Equations with radicals — squaring both sides, extraneous solutions

2.1.23 · Maths › Algebra — Introduction & Intermediate

Overview

Jab hum radicals (especially square roots) wale equations solve karte hain, toh aksar hume radical ko hatane ke liye square both sides karna padta hai. Lekin yeh operation extraneous solutions introduce kar sakta hai — aisi solutions jo squared equation ko satisfy karti hain lekin original equation ko nahi. Yeh note explain karta hai ki aisa kyun hota hai, radical equations ko sahi tarike se kaise solve karein, aur extraneous solutions ko kaise identify karke reject karein.

Figure — Equations with radicals — squaring both sides, extraneous solutions

Core Concepts

Intuition Squaring Fake Solutions Kyun Create Kar Sakta Hai

Squaring ek reversible operation nahi hai jaise addition ya non-zero numbers se multiplication hoti hai. Jab tum ek equation ke dono sides square karte ho, tum pooch rahe ho "kaunsi values hain jo squares ko equal banati hain?" — lekin squaring sign ke baare mein information destroy kar deti hai.

Key insight: Agar $a = b$ , toh $a^{2} = b^{2}$ . LEKIN agar $a^{2} = b^{2}$ , toh hum sirf yeh conclude kar sakte hain ki $a = \pm b$ , necessarily $a = b$ nahi. Example: $x = 3$ implies $x^{2} = 9$ , lekin $x^{2} = 9$ se hume $x = 3$ YA $x = - 3$ milta hai. Squaring karne se humne ek extra solution gain kar liya!

Radicals ke liye yeh kyun matter karta hai: Square root function $x$ sirf principal (non-negative) root return karne ke liye defined hai. Toh $9 = 3$ hota hai, $\pm 3$ nahi. Jab hum $x = k$ ko square karte hain, toh hume $x = k^{2}$ milta hai, lekin humne woh constraint kho diya ki original left side non-negative honi chahiye.

Step-by-Step Solution Method

Har Step Kyun Zaroori Hai

Pehle isolate kyun karein? Agar tumhare paas $x + 3 = 5$ hai aur tum turant square karo, toh tumhe $(x)^{2} + 2 (3) (x) + 9 = 25$ milega, jisse $x + 6 x + 9 = 25$ ban jaata hai. Ab bhi radical baaki hai! Isolate karne se $x = 2$ milta hai, phir seedha $x = 4$ .

Check kyun karna zaroori hai? Kyunki squaring ek "if and only if" operation nahi hai. Yeh ek one-way implication hai: original equation true → squared equation true, lekin ulta nahi.

Worked Examples

Worked example Example 1: Basic Radical Equation (Koi Extraneous Solution Nahi)

Solve karo: $2 x + 3 = 5$

Solution: Step 1: Radical already isolated hai. ✓

Step 2: Dono sides square karo: $(2 x + 3)^{2} = 5^{2}$ $2 x + 3 = 25$

Yeh step kyun? Squaring square root ko eliminate kar deti hai kyunki $(a)^{2} = a$ for $a \geq 0$ .

Step 3: Solve karo: $2 x = 22$ $x = 11$ Step 4: Original equation mein check karo:

2 (11) + 3 = 22 + 3 = 25 = 5 ✓

The left side equals the right side, so x = 11 is valid.

$No extraneous solution here because our solution satisfies the original equation.$

$Example 2: Equation with One Extraneous Solution$ $Solve: x = x - 6$

$Solution:$

$Step 1: Already isolated. ✓$ $Step 2: Square both sides:$ $(x)^{2} = (x - 6)^{2}$ $x = x^{2} - 12 x + 36$

Yeh step kyun? Hume $x$ ke liye solve karne ke liye radical ko eliminate karna hai. Step 3: Standard form mein rearrange karo: $0 = x^{2} - 12 x - x + 36$ $0 = x^{2} - 13 x + 36$

Factor karo: $0 = (x - 4) (x - 9)$ Factor kyun? Isse humein potential solutions milte hain: x = 4 ya x = 9. Step 4: Dono solutions check karo: Check x = 4: $4 = 4 - 6$ $2 = - 2$ ✗

Yeh galat hai! Left side positive hai, right side negative hai. $x = 4$ extraneous hai.

Check $x = 9$ : $9 = 9 - 6$ $3 = 3$ ✓

Yeh kaam karta hai! x = 9 ek maatra valid solution hai.

x = 4 kyun aaya? Jab humne square kiya, humne equation ko $x = x - 6$ (jo require karta hai $x - 6 \geq 0$ , yani $x \geq 6$ ) se $x = (x - 6)^{2}$ mein transform kar diya, jisme aisi koi restriction nahi thi. Value $x = 4$ , $x = - (x - 6) = 6 - x$ ko satisfy karta hai, jo equation $4 = 2$ hai, lekin hamare original equation mein demand thi $4 = - 2$ .

[!example] Example 3: Multiple Radicals

Solve karo: $x + 7 = 1 + x$

Solution:

Step 1: Already isolated hai (dono radicals present hain, lekin ek isolated hai).

Step 2: Dono sides square karo: $(x + 7)^{2} = (1 + x)^{2}$ Left side: $x + 7$

Right side: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ use karke expand karo: $(1)^{2} + 2 (1) (x) + (x)^{2} = 1 + 2 x + x$

Dhyan se expand kyun karein? Middle term $2 x$ bahut zaroori hai — ise bhoolna ek common mistake hai.

Toh hamare paas hai: $x + 7 = 1 + 2 x + x$ Step 3: Simplify karo (dono sides se x subtract karo): $7 = 1 + 2 x$ $6 = 2 x$ $3 = x$

Yeh step kyun? Humne baaki bacha radical isolate kar diya hai.

Step 4: Phir se square karo: $9 = x$

Step 5: Original equation mein check karo: $9 + 7 = 1 + 9$ $16 = 1 + 3$ $4 = 4$ ✓

Valid hai! x = 9 solution hai.

[!example] Example 4: Negative Right Side se Extraneous Solution (Koi Valid Solution Nahi)

Solve karo: $3 x + 4 = - 2$

Analysis:

Square karne se pehle, domain ke baare mein socho: Ek square root (principal root) hamesha non-negative hoti hai. Right side -2 hai, jo negative hai.

Is equation ka koi solution nahi hai kyunki $3 x + 4 \geq 0$ sabhi valid x ke liye, lekin $- 2 < 0$ .

Agar hum ise ignore karke phir bhi square karein toh? $3 x + 4 = 4$ $3 x = 0$ $x = 0$

Check karo: $3 (0) + 4 = 4 = 2 \neq = - 2$ ✗

Akela candidate $x = 0$ extraneous hai. Final answer: Koi solution nahi.

Extraneous candidate aaya hi kyun? Dono sides ko square karne se $- 2$ , $(- 2)^{2} = 4$ ban gaya, sign erase ho gaya. Squared equation $3 x + 4 = 4$ "dekh" hi nahi sakta ki original right side negative thi, isliye woh ek aisa candidate produce karta hai jo original mein sign check fail kar deta hai.

Common Mistakes

Domain Considerations

Advanced: Multiple Squaring Steps

Jab kisi equation mein multiple radicals hain ya pehli squaring ke baad bhi radicals baach jaate hain, toh tumhe multiple times square karna pad sakta hai. Har squaring potentially extraneous solutions introduce karti hai, isliye checking aur bhi zaroori ho jaati hai.

Strategy:

Ek radical isolate karo
Square karo
Agar radicals baaki hain, phir isolate karo aur phir square karo
Resulting polynomial solve karo
Sabhi candidates ko original equation mein check karo

Memory Aids

Recall Ek 12-Saal Ke Bachche Ko Explain Karo

Socho tumhare paas ek magic box hai jo sirf positive numbers dikhata hai. Agar main tumse kahun "box mein 3 dikh raha hai," toh tum jaante ho andar 3 hai. Lekin agar main box ke dono sides square kar dun, toh main pooch raha hun "kaunsa number, jab square ho, 9 deta hai?" Well, 3 aur -3 dono kaam karte hain! Lekin -3 kabhi magic box mein nahi tha kyunki box sirf positive numbers dikhata hai.

Yahi square root equations mein hota hai. Square root woh magic positive-only box ki tarah hai. Jab hum square root hatane ke liye dono sides square karte hain, hum ek naya puzzle solve kar rahe hote hain jisme extra answers ho sakte hain jo original magic box mein fit nahi honge. Isliye hume har answer ko original magic box mein rakhke check karna hota hai ki woh actually kaam karta hai ya nahi.

Agar koi answer original equation mein kaam nahi karta, toh hum use "extraneous" kehte hain — fancy word for "yeh galti se ghus aaya hai aur yahan belong nahi karta!"

Connections

Quadratic Equations — radical equations ko square karne se aksar quadratics bante hain
Domain and Range of Functions — samajhna ki $x \geq 0$ kyun hai, extraneous solutions explain karta hai
Inverse Operations — squaring square root ka perfect inverse nahi hai
Absolute Value Equations — squaring se yeh bhi extraneous solutions produce karte hain
Rational Exponents — $x = x^{1/2}$ , power rules se connection hai
Function Transformations — $y = x$ aur uski properties

Flashcards

#flashcards/maths

Extraneous solution kya hoti hai?

Ek aisi value jo algebraic manipulation (jaise squaring) ke baad equation ko satisfy karti hai lekin original equation ko satisfy nahi karti. Yeh isliye arise hoti hai kyunki manipulation ne naye solutions introduce kar diye jo original problem mein the hi nahi.

Kisi equation ke dono sides square karne se extraneous solutions introduce hone ka risk kyun hota hai?

Kyunki squaring sign ke terms mein reversible nahi hoti. Agar

a = b

toh

a^{2} = b^{2}

, lekin agar

a^{2} = b^{2}

toh hum sirf yeh conclude kar sakte hain ki

a = \pm b

. Squaring aisi solutions introduce kar sakti hai jo squared equation ko true banati hain lekin original ko nahi.

Radical equations mein squaring ke baad mandatory step kya hai?

Har candidate solution ko original equation mein substitute karke check karo. Jo solutions original equation satisfy nahi karti woh extraneous hain aur reject karni padti hain.

x + 3 = 7

jaise equations mein squaring se pehle radical isolate karna kyun zaroori hai?

Agar hum isolate kiye bina square karein, toh

(x + 3)^{2} = x + 6 x + 9

expand hoga, jisme radical term baach jaata hai. Pehle isolate karne se

x = 4

milta hai, toh squaring se

x = 16

aata hai bina kisi baaki radical ke.

f (x) = g (x)

equation ke liye valid solution hone ke liye kaunsi do domain conditions honi chahiye?

(1)

f (x) \geq 0

kyunki real numbers mein negative number ka square root nahi le sakte, aur (2)

g (x) \geq 0

kyunki principal square root hamesha non-negative hoti hai.

x = x - 6

solve karo aur koi bhi extraneous solutions identify karo.

Dono sides square karo:

x = (x - 6)^{2} = x^{2} - 12 x + 36

. Rearrange karo:

x^{2} - 13 x + 36 = 0

, factor karo:

(x - 4) (x - 9) = 0

. Candidates:

x = 4, 9

. Check

x = 4

4 = 2 \neq = - 2 = 4 - 6

✗ extraneous. Check

x = 9

9 = 3 = 9 - 6

✓. Solution: sirf

x = 9

3 x + 4 = - 2

ka solve karne se pehle hi koi solution kyun nahi hai?

Principal square root

3 x + 4

hamesha non-negative (

\geq 0

) hoti hai, lekin right side

- 2 < 0

hai. Ek non-negative quantity kabhi negative quantity ke equal nahi ho sakti, isliye koi solution exist nahi karta.

x + 7 = 1 + x

solve karte waqt dono sides square karne par kya hota hai?

Left side

x + 7

ban jaata hai. Right side:

(1 + x)^{2} = 1 + 2 x + x

. Isse

x + 7 = 1 + 2 x + x

milta hai, jo simplify hokar

6 = 2 x

banta hai, phir

x = 3

, toh phir se square karne par

x = 9

milta hai.

3 x + 4 = - 2

mein, squaring se ek candidate

x = 0

kyun produce hota hai jabki koi real solution nahi hai?

Squaring

- 2

(- 2)^{2} = 4

mein badal deti hai, sign erase ho jaata hai. Squared equation

3 x + 4 = 4

detect hi nahi kar sakta ki original right side negative thi, isliye woh

x = 0

deta hai, jo check

4 = 2 \neq = - 2

mein fail karta hai aur isliye extraneous hai.