebraically: $- b < A < b$

Kyun? Agar $A \geq 0$ : $∣ A ∣ = A < b$ . Agar $A < 0$ : $∣ A ∣ = - A < b ⟹ A > - b$ . Dono cases combine karne par: $- b < A < b$ .

Solution: $- 5 \leq x + 2 \leq 5$

Ye step kyun? Hum woh saare $x$ dhundh rahe hain jahan $x + 2$ zero se zyada se zyada 5 units door ho, matlab $- 5$ aur $5$ ke beech.

Sabhi parts se 2 minus karo: $- 7 \leq x \leq 3$

Interval notation: $[- 7, 3]$

Geometric check: Center $x = - 2$ par hai (jahan $x + 2 = 0$ ). Hum 5 units dono directions mein jaate hain: $- 2 - 5 = - 7$ aur $- 2 + 5 = 3$ . ✓

Greater Than Inequalities

$∣ A ∣ > b$ solve karne ke liye jahan $b > 0$ :

Derivation:

$∣ A ∣ > b$ ka matlab hai " $A$ ki zero se distance $b$ se zyada hai"
Ye tab hota hai jab $A$ interval $[- b, b]$ ke bahar ho
Toh ya toh $A > b$ (far right) OR $A < - b$ (far left)

OR kyun? Number line par zero se door do regions hain: positive side aur negative side. $b$ units se zyada door hona matlab kisi ek region mein hona, dono mein ek saath nahi.

Solution: Do cases mein split karo:

Case 1: $3 x - 1 > 8$
- Kyun? Expression far right hai
- $3 x > 9$
- $x > 3$
Case 2: $3 x - 1 < - 8$
- Kyun? Expression far left hai
- $3 x < - 7$
- $x < - \frac{7}{3}$

Answer: $x \in (- \infty, - \frac{7}{3}) \cup (3, \infty)$

Geometric interpretation: Hum woh $x$ values chahte hain jahan $3 x - 1$ zero se 8 units se zyada door ho. Center $3 x - 1 = 0 ⟹ x = \frac{1}{3}$ par hai. $\frac{1}{3}$ ke $\frac{8}{3}$ units ke andar ki sab cheez exclude hai, bahar ki sab cheez include hai.

Advanced Techniques

Compound Absolute Value Equations

Solution: Critical points identify karo jahan andar ke expressions ka sign change ho: $x = 1$ aur $x = - 2$ . Ye number line ko teen regions mein divide karte hain.

Critical points kyun? Inn points par expressions zero ke equal hote hain, aur absolute value ka behavior switch ho jaata hai.

Region 1: $x < - 2$ (dono expressions negative)
- $∣ x - 1∣ = - (x - 1) = 1 - x$
- $∣ x + 2∣ = - (x + 2) = - x - 2$
- Equation: $(1 - x) + (- x - 2) = 5$
- $- 2 x - 1 = 5 ⟹ x = - 3$
- Check: $x = - 3 < - 2$ ✓
- Verify: $∣ - 3 - 1∣ + ∣ - 3 + 2∣ = 4 + 1 = 5$ ✓
Region 2: $- 2 \leq x < 1$ (pehla negative, doosra positive)
- $∣ x - 1∣ = 1 - x$
- $∣ x + 2∣ = x + 2$
- Equation: $(1 - x) + (x + 2) = 5$
- $3 = 5$ ✗
- Is region mein koi solution nahi!
Region 3: $x \geq 1$ (dono positive)
- $∣ x - 1∣ = x - 1$
- $∣ x + 2∣ = x + 2$
- Equation: $(x - 1) + (x + 2) = 5$
- $2 x + 1 = 5 ⟹ x = 2$
- Check: $x = 2 \geq 1$ ✓
- Verify: $∣2 - 1∣ + ∣2 + 2∣ = 1 + 4 = 5$ ✓

Answer: $x \in {- 3, 2}$

Common Mistakes and Fixes

Kyun sahi lagta hai: Hum ek equation dekhte hain aur seedha solve karte hain, yeh bhool jaate hain ki absolute value ke do "branches" hote hain.

Fix: Hamesha dono cases likho: $x - 2 = 5$ AND $x - 2 = - 5$ . Geometric intuition help karta hai: "2 se 5 units door kya hai?" ke do jawaab hain: $2 + 5 = 7$ aur $2 - 5 = - 3$ .

Kyun sahi lagta hai: Students " $∣ x ∣ < 3$ ka matlab $- 3 < x < 3$ hai" yaad kar lete hain aur same pattern apply karne ki koshish karte hain.

Fix: Greater-than ka matlab interval ke bahar hai, toh OR use karo: $x > 3$ ya $x < - 3$ . Yaad rakho: "less than" AND hai (beech mein), "greater than" OR hai (bahar). Union symbol $\cup$ interval notation humein yaad dilata hai ki ye alag regions hain.

Kyun sahi lagta hai: Hum apni algebra par trust karte hain aur assume karte hain ki dono case solutions valid hain.

Fix: Jab equation $∣ A ∣ = B$ ho aur $B$ mein variables hon, hamesha verify karo ki tumhare solution ke liye $B \geq 0$ ho, AUR ye bhi ki tumne $A$ ke baare mein jo sign assumption ki thi woh hold kare. Agar tumhare $x$ -value par $B < 0$ ho, toh woh solution extraneous hai. Upar Example 2 dono checks ko action mein dikhata hai.

Kyun sahi lagta hai: Students equations ke "split into cases" approach ko inequality rules se confuse kar lete hain.

Fix: $∣ A ∣ < b$ ke liye, single compound inequality $- b < A < b$ likho. Split mat karo! $∣ A ∣ > b$ ke liye, tab split karo $A > b$ OR $A < - b$ mein. Pattern: "less than" saath rehta hai (AND), "greater than" alag ho jaata hai (OR).

Memory Aids

$∣ x ∣ < 5$ : $- 5$ aur $5$ ke beech LANDS karta hai → $- 5 < x < 5$ (ek interval, AND)
$∣ x ∣ > 5$ : Extremes par GOES karta hai → $x < - 5$ ya $x > 5$ (do intervals, OR)

Visual: Number line draw karo. "Less than" beech ko shade karta hai (landing zone). "Greater than" ends ko shade karta hai (extremes).

Active Recall Practice

#flashcards/maths

$∣ x ∣ = a$ cases ke terms mein kya translate karta hai? :: Do equations: $x = a$ ya $x = - a$ (provided $a \geq 0$ ho). Geometric: zero se exactly $a$ units door numbers.

a > 0

ke liye

∣ x ∣ < a

kya translate karta hai?

Single compound inequality:

- a < x < a

. Geometric: zero ke

a

units ke andar ke saare points (beech mein).

a > 0

ke liye

∣ x ∣ > a

kya translate karta hai?

Do alag inequalities:

x > a

x < - a

. Geometric: zero se

a

units se zyada door ke saare points (bahar).

∣ A ∣ = B

solve karte waqt jahan

B

mein variables hon, kya check karna padta hai?

Do cheezein: (1) ki har solution ke liye

B \geq 0

ho, aur (2) ki har case ke liye

A

ke baare mein sign assumption hold kare. Agar dono mein se koi fail ho, toh solution extraneous hai.

Multiple absolute values wale equations jaise

∣ x - 1∣ + ∣ x + 2∣ = 5

kaise solve karte hain?

Critical points dhundho (jahan har inner expression

= 0

ho), number line ko regions mein split karo, har region mein sign ke hisaab se absolute value bars hatao, resulting equation har region mein solve karo, check karo ki solution apne region mein hai.

∣ x - 3∣ < 5

ka geometric meaning kya hai?

Woh saare numbers jinki

3

se distance

5

se kam ho. Solution:

3 - 5 < x < 3 + 5

, toh

- 2 < x < 8

∣ x ∣ = - 4

ka koi solution kyun nahi ho sakta?

Absolute value distance represent karta hai, jo hamesha non-negative hoti hai. Kisi bhi real number ki zero se negative distance nahi hoti.

Recall Ek 12-saal ke bachche ko explain karo

Socho tum number line par zero par khade ho. Absolute value aise poochna hai "main zero se kitne steps door hun?" chahe tum left gaaye ho ya right. Agar tum $5$ par ho ya $- 5$ par ho, tum phir bhi zero se $5$ steps door ho, toh $∣5∣ = ∣ - 5∣ = 5$ .

Jab hum $∣ x ∣ = 3$ solve karte hain, hum pooch rahe hain: "Main exactly $3$ steps zero se door hone ke liye kahan khad sakta hun?" Do jagah: $3$ par (teen steps right) aur $- 3$ par (teen steps left).

$∣ x ∣ < 3$ ke liye, hum chahte hain: "Main zero se $3$ se kam steps door hone ke liye kahan khad sakta hun?" Jawab: $- 3$ aur $3$ ke beech kahin bhi, kyunki wahi sirf itne paas wale spots hain.

$∣ x ∣ > 3$ ke liye, hum chahte hain: "Main zero se $3$ se zyada steps door hone ke liye kahan khad sakta hun?" Jawab: ya toh right mein $3$ ke aage, OR left mein $- 3$ ke aage — dono taraf ke door wale jagah!

Tricky part: jab problem kuch aisi ho jaise $∣ x - 2∣ = 5$ , ab zero shift ho gaya! Hum $0$ ki jagah $2$ se $5$ steps measure kar rahe hain. Toh hamare answers hain $2 + 5 = 7$ aur $2 - 5 = - 3$ .

Connections

Algebraic equations — absolute value equations special cases hain jinmein analysis chahiye
Linear inequalities — absolute value inequalities linear inequalities ko OR/AND logic ke saath combine karti hain
Number line and intervals — absolute value ki distance ke roop mein geometric interpretation
Function transformations — $∣ f (x) ∣$ negative parts of $f (x)$ ko upar reflect karta hai
Piecewise functions — absolute value piecewise define hota hai: $∣ x ∣ = x$ agar $x \geq 0$ ho, $- x$ agar $x < 0$ ho
Distance formula — $∣ x - a ∣$ number line par $x$ aur $a$ ke beech ki distance hai, jo 2D mein $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ tak generalize hoti hai
Quadratic equations — $∣ A ∣ = b$ ke dono sides square karne par $A^{2} = b^{2}$ milta hai, ek aur solution method

Absolute value master karo hamesha ye sochte hue: geometric intuition ke liye "zero se distance", phir systematically cases mein translate karo.